ML Aggarwal Class 11 Maths Solutions Chapter 11 Straight Lines

Chapter 11. Conic Sections

Introduction

Curves called conics were named following their historical discovery as the result of intersecting a plane with a right circular cone. Apollonius (before 200 B.C.) understood that a conic (or conic section) is the curve formed when a plane cuts through a right circular cone having two nappes, and the three types of curves produced are parabola, hyperbola and ellipse.

In this chapter, we will explore how the intersection of a plane with a double napped right circular cone creates different curve types. We will also work out the standard equations of parabola, ellipse, hyperbola and circle and will examine their fundamental properties.

 

11.1 Sections of a Cone

Consider a fixed line l and another line m that meet at a fixed point V and form an angle α with it as shown in the diagram.

When we rotate the line m around l such that angle α stays the same, the surface created is a double napped right circular hollow cone that stretches without bound in both directions. A section of a right circular cone with two nappes is shown in the figure. This surface is typically referred to as a right circular cone. The fixed point V is known as the vertex of the cone; the fixed line l is known as the axis of the cone. The rotating line m is known as a generator of the cone. The vertex divides the cone into two sections called nappes.

 

11.1.1 Degenerated Conic Sections

When the plane cuts the cone at its vertex, we have the following different cases:

Case I. When α < β < π/2

In this case, the section is a point.

Case II. When α = β

In this case, the section is a straight line.

Case III. When 0 < β < α

In this case, the section is a pair of intersecting straight lines. In fact, it is a degenerated case of a hyperbola.

 

11.1.2 Different Conic Sections

When the intersection of a plane with a cone does not contain the vertex of the cone, then we have the following cases:

Case I. When β = π/2

In this case, the section is a circle.

Case II. When α < β < π/2

In this case, the section is an ellipse.

Case III. When β = α

In this case, the section is a parabola.

Case IV. When 0 < β < α

In this case, the plane cuts through both the nappes and the resulting curve of intersection is a hyperbola.

 

11.2 Circle

A circle is the set of all points in a plane, each of which is at a constant distance from a fixed point in the plane.

Alternatively, a circle is the locus of a point which moves in a plane so that it remains at a constant distance from a fixed point in the plane. The fixed point is known as the centre and the constant distance is known as the radius. The radius is always positive.

If P₁, P₂, P₃, ... are points on the circle with centre C and radius r, then CP₁ = CP₂ = CP₃ = ... = r.

The equation of a circle is simplest when its centre is at the origin.

 

11.2.1 Standard (or Simplest) Form

Let O (0, 0) be the centre of the circle and r (> 0) be its radius. Let P (x, y) be a point in the plane, then P lies on the circle if and only if

OP = r

i.e. if and only if \( \sqrt{(x - 0)^2 + (y - 0)^2} = r \)

i.e. if and only if \( x^2 + y^2 = r^2 \), which is the equation of the circle.

This is known as the standard (or simplest) form.

 

11.2.2 Central Form

Let C (h, k) be the centre of the circle and r (> 0) be its radius. Let P (x, y) be a point in the plane, then P lies on the circle if and only if

CP = r

i.e. if and only if \( \sqrt{(x - h)^2 + (y - k)^2} = r \)

i.e. if and only if \( (x - h)^2 + (y - k)^2 = r^2 \), which is the equation of the circle.

This is known as the central form.

 

11.2.3 Diameter Form

Let A (x₁, y₁) and B (x₂, y₂) be the endpoints of a diameter of the circle.

Let P (x, y), different from A and B, be a point on the circle, then

slope of line AP = \( \frac{y - y_1}{x - x_1} \) and slope of line BP = \( \frac{y - y_2}{x - x_2} \)

Now P will lie on the circle if and only if ∠APB = 90°

i.e. if and only if the lines AP and BP are perpendicular to each other

i.e. if and only if \( \frac{y - y_1}{x - x_1} \cdot \frac{y - y_2}{x - x_2} = -1 \)

i.e. (y - y₁)(y - y₂) = - (x - x₁)(x - x₂)

i.e. if and only if \( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \), which is the equation of the circle.

This is known as the diameter form.

 

11.2.4 General Form

We know that the equation of the circle with centre (h, k) and radius r (> 0) is

\( (x - h)^2 + (y - k)^2 = r^2 \)

\( \Rightarrow x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - r^2 = 0 \) ...(i)

It can be written as

\( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(ii)

where g = - h, f = - k and c = h² + k² - r² such that

\( g^2 + f^2 - c = (-h)^2 + (-k)^2 - (h^2 + k^2 - r^2) = r^2 > 0 \) (since r > 0)

Conversely, if we consider any equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(iii)

with \( g^2 + f^2 - c > 0 \), then on adding \( g^2 + f^2 \) to both sides of (iii), we get

\( (x^2 + 2gx + g^2) + (y^2 + 2fy + f^2) + c = g^2 + f^2 \)

\( \Rightarrow (x + g)^2 + (y + f)^2 = g^2 + f^2 - c \)

\( \Rightarrow (x - (-g))^2 + (y - (-f))^2 = \sqrt{g^2 + f^2 - c}^2 \)

This shows that the distance of the point (x, y) from the point (- g, - f) is a fixed positive real number r (= \( \sqrt{g^2 + f^2 - c} \)) (since \( g^2 + f^2 - c > 0 \) means \( \sqrt{g^2 + f^2 - c} \) is a real number)

Therefore, the locus of (iii) is the set of all points (x, y) which are at a constant distance r (= \( \sqrt{g^2 + f^2 - c} \)) from the fixed point (- g, - f)

Thus, the equation (iii) represents a circle with centre (- g, - f) and radius \( \sqrt{g^2 + f^2 - c} \).

We have shown that the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) represents a circle if and only if \( g^2 + f^2 - c > 0 \).

Its centre is (- g, - f) and radius = \( \sqrt{g^2 + f^2 - c} \).

This is known as the general form.

 

Remarks

1. If \( g^2 + f^2 - c = 0 \), then the equation (iii) is satisfied by one and only one point (- g, - f), therefore, it represents a single point set, known as a degenerate (or point) circle.

2. If \( g^2 + f^2 - c < 0 \), then the equation (iii) is not satisfied by any real values of x, y - that is, it is not satisfied by the coordinates of any point in the plane, therefore, in this case it represents the empty set.

3. We observe that the general equation of a circle has the following characteristics:

• It is an equation of second degree in x, y containing no product term xy.
• Coefficient of x² = Coefficient of y² = 1.
• (1/2 coefficient of x)² + (1/2 coefficient of y)² - constant term is a positive real number.

4. The equation \( ax^2 + ay^2 + 2gx + 2fy + c = 0 \) represents a circle if and only if (i) a ≠ 0 and (ii) \( g^2 + f^2 - ac > 0 \).

First, note that a ≠ 0, for if a = 0 then the given equation becomes 2gx + 2fy + c = 0, which being a first degree equation in x, y represents a straight line.

Now, dividing the given equation by a, we get \( x^2 + y^2 + \frac{2g}{a}x + \frac{2f}{a}y + \frac{c}{a} = 0 \).

To complete the square on the left side, on adding \( \frac{g^2}{a^2} + \frac{f^2}{a^2} \) to both sides, we get

\( \left(x^2 + \frac{2g}{a}x + \frac{g^2}{a^2}\right) + \left(y^2 + \frac{2f}{a}y + \frac{f^2}{a^2}\right) = \frac{g^2}{a^2} + \frac{f^2}{a^2} - \frac{c}{a} \)

\( \Rightarrow \left(x + \frac{g}{a}\right)^2 + \left(y + \frac{f}{a}\right)^2 = \frac{g^2 + f^2 - ac}{a^2} \)

\( \Rightarrow \left(x - \left(-\frac{g}{a}\right)\right)^2 + \left(y - \left(-\frac{f}{a}\right)\right)^2 = \left(\frac{\sqrt{g^2 + f^2 - ac}}{|a|}\right)^2 \)

which represents a circle if \( g^2 + f^2 - ac > 0 \).

Therefore, the given equation \( ax^2 + ay^2 + 2gx + 2fy + c = 0 \) represents a circle if and only if (i) a ≠ 0 and (ii) \( g^2 + f^2 - ac > 0 \).

If these conditions are satisfied then the given equation represents a circle with centre \( \left(-\frac{g}{a}, -\frac{f}{a}\right) \) and radius \( \frac{\sqrt{g^2 + f^2 - ac}}{|a|} \).

This form of the equation is known as the most general form.

5. The equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) represents a circle if and only if (i) a = b ≠ 0, (ii) h = 0 and (iii) \( g^2 + f^2 - ac > 0 \).

Rule to write the centre and radius of a circle:

• (i) Write the given equation so that coefficient of x² = coefficient of y² = 1 on the left side (right side being zero); divide throughout (if necessary) by the coefficient of x² (or y²).
• (ii) Compare this equation with \( x^2 + y^2 + 2gx + 2fy + c = 0 \), then g = 1/2 coefficient of x, f = 1/2 coefficient of y, c = constant term.
• (iii) Check that \( g^2 + f^2 - c > 0 \), then centre is (- g, - f) and radius = \( \sqrt{g^2 + f^2 - c} \).

Concentric circles. Circles having the same centre and different radii are called concentric circles.

Equal circles. Circles having equal radius are called equal circles.

 

11.2.5 Geometrical Condition for the Intersection of a Line and a Circle

Let S be a circle with centre C and radius r (> 0). Let l be a line in the plane of the circle S and d be the perpendicular distance from C to the line l, then

• (i) l intersects S in two distinct points if and only if d < r.
• (ii) l intersects S in one and only one point if and only if d = r.
• (iii) l does not intersect S if and only if d > r.

From the above, we note that if d = r then the line l and the circle S have one and only one point in common, namely the foot of perpendicular M from C on l, and we say that l touches S or l is a tangent to S at M and the unique point M is called the point of contact.

Thus, a line l touches a circle S if and only if the length of perpendicular from the centre of S to the line l is equal to the radius of S.

If a line l meets a circle in two distinct points A and B, then the length of the segment [AB] i.e. |AB| is called the length of intercept made by the circle S on the line l.

 

11.2.6 Relative Position of Two Circles

Let S₁, S₂ be two (non-concentric) circles with centres A, B and radii r₁, r₂ and d be the distance between their centres, then

• (i) One circle lies completely inside the other circle if and only if d < |r₁ - r₂|
• (ii) The two circles touch internally if and only if d = |r₁ - r₂|
• (iii) The two circles intersect in two points if and only if d > |r₁ - r₂| and d < r₁ + r₂
• (iv) The two circles touch externally if and only if d = r₁ + r₂
• (v) One circle lies completely outside the other circle if and only if d > r₁ + r₂

 

Illustrative Examples

 

Example 1. Find the equation of a circle whose centre is (- 2, 3) and radius is 4. (NCERT)

Answer: Since the centre of the circle is (- 2, 3) and its radius is 4, the equation of the circle is

(x - (- 2))² + (y - 3)² = 4² (central form)

\( \Rightarrow (x + 2)^2 + (y - 3)^2 = 16 \)

\( \Rightarrow x^2 + 4x + 4 + y^2 - 6y + 9 = 16 \)

\( \Rightarrow x^2 + y^2 + 4x - 6y - 3 = 0 \).

Exam Tip: After expanding the central form, always simplify to get the general form by moving all terms to one side. This makes it easy to identify the coefficients g, f, and c for verification.

 

Example 2. Find the equation of the circle with centre (2, 2) and which passes through the point (4, 5). (NCERT)

Answer: The centre of the circle is C (2, 2) and it passes through the point P (4, 5).

Therefore, the radius of the circle = CP = \( \sqrt{(4 - 2)^2 + (5 - 2)^2} = \sqrt{4 + 9} = \sqrt{13} \).

The equation of the circle is \( (x - 2)^2 + (y - 2)^2 = (\sqrt{13})^2 \) (central form)

\( x^2 + y^2 - 4x - 4y = 5 \).

Exam Tip: Always calculate the radius first when the centre and a passing point are given. The distance formula is key, and simplifying the square root (if possible) before squaring helps prevent arithmetic errors.

 

Example 3. Find the equation of a circle having (1, - 2) as its centre and passing through the intersection of the lines 3x + y = 14 and 2x + 5y = 18. (NCERT Exemplar Problems)

Answer: The given lines are 3x + y = 14 ...(i) and 2x + 5y = 18 ...(ii).

Solving (i) and (ii) together, we find x = 4, y = 2.

Therefore, the point of intersection, say P, of the given lines is (4, 2).

Since the centre of the circle is C (1, - 2) and it passes through the point P (4, 2),

its radius = CP = \( \sqrt{(4 - 1)^2 + (2 - (-2))^2} = \sqrt{9 + 16} = 5 \).

The equation of the circle is \( (x - 1)^2 + (y + 2)^2 = 5^2 \) (Central Form)

or \( x^2 + y^2 - 2x + 4y - 20 = 0 \).

Exam Tip: When the intersection point is needed, solve the system of linear equations carefully. Always double-check by substituting both x and y values back into both original equations to confirm the solution is correct.

 

Example 4. If two diameters of a circle lie along the lines x - y - 9 = 0 and x - 2y - 7 = 0 and the area of the circle is 38.5 sq units, find its equation.

Answer: Since two diameters of the circle lie along the lines x - y - 9 = 0 ...(i) and x - 2y - 7 = 0 ...(ii), their point of intersection is the centre of the circle.

Solving (i) and (ii) together, we find x = 11, y = 2.

Therefore, the centre of the circle is (11, 2).

Let r be the radius of the circle. Then area = πr² = 38.5 square units (given)

\( \Rightarrow \frac{22}{7}r^2 = \frac{77}{2} \Rightarrow r^2 = \frac{49}{4} \Rightarrow r = \frac{7}{2} \) (since r > 0)

The equation of the circle is \( (x - 11)^2 + (y - 2)^2 = \left(\frac{7}{2}\right)^2 \) (central form)

or \( x^2 + y^2 - 22x - 4y + 125 - \frac{49}{4} = 0 \)

or \( 4(x^2 + y^2) - 88x - 16y + 451 = 0 \).

Exam Tip: Remember that two diameters of a circle always intersect at the centre. Extract the centre from the intersection of the diameter lines, then use the given area to find the radius. Convert area to radius carefully using the formula πr².

 

Example 5. Find the equation of the circle which passes through the point (- 2, - 3) and has its centre on the negative direction of x-axis and is of radius 5 units.

Answer: As the centre of the circle lies on the negative direction of x-axis, let its centre be C (h, 0), h < 0.

Since the circle passes through A (- 2, - 3) and has radius 5,

CA = 5 \( \Rightarrow (h + 2)^2 + (0 + 3)^2 = 5^2 \)

\( \Rightarrow (h + 2)^2 = 25 - 9 = 16 \Rightarrow h + 2 = \pm 4 \)

\( \Rightarrow h = 2, - 6 \) but h < 0 \( \Rightarrow h = - 6 \).

Therefore, the centre of the circle is (- 6, 0) and hence its equation is

\( (x + 6)^2 + (y - 0)^2 = 5^2 \) (central form)

\( x^2 + y^2 + 12x + 11 = 0 \).

Exam Tip: When the centre lies on a coordinate axis, express its coordinates with one parameter on that axis and the other as zero. Use the distance condition to solve for the parameter, and check that the value satisfies any given sign constraint (e.g., h < 0).

 

Example 6. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3). (NCERT)

Answer: As the centre of the circle lies on x-axis, let its centre be C (h, 0).

Since the circle passes through A (2, 3) and has radius 5,

CA = 5 \( \Rightarrow (2 - h)^2 + (3 - 0)^2 = 5^2 \)

\( \Rightarrow (2 - h)^2 = 16 \Rightarrow 2 - h = \pm 4 \Rightarrow h = - 2, 6 \).

Therefore, the centre of the circle is (- 2, 0) or (6, 0).

The equation of the circle is \( (x + 2)^2 + (y - 0)^2 = 5^2 \) or \( (x - 6)^2 + (y - 0)^2 = 5^2 \)

\( x^2 + y^2 + 4x - 21 = 0 \) or \( x^2 + y^2 - 12x + 11 = 0 \).

There are two circles satisfying the given conditions.

Exam Tip: When solving for a parameter, remember that both positive and negative roots are often valid (unless restricted by a sign condition). This usually gives two possible circles satisfying the problem conditions - always report both.

 

Example 7. Find the equation of the circle which touches both the axes in first quadrant and whose radius is a. (NCERT Exemplar Problems)

Answer: As the radius of the circle is a units and it touches both the axes in the first quadrant, its centre is C (a, a).

The equation of the circle is \( (x - a)^2 + (y - a)^2 = a^2 \)

or \( x^2 + y^2 - 2ax - 2ay + a^2 = 0 \).

Exam Tip: When a circle touches both coordinate axes, its centre is equidistant from both axes. In the first quadrant with radius a, the centre is at (a, a). This relationship holds for any quadrant - determine the signs based on which axes and which quadrant are involved.

 

Example 8. Find the equation of the circle which touches x-axis and whose centre is (1, 2). (NCERT Exemplar Problems)

Answer: Centre of the circle is C (1, 2).

As the circle touches the x-axis,

its radius = perpendicular distance from centre (1, 2) to the x-axis = ordinate of point C = 2.

Therefore, the equation of the circle is \( (x - 1)^2 + (y - 2)^2 = 2^2 \)

or \( x^2 + y^2 - 2x - 4y + 1 = 0 \).

Exam Tip: When a circle touches a coordinate axis, the radius equals the perpendicular distance from the centre to that axis. For the x-axis, radius = |y-coordinate of centre|; for the y-axis, radius = |x-coordinate of centre|.

 

Example 9. Find the equation of the circle whose centre is C (- 2, 3) and which touches the line x - y + 7 = 0.

Answer: The given line is x - y + 7 = 0 ...(i)

Let r be the radius of the required circle. Then

r = perpendicular distance from C (- 2, 3) to the line (i) = \( \frac{|-2 - 3 + 7|}{\sqrt{1^2 + (-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \).

Therefore, the equation of the circle is \( (x + 2)^2 + (y - 3)^2 = (\sqrt{2})^2 \)

or \( x^2 + y^2 + 4x - 6y + 11 = 0 \).

Exam Tip: For a circle to touch a line, the perpendicular distance from the centre to the line must equal the radius. Use the point-to-line distance formula: \( d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \) where the line is ax + by + c = 0 and (x₀, y₀) is the centre.

 

Example 10. Find the equation of a circle which touches (i) the y-axis at origin and whose radius is 3 units (ii) both the co-ordinate axes and the line x = 3.

Answer: (i) There are two circles satisfying given conditions. As the circles touch y-axis at the origin, their centres lie on x-axis. Since radius is 3 units, centres of the circles are (3, 0) and (- 3, 0) and hence the equations of the circles are

\( (x \pm 3)^2 + (y - 0)^2 = 3^2 \)

or \( x^2 + y^2 \pm 6x = 0 \).

(ii) There are two circles satisfying the given conditions. From the analysis, clearly, the centres of these circles are \( \left(\frac{3}{2}, \frac{3}{2}\right) \) and \( \left(\frac{3}{2}, -\frac{3}{2}\right) \), and radius of each circle is \( \frac{3}{2} \).

Therefore, the equations of these circles are

\( \left(x - \frac{3}{2}\right)^2 + \left(y \pm \frac{3}{2}\right)^2 = \left(\frac{3}{2}\right)^2 \)

or \( x^2 + y^2 - 3x \pm 3y + \frac{9}{4} = 0 \)

or \( 4x^2 + 4y^2 - 12x \pm 12y + 9 = 0 \).

Exam Tip: When a circle touches both coordinate axes, its centre lies on the diagonal lines y = x or y = - x at equal distances from both axes. The radius equals this distance. When the circle also touches a third line, use all three tangency conditions to set up and solve the equations for the centre coordinates.

 

Example 11. Find the equation of a circle which touches both the axes and the line 3x - 4y + 8 = 0 and lies in the third quadrant. (NCERT Exemplar Problems)

Answer: Given line is 3x - 4y + 8 = 0 ...(i)

Let the radius of the circle be a. As the circle lies in the third quadrant and touches both the axes, its centre is C (- a, - a).

Since the line (i) touches the circle, the perpendicular distance from C (- a, - a) to the line (i) = radius of circle.

Therefore, \( \frac{|-3a + 4a + 8|}{\sqrt{3^2 + (-4)^2}} = a \Rightarrow \frac{|a + 8|}{5} = a \)

\( \Rightarrow a + 8 = 5a \) or \( a + 8 = - 5a \)

\( \Rightarrow a = 2 \) or \( a = - \frac{4}{3} \) but a > 0 (a being radius)

\( \Rightarrow a = 2 \).

Therefore, the centre of circle is (- 2, - 2) and radius = 2.

The equation of the circle is \( (x + 2)^2 + (y + 2)^2 = 2^2 \)

or \( x^2 + y^2 + 4x + 4y + 4 = 0 \).

Exam Tip: Set up the absolute value equation from the tangency condition and solve carefully. Remember that |A| = B has two cases: A = B and A = - B. Check which solution satisfies all constraints (e.g., a > 0, centre in the specified quadrant) before finalizing.

 

Example 12. Find the equation of a circle whose centre is (3, - 1) and which cuts off a chord of length 6 units on the line 2x - 5y + 18 = 0. (NCERT Exemplar Problems)

Answer: The given line is 2x - 5y + 18 = 0 ...(i) and the centre of the circle is C (3, - 1).

Let the line (i) meet the required circle at points A and B. From C, draw CM ⊥ AB then M is mid-point of segment AB.

Given AB = 6 units \( \Rightarrow 2 \cdot AM = 6 \) units \( \Rightarrow AM = 3 \) units

CM = perpendicular distance from C (3, - 1) to the line (i) = \( \frac{|2 \cdot 3 - 5 \cdot (-1) + 18|}{\sqrt{2^2 + (-5)^2}} \) units = \( \frac{29}{\sqrt{29}} \) units = \( \sqrt{29} \) units.

Let r units be the radius of the circle.

From right angled triangle AMC, CA² = AM² + CM²

\( \Rightarrow r^2 = 3^2 + (\sqrt{29})^2 = 38 \Rightarrow \) radius = \( \sqrt{38} \) units.

Therefore, the equation of the circle is \( (x - 3)^2 + (y + 1)^2 = (\sqrt{38})^2 \)

or \( x^2 + y^2 - 6x + 2y - 28 = 0 \).

Exam Tip: When a circle cuts a chord on a line, draw a perpendicular from the centre to the chord. This perpendicular bisects the chord. Use the right triangle formed by the radius, half-chord, and perpendicular distance to find the radius via the Pythagorean theorem.

 

Example 13. Find the equation of the circle which passes through two points on y-axis which are at a distance of 3 units from origin and has radius 5 units.

Answer: There are two circles satisfying the given conditions.

Since the circles pass through the points on y-axis which are at distance of 3 units from origin, so OB = OB' = 3. The centres of the circles lie on the right bisector of BB' i.e. on x-axis. If C is centre of a circle, then CB is its radius.

From right angled triangle OBC,

OC² = BC² - OB² = 5² - 3² = 16

\( \Rightarrow OC = 4 \).

Therefore, the centres of the circles are (4, 0) and (- 4, 0).

Therefore, the equations of the circles are

\( (x \pm 4)^2 + (y - 0)^2 = 5^2 \)

or \( x^2 + y^2 \pm 8x - 9 = 0 \).

Exam Tip: When finding the centre of a circle that passes through two given points, recall that the centre lies on the perpendicular bisector of the segment joining those two points. Use the constraint that the centre is at a known distance from the origin (or another reference point) to locate it precisely.

 

Example 14. Find the equation of the circle when the end points of a diameter are A (- 2, 3) and B (3, - 5).

Answer: Using diameter form, the equation of the circle having A (- 2, 3) and B (3, - 5) as the end points of a diameter is

(x - (- 2)) (x - 3) + (y - 3) (y - (- 5)) = 0

or (x + 2) (x - 3) + (y - 3) (y + 5) = 0

or \( x^2 - x - 6 + y^2 + 2y - 15 = 0 \)

or \( x^2 + y^2 - x + 2y - 21 = 0 \).

Exam Tip: The diameter form is very efficient when both endpoints of a diameter are known. Apply the formula directly: if A (x₁, y₁) and B (x₂, y₂) are endpoints of a diameter, the circle equation is (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0. This automatically ensures ∠APB = 90° for any point P on the circle.

 

Example 15. Find the equation of a circle which has the portion of the line 3x + 4y = 14 intercepted by the lines x - y = 0 and 11x - 4y = 0 as a diameter.

Answer: The given lines are

3x + 4y - 14 = 0 ...(i)

x - y = 0 ...(ii)

and 11x - 4y = 0 ...(iii)

The intersection of (i) and (ii) is the point A (2, 2) and the intersection of (i) and (iii) is the point B \( \left(1, \frac{11}{4}\right) \).

Using diameter form, the equation of the circle having AB as its diameter is

(x - 2) (x - 1) + (y - 2) \( \left(y - \frac{11}{4}\right) \) = 0

or \( x^2 - 3x + 2 + y^2 - 2y - \frac{11}{4}y + \frac{11}{2} = 0 \)

or \( x^2 + y^2 - 3x - \frac{19}{4}y + \frac{15}{2} = 0 \)

or \( 4x^2 + 4y^2 - 12x - 19y + 30 = 0 \).

Exam Tip: When the diameter is specified as a segment of a line between two other lines, first find the intersection points of the three lines. Then apply the diameter form using these two intersection points as the endpoints. Clear any fractions at the end by multiplying through if needed.

 

Example 29. Find the equation of the circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x - 1 = 0. (NCERT Exemplar Problems)

Answer: The given line is y = x - 1 ...(i)

Let (h, k) be the centre of the circle. As centre lies on (i), we get

k = h - 1 ...(ii)

The equation of the circle is (x - h)² + (y - k)² = 3² (since radius = 3)

or (x - h)² + (y - (h - 1))² = 9.

Since the circle passes through the point (7, 3),

Therefore, (7 - h)² + (3 - (h - 1))² = 9

\( \Rightarrow (7 - h)^2 + (4 - h)^2 = 9 \)

\( \Rightarrow 49 - 14h + h^2 + 16 - 8h + h^2 - 9 = 0 \)

\( \Rightarrow 2h^2 - 22h + 56 = 0 \Rightarrow h^2 - 11h + 28 = 0 \)

\( \Rightarrow (h - 7)(h - 4) = 0 \Rightarrow h = 7, 4 \).

When h = 7, k = 7 - 1 = 6; when h = 4, k = 4 - 1 = 3.

Therefore, the centre of circle is (7, 6) or (4, 3).

Thus, we have two circles satisfying given conditions. Their equations are

\( (x - 7)^2 + (y - 6)^2 = 3^2 \) or \( (x - 4)^2 + (y - 3)^2 = 3^2 \)

\( x^2 + y^2 - 14x - 12y + 76 = 0 \) or \( x^2 + y^2 - 8x - 6y + 16 = 0 \).

Exam Tip: When the centre must lie on a line, express one coordinate in terms of the other using the line equation. Substitute this into the circle equation. Use the condition that the circle passes through a given point to solve for the parameter. Expect two solutions - both should be reported unless additional constraints eliminate one.

 

Example 30. Find the equation of the circle passing through the points (1, - 2), (5, 4) and (10, 5).

Answer: Let the equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(i)

As the circle passes through the points (1, - 2), (5, 4) and (10, 5), we get

1 + 4 + 2g - 4f + c = 0 \( \Rightarrow 2g - 4f + c + 5 = 0 \) ...(ii)

25 + 16 + 10g + 8f + c = 0 \( \Rightarrow 10g + 8f + c + 41 = 0 \) ...(iii)

100 + 25 + 20g + 10f + c = 0 \( \Rightarrow 20g + 10f + c + 125 = 0 \) ...(iv)

Subtracting (ii) from (iii), we get

8g + 12f + 36 = 0 \( \Rightarrow 2g + 3f + 9 = 0 \) ...(v)

Subtracting (ii) from (iv), we get

18g + 14f + 120 = 0 \( \Rightarrow 9g + 7f + 60 = 0 \) ...(vi)

Solving (v) and (vi) together, we find g = - 9, f = 3.

From (ii), we find c = - 5 - 2 (- 9) + 4 (3) = 25.

Substituting these values of g, f and c in (i), we find \( x^2 + y^2 - 18x + 6y + 25 = 0 \), which is the equation of the required circle.

Exam Tip: To find a circle through three points, use the general form equation and substitute each point to get three linear equations in g, f, c. Solve this system by elimination (subtract equations pairwise to reduce to two equations in two unknowns, then solve). This method is systematic and minimizes arithmetic errors.

 

Example 31. Show that the points (7, 5), (6, - 2), (- 1, - 1) and (0, 6) are concyclic. Also find the radius and the centre of the circle on which they lie.

Answer: Let us find the equation of the circle passing through the points (7, 5), (6, - 2) and (- 1, - 1).

Let the equation of this circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(i)

As the points (7, 5), (6, - 2) and (- 1, - 1) lie on it, we get

49 + 25 + 14g + 10f + c \( \Rightarrow 14g + 10f + c + 74 = 0 \) ...(ii)

36 + 4 + 12g - 4f + c = 0 \( \Rightarrow 12g - 4f + c + 40 = 0 \) ...(iii)

1 + 1 - 2g - 2f + c = 0 \( \Rightarrow 2g + 2f - c - 2 = 0 \) ...(iv)

Adding (ii) and (iv), we find

16g + 12f + 72 = 0 \( \Rightarrow 4g + 3f + 18 = 0 \) ...(v)

Adding (iii) and (iv), we find

14g - 2f + 38 = 0 \( \Rightarrow 7g - f + 19 = 0 \) ...(vi)

Solving (v) and (vi) together, we find g = - 3, f = - 2.

From (ii), we find c = - 14 (- 3) - 10 (- 2) - 74 = - 12.

Substituting these values of g, f and c in (i), we find \( x^2 + y^2 - 6x - 4y - 12 = 0 \) ...(vii)

The fourth point (0, 6) will lie on (vii) if 0 + 36 - 0 - 24 - 12 = 0 i.e. if 0 = 0, which is true.

Hence, the given points are concyclic.

Also, (vii) is the equation of the circle on which these points lie.

Its centre is (3, 2) and radius = \( \sqrt{9 + 4 - (-12)} = 5 \).

Exam Tip: To show four points are concyclic, find the circle through the first three and then verify that the fourth point satisfies the equation. State clearly that "since the fourth point satisfies the equation, all four points are concyclic." Always compute the centre and radius at the end using the general form relationships.

 

Example 32. Find the equation of the circle circumscribing the triangle formed by the straight lines x + y = 6, 2x + y = 4 and x + 2y = 5.

Answer: Let the equations of the sides AB, BC and CA of triangle ABC be

x + y = 6 ...(i)

2x + y = 4 ...(ii)

and x + 2y = 5 ...(iii)

respectively. Solving (i) and (iii); (i) and (ii); (ii) and (iii) together, we find the coordinates of the points A, B and C as (7, - 1), (- 2, 8) and (1, 2) respectively.

Let the equation of the circle be \( x^2 + y^2 + 2gx + 2fy + c = 0 \) ...(iv)

As the circle passes through the points A (7, - 1), B (- 2, 8) and C (1, 2), we get

49 + 1 + 14g - 2f + c = 0 i.e. 14g - 2f + c + 50 = 0 ...(v)

4 + 64 - 4g + 16f + c = 0 i.e. - 4g + 16f + c + 68 = 0 ...(vi)

1 + 4 + 2g + 4f + c = 0 i.e. 2g + 4f + c + 5 = 0 ...(vii)

Subtracting (vi) from (v), we find

18g - 18f - 18 = 0 i.e. g - f - 1 = 0 ...(viii)

Subtracting (vii) from (v), we find

12g - 6f + 45 = 0 i.e. 4g - 2f + 15 = 0 ...(ix)

Solving (viii) and (ix), we find g = - 17/2, f = - 19/2.

Putting these values of g and f in (vii), we find c = 50.

Substituting these values of g, f and c in (iv), we find \( x^2 + y^2 - 17x - 19y + 50 = 0 \), which is the equation of the required circle.

Exam Tip: To find the circumcircle of a triangle, first find the three vertices by solving pairs of the given line equations. Then use the general form method: substitute the three vertex coordinates into the general circle equation to get three equations, and solve for g, f, and c using elimination. Always verify that your solution is correct by checking that all three vertices satisfy the final equation.

 

Exercise 11.1

Very Short Answer Type Questions (1 to 23):

 

Question 1. Find the equation of the circle whose :
(i) centre is at the origin and the radius is 5 units
(ii) centre is (0, 2) and radius 2
(iii) centre is (- 3, 2) and radius 4
(iv) centre is (1, 1) and radius 2
(v) centre is 1/2, 1/4 and radius 1/12
(vi) centre is (- a, - b) and radius \( \sqrt{a^2 - b^2} \)

Answer:
(i) Since the centre is at the origin and the radius is 5 units, the circle equation using standard form is \( x^2 + y^2 = 5^2 \), which gives \( x^2 + y^2 = 25 \).
(ii) With centre at (0, 2) and radius 2, the central form gives \( (x - 0)^2 + (y - 2)^2 = 2^2 \), so \( x^2 + (y - 2)^2 = 4 \) or \( x^2 + y^2 - 4y = 0 \).
(iii) Centre (- 3, 2) and radius 4 give \( (x + 3)^2 + (y - 2)^2 = 16 \) or \( x^2 + y^2 + 6x - 4y - 3 = 0 \).
(iv) Centre (1, 1) and radius 2 give \( (x - 1)^2 + (y - 1)^2 = 4 \) or \( x^2 + y^2 - 2x - 2y - 2 = 0 \).
(v) Centre 1/2, 1/4 and radius 1/12 give \( \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144} \).
(vi) Centre (- a, - b) and radius \( \sqrt{a^2 - b^2} \) give \( (x + a)^2 + (y + b)^2 = a^2 - b^2 \).

Exam Tip: For each sub-part, identify the centre and radius, then apply the appropriate circle form (standard for origin centre, central form for general centre). Expand if asked, but always verify by checking coefficients match the general form characteristics.

 

Question 2. Write the equation of a circle whose centre is origin and which passes through the point (3, - 4).

Answer: Since the centre is at the origin and the circle passes through (3, - 4), the radius equals the distance from the origin to this point:

r = \( \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

Therefore, the circle equation in standard form is \( x^2 + y^2 = 25 \).

Exam Tip: When the centre is at the origin and a point on the circle is given, always compute the radius first as the distance from the origin to that point. This is the quickest route to the standard form \( x^2 + y^2 = r^2 \).

 

Question 3. Determine the equation of a circle whose centre is (8, - 6) and which passes through the point (5, - 2).

Answer: The centre of the circle is (8, - 6) and it passes through (5, - 2).

The radius is the distance from the centre to this point:

r = \( \sqrt{(5 - 8)^2 + (-2 - (-6))^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

Using the central form, the equation is \( (x - 8)^2 + (y + 6)^2 = 25 \).

Exam Tip: Always use the distance formula to find the radius when both the centre and a passing point are given. Double-check your arithmetic on the coordinate differences before squaring.

 

Question 4. Show that the points (- 3, 2) and (4, 3) lie on a circle with centre at the point (1, - 1).

Answer: To show both points lie on the same circle with centre (1, - 1), we check that they are equidistant from the centre.

Distance from (1, - 1) to (- 3, 2) = \( \sqrt{(-3 - 1)^2 + (2 - (-1))^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)

Distance from (1, - 1) to (4, 3) = \( \sqrt{(4 - 1)^2 + (3 - (-1))^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

Since both distances equal 5, both points lie on a circle with centre (1, - 1) and radius 5.

Exam Tip: To verify that two (or more) points lie on a circle with a given centre, compute the distance from each point to the centre. If all distances are equal, they lie on the same circle.

 

Question 5. Find the equation of a circle of radius 6 units and whose centre lies on the negative direction of x-axis at a distance of 4 units from the origin.

Answer: Since the centre lies on the negative direction of x-axis at distance 4 units from the origin, the centre is at (- 4, 0).

With radius 6, the equation using central form is \( (x + 4)^2 + y^2 = 36 \) or \( x^2 + y^2 + 8x - 20 = 0 \).

Exam Tip: On the negative x-axis at distance d from the origin, the centre is at (- d, 0). On the negative y-axis at distance d, the centre is at (0, - d). Always respect the sign and direction specified in the problem.

 

Question 6. Find the equation of the circle whose centre lies on the negative direction of y-axis at a distance 3 units from origin and whose radius is 4 units.

Answer: Since the centre lies on the negative direction of y-axis at distance 3 units from origin, the centre is at (0, - 3).

With radius 4, the equation is \( x^2 + (y + 3)^2 = 16 \) or \( x^2 + y^2 + 6y - 7 = 0 \).

Exam Tip: Identify the axis (x or y) and the sign direction (positive or negative) first. This determines the centre's coordinates immediately. Then apply the radius to form the circle equation.

 

Question 7. Find the equation of the circle which has A (1, 3) and B (4, 5) as opposite ends of a diameter.

Answer: Using diameter form, the equation is

(x - 1)(x - 4) + (y - 3)(y - 5) = 0

or \( x^2 - 5x + 4 + y^2 - 8y + 15 = 0 \)

or \( x^2 + y^2 - 5x - 8y + 19 = 0 \).

Exam Tip: The diameter form is the most direct when endpoints of a diameter are known: (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0. No need to find the centre or radius separately.

 

Question 8. Find the equation of the circle which has the points (- 2, 3) and (0, - 1) as opposite ends of a diameter.

Answer: Using diameter form, the equation is

(x - (- 2))(x - 0) + (y - 3)(y - (- 1)) = 0

or (x + 2)x + (y - 3)(y + 1) = 0

or \( x^2 + 2x + y^2 - 2y - 3 = 0 \).

Exam Tip: Be careful with signs when substituting negative coordinates into the diameter form. Double-check by expanding carefully.

 

Question 9. Find the equation of the circle which passes through the origin and cuts off intercepts - 2 and 3 from the co-ordinate axes.

Answer: Since the circle passes through the origin and cuts off intercepts - 2 and 3 from the coordinate axes, it passes through the points (0, 0), (- 2, 0) and (0, 3).

The endpoints of the diameter on the x-axis are (- 2, 0) and (0, 0). Therefore, the diameter along the x-axis has length 2.

The endpoints of the diameter on the y-axis are (0, 0) and (0, 3). Therefore, the diameter along the y-axis has length 3.

Using diameter form with endpoints (- 2, 0) and (0, 0) on the x-axis:

(x - (- 2))(x - 0) + (y - 0)(y - 0) = 0

or (x + 2)x + y² = 0

or \( x^2 + 2x + y^2 = 0 \).

Verify: At (0, 3), we have 0 + 0 + 9 ≠ 0, so this doesn't pass through (0, 3).

Let the circle equation be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

Through (0, 0): c = 0.

Through (- 2, 0): 4 - 4g + 0 = 0, so g = 1.

Through (0, 3): 9 + 6f = 0, so f = - 3/2.

Therefore, the equation is \( x^2 + y^2 + 2x - 3y = 0 \).

Exam Tip: When a circle passes through the origin and cuts off intercepts a and b on the axes, the circle also passes through the points (a, 0) and (0, b). Substitute all three points into the general form to find the coefficients.

 

Question 10. Find the equation of the circle which passes through origin and cuts off intercepts 3 and - 2 from the coordinate axes.

Answer: The circle passes through (0, 0), (3, 0) and (0, - 2).

Let the equation be \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

Through (0, 0): c = 0.

Through (3, 0): 9 + 6g = 0, so g = - 3/2.

Through (0, - 2): 4 - 4f = 0, so f = 1.

Therefore, the equation is \( x^2 + y^2 - 3x + 2y = 0 \).

Exam Tip: Same method as Question 9 - substitute the three known points into the general form, solve for g, f, c, and write the final equation.

 

Question 11. Does the point (- 2.5, 3.5) lie inside, outside or on the circle x² + y² = 25? (NCERT)

Answer: The centre of the circle is O (0, 0) and its radius is 5.

Distance from the centre to the given point = \( \sqrt{(-2.5)^2 + (3.5)^2} = \sqrt{6.25 + 12.25} = \sqrt{18.5} \), which is less than the radius 5.

Since the distance is less than the radius, the point (- 2.5, 3.5) lies inside the circle.

Exam Tip: For a circle with centre C and radius r, a point P lies inside if distance CP < r, on the circle if CP = r, and outside if CP > r. Always compute the distance carefully.

 

Question 12. Does the (- 3, 7) lie inside, outside or on the circle x² + y² = 49?

Answer: The circle has centre O (0, 0) and radius 7.

Distance from the centre to (- 3, 7) = \( \sqrt{9 + 49} = \sqrt{58} \), which is greater than 7 (since 58 > 49).

Therefore, the point (- 3, 7) lies outside the circle.

Exam Tip: You can compare the squared distances to avoid computing square roots if exact comparison is difficult: \( (-3)^2 + 7^2 = 58 > 49 = 7^2 \), so the point is outside.

 

Question 13. Which of the following equations represent a circle?
(i) x² + y² + 6x - 14y + 5 = 0
(ii) x² + y² + 3x - 2y + 7 = 0
(iii) 2x² + 3y² - 4x - 6y - 5 = 0
(iv) x² + y² - 10x + 2y + 20 = 0

Answer:
(i) Here, the coefficients of x² and y² are equal (both 1), and there is no xy term. Checking the discriminant: g² + f² - c = 3² + (- 7)² - 5 = 9 + 49 - 5 = 53 > 0. Therefore, this represents a circle.
(ii) Coefficients of x² and y² are equal (both 1), and there is no xy term. But g² + f² - c = (3/2)² + (- 1)² - 7 = 9/4 + 1 - 7 = - 19/4 < 0. Therefore, this does not represent a circle (it represents an empty set).
(iii) Coefficients of x² and y² are not equal (2 and 3), so this does not represent a circle (it represents an ellipse or other conic).
(iv) Coefficients of x² and y² are equal (both 1). Checking: g² + f² - c = (- 5)² + 1² - 20 = 25 + 1 - 20 = 6 > 0. Therefore, this represents a circle.

Exam Tip: To check if an equation represents a circle, verify: (a) coefficients of x² and y² are equal and non-zero, (b) no xy term, and (c) g² + f² - c > 0 (where you find g, f, c from the equation). The first two are obvious visual checks; the third is the discriminant test.

 

Question 14. What does the equation x² + y² + 4x + 6y + 13 = 0 represent?

Answer: Rewriting the equation by completing the square:

(x² + 4x + 4) + (y² + 6y + 9) + 13 - 4 - 9 = 0

(x + 2)² + (y + 3)² = 0

This is satisfied only when x + 2 = 0 and y + 3 = 0, i.e., only at the point (- 2, - 3).

Therefore, this equation represents a single point or a degenerate circle (point circle) at (- 2, - 3).

Exam Tip: Complete the square to check the constant term on the right. If it equals zero, you have a point circle. If it's negative, you have an empty set.

 

Question 15. Find the centre and the radius of the circle :
(i) (x + 5)² + (y - 3)² = 36
(ii) x² + y² - 2x + 4y = 8
(iii) x² + y² + 8x + 10y - 8 = 0
(iv) x² + y² - 4x - 8y - 45 = 0
(v) 2x² + 2y² - 3x + 5y - 7 = 0
(vi) x² + y² - ax - by = 0

Answer:
(i) From (x + 5)² + (y - 3)² = 36, we have centre = (- 5, 3) and radius = 6.
(ii) Rearranging: x² - 2x + y² + 4y = 8. Completing the square: (x - 1)² + (y + 2)² = 8 + 1 + 4 = 13. Centre = (1, - 2), radius = √13.
(iii) Completing the square: (x + 4)² + (y + 5)² = 8 + 16 + 25 = 49. Centre = (- 4, - 5), radius = 7.
(iv) Completing the square: (x - 2)² + (y - 4)² = 45 + 4 + 16 = 65. Centre = (2, 4), radius = √65.
(v) Dividing by 2: x² + y² - (3/2)x + (5/2)y - 7/2 = 0. Completing the square: (x - 3/4)² + (y + 5/4)² = 7/2 + 9/16 + 25/16 = (56 + 9 + 25)/16 = 90/16 = 45/8. Centre = (3/4, - 5/4), radius = √(45/8) = (3√10)/(2√2) = (3√5)/2.
(vi) Completing the square: (x - a/2)² + (y - b/2)² = (a² + b²)/4. Centre = (a/2, b/2), radius = √(a² + b²)/2.

Exam Tip: For central form, read centre and radius directly. For general form, complete the square on both x and y terms. For equations with a coefficient other than 1 in front of x² + y², divide first to normalize.

 

Question 16. Which of the following equations represent a circle? If so, determine its centre and radius :
(i) x² + y² + x - y = 0
(ii) x² + y² - 3x + 3y + 10 = 0
(iii) 2x² + 2y² = 5x + 7y + 3
(iv) x² + y² + 2x + 10y + 26 = 0

Answer:
(i) g = 1/2, f = - 1/2, c = 0. Check: g² + f² - c = 1/4 + 1/4 - 0 = 1/2 > 0. This represents a circle with centre (- 1/2, 1/2) and radius √(1/2) = 1/√2.
(ii) g = - 3/2, f = 3/2, c = 10. Check: g² + f² - c = 9/4 + 9/4 - 10 = 18/4 - 10 = - 11/2 < 0. This does not represent a circle (empty set).
(iii) Dividing by 2: x² + y² - (5/2)x - (7/2)y - 3/2 = 0. g = - 5/4, f = - 7/4, c = - 3/2. Check: g² + f² - c = 25/16 + 49/16 + 3/2 = (25 + 49 + 24)/16 = 98/16 > 0. This represents a circle with centre (5/4, 7/4) and radius √(98/16) = (7√2)/4.
(iv) g = 1, f = 5, c = 26. Check: g² + f² - c = 1 + 25 - 26 = 0. This represents a point circle at (- 1, - 5).

Exam Tip: For each equation, first normalize so coefficients of x² and y² are equal. Then extract g, f, c values and check the discriminant g² + f² - c. The sign of this discriminant tells you if it's a circle (>0), point circle (=0), or empty set (<0).

 

Question 17. Find the value of p so that x² + y² + 8x + 10y + p = 0 is the equation of a circle of radius 7 units.

Answer: From the given equation, g = 4, f = 5, c = p.

For a circle of radius 7, we have g² + f² - c = r² = 49.

16 + 25 - p = 49

41 - p = 49

p = - 8.

Exam Tip: Use the relationship r² = g² + f² - c. Given the radius, set up an equation for the unknown parameter and solve.

 

Question 18. Find the value(s) of k so that the equation x² + y² - 2kx + 4y - 12 = 0 may represent a circle of radius 5 units.

Answer: From the given equation, g = - k, f = 2, c = - 12.

For a circle of radius 5, g² + f² - c = 25.

k² + 4 - (- 12) = 25

k² + 4 + 12 = 25

k² = 9

k = ± 3.

Exam Tip: Set up the radius condition g² + f² - c = r² and solve for the parameter. Remember that the result may yield multiple values (here, ± 3), and both are valid unless further constraints are given.

 

Question 19. Find the equation of a circle concentric with the circle x² + y² - 8x + 2y + 3 = 0 and of radius 3 units.

Answer: For the given circle, g = - 4, f = 1, c = 3.

The centre is (- g, - f) = (4, - 1).

A concentric circle with the same centre and radius 3 has equation

(x - 4)² + (y + 1)² = 9

or x² + y² - 8x + 2y + 8 = 0.

Exam Tip: Extract the centre from the given circle using (- g, - f). Use this centre with the new radius to form the equation of the concentric circle.

 

Question 20. Find the value of k for which the circles x² + y² - 3x + ky - 5 = 0 and 4x² + 4y² - 12x - y - 9 = 0 are concentric.

Answer: For the first circle: g₁ = - 3/2, f₁ = k/2. Centre₁ = (3/2, - k/2).

For the second circle, divide by 4: x² + y² - 3x - (1/4)y - 9/4 = 0. Here g₂ = - 3/2, f₂ = - 1/8. Centre₂ = (3/2, 1/8).

For concentric circles, the centres must be equal:

3/2 = 3/2 (always true)

- k/2 = 1/8

k = - 1/4.

Exam Tip: Concentric circles have the same centre, so equate the centre coordinates of both circles. Normalize each equation first (divide if needed) to extract g and f values.

 

Question 21. Find the equation of the circle which passes through the point (1, - 2) and is concentric with the circle x² + y² - 4x + 5y - 7 = 0.

Answer: For the given circle: g = - 2, f = 5/2. Centre = (2, - 5/2).

The required circle is concentric, so it has the same centre (2, - 5/2).

The radius is the distance from the centre to the point (1, - 2):

r = \( \sqrt{(1 - 2)^2 + (-2 - (-5/2))^2} = \sqrt{1 + (1/2)^2} = \sqrt{1 + 1/4} = \sqrt{5/4} = \sqrt{5}/2 \)

The equation is (x - 2)² + (y + 5/2)² = 5/4 or 4(x² + y²) - 16x + 20y + 41 = 0.

Exam Tip: Extract the centre from the given concentric circle. Compute the radius using the distance from the centre to the given passing point. Form the new circle equation using the central form.

 

Question 22. Find the shortest distance of the point (8, 1) from the circle (x + 2)² + (y - 1)² = 25.

Answer: The circle has centre C (- 2, 1) and radius 5.

Distance from the point P (8, 1) to the centre = \( \sqrt{(8 - (-2))^2 + (1 - 1)^2} = \sqrt{100} = 10 \)

Since the distance from P to C is 10, which is greater than the radius 5, the point is outside the circle.

The shortest distance from P to the circle = distance from P to C - radius = 10 - 5 = 5 units.

Exam Tip: For a point outside a circle, the shortest distance is along the line joining the point to the centre. It equals (distance from point to centre) - (radius). For a point inside, it equals (radius) - (distance from point to centre).

 

Question 41. Find the equation of the circle which passes through the centre of the circle x² + y² + 8x + 10y + 7 = 0 and is concentric with the circle 2x² + 2y² - 8x - 12y - 9 = 0.
Answer: First, find the centre of the given circle x² + y² + 8x + 10y + 7 = 0 by rewriting it as (x + 4)² + (y + 5)² = 16. The centre is (-4, -5). Next, rewrite the second circle 2x² + 2y² - 8x - 12y - 9 = 0 as x² + y² - 4x - 6y - 4.5 = 0, which gives centre (2, 3). Since the required circle is concentric with the second circle, its centre is also (2, 3). The required circle passes through (-4, -5), so its radius is the distance from (2, 3) to (-4, -5): \( r = \sqrt{(-4-2)^2 + (-5-3)^2} = \sqrt{36 + 64} = 10 \). Therefore, the equation is \( (x - 2)^2 + (y - 3)^2 = 100 \), or x² + y² - 4x - 6y - 87 = 0.
In simple words: A concentric circle has the same centre but a different radius. Find where the first circle's centre is, then use that point's distance from the new centre to get the radius.

Exam Tip: Always convert the circle equations to standard form to identify the centre quickly. The key is that concentric circles share a centre.

 

Question 42. Find the equation of the circle concentric with the circle x² + y² + 4x - 8y - 6 = 0 and having radius double of its radius.
Answer: Rewrite the given circle as (x + 2)² + (y - 4)² = 26, so the centre is (-2, 4) and radius is \( \sqrt{26} \). The new circle has the same centre (-2, 4) but radius \( 2\sqrt{26} \). Its equation is \( (x + 2)^2 + (y - 4)^2 = 4 \times 26 = 104 \), which expands to x² + y² + 4x - 8y - 84 = 0.
In simple words: Keep the centre the same but make the radius twice as big. Then expand the equation.

Exam Tip: When the radius doubles, the radius-squared (used in the equation) becomes four times larger.

 

Question 43. Find the equation of the circle concentric with the circle 2x² + 2y² + 8x + 10y - 35 = 0 and with area 16π square units.
Answer: Simplify the given circle to x² + y² + 4x + 5y - 17.5 = 0, giving centre (-2, -2.5). From area 16π, we get \( \pi r^2 = 16\pi \), so r² = 16, meaning r = 4. The new circle's equation is \( (x + 2)^2 + (y + 2.5)^2 = 16 \), which expands to 4x² + 4y² + 16x + 20y - 23 = 0.
In simple words: Use the area to find the radius. The concentric circle uses the same centre but this new radius.

Exam Tip: Remember that area = πr², so always solve for r first before writing the circle equation.

 

Question 44. Find the equation of the circle which is concentric with the circle x² + y² - 4x + 6y - 3 = 0 and of double its
(i) circumference
(ii) area

Answer: The given circle rewrites as (x - 2)² + (y + 3)² = 16, so centre is (2, -3) and radius is 4.
(i) If circumference doubles, then 2πr becomes 4πr, so new radius = 8. Equation: (x - 2)² + (y + 3)² = 64, or x² + y² - 4x + 6y - 51 = 0.
(ii) If area doubles, then πr² becomes 2πr², so new r² = 32. Equation: (x - 2)² + (y + 3)² = 32, or x² + y² - 4x + 6y - 19 = 0.
In simple words: Double circumference means double radius. Double area means radius times \( \sqrt{2} \).

Exam Tip: Distinguish carefully: doubling circumference ≠ doubling area. Circumference involves r linearly; area involves r².

 

Question 45. Prove that the centres of three circles x² + y² - 4x - 6y - 14 = 0, x² + y² + 2x + 4y - 5 = 0 and x² + y² - 10x - 16y + 7 = 0 are collinear.
Answer: Rewrite each circle in standard form to find centres. The first circle (x - 2)² + (y - 3)² = 27 has centre C₁(2, 3). The second circle (x + 1)² + (y + 2)² = 10 has centre C₂(-1, -2). The third circle (x - 5)² + (y - 8)² = 82 has centre C₃(5, 8). To check collinearity, use the slope method: slope of C₁C₂ = \( \frac{-2-3}{-1-2} = \frac{-5}{-3} = \frac{5}{3} \) and slope of C₂C₃ = \( \frac{8-(-2)}{5-(-1)} = \frac{10}{6} = \frac{5}{3} \). Since both slopes are equal, the three centres are collinear.
In simple words: Three points are collinear if the slope between any two pairs matches. Find each circle's centre, then check the slopes.

Exam Tip: Always convert to standard form (x - h)² + (y - k)² = r² first to extract the centre (h, k) clearly.

 

Question 46. Prove that the radii of the circles x² + y² = 1, x² + y² - 2x - 6y - 6 = 0 and x² + y² - 4x - 12y - 9 = 0 are in A.P.
Answer: For the first circle x² + y² = 1, the radius is r₁ = 1. For the second circle, rewrite as (x - 1)² + (y - 3)² = 16, so r₂ = 4. For the third, rewrite as (x - 2)² + (y - 6)² = 49, so r₃ = 7. Check if r₁, r₂, r₃ form an A.P.: the common difference should be constant. r₂ - r₁ = 4 - 1 = 3 and r₃ - r₂ = 7 - 4 = 3. Since the common difference is 3, the radii 1, 4, 7 are in A.P.
In simple words: Three numbers form an arithmetic sequence if the gap between the first and second equals the gap between the second and third.

Exam Tip: For an A.P., always verify that the middle term equals half the sum of the first and third: \( r_2 = \frac{r_1 + r_3}{2} \).

 

Question 47. Find the equation of the circle passing through the points (4, 1), (6, 5) and whose centre is on the line 4x + y = 16. (NCERT)
Answer: Let the centre be (h, k). Since it lies on 4x + y = 16, we have 4h + k = 16, so k = 16 - 4h. The circle passes through (4, 1) and (6, 5), so the distances from the centre to each point are equal: \( (h - 4)^2 + (k - 1)^2 = (h - 6)^2 + (k - 5)^2 \). Expanding: h² - 8h + 16 + k² - 2k + 1 = h² - 12h + 36 + k² - 10k + 25. Simplifying: -8h - 2k + 17 = -12h - 10k + 61, which gives 4h + 8k = 44, or h + 2k = 11. Substituting k = 16 - 4h: h + 2(16 - 4h) = 11, so h + 32 - 8h = 11, giving 7h = 21, thus h = 3. Then k = 16 - 12 = 4. The radius is \( \sqrt{(3-4)^2 + (4-1)^2} = \sqrt{1 + 9} = \sqrt{10} \). The equation is (x - 3)² + (y - 4)² = 10, or x² + y² - 6x - 8y + 15 = 0.
In simple words: Use the constraint that the centre lies on a given line, and the equal-distance property for two points.

Exam Tip: Always equate the distances from centre to the two given points; this eliminates the radius variable.

 

Question 48. Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the line y - 4x + 3 = 0. (NCERT Exemplar Problems)
Answer: Let the centre be (h, k). Since it lies on y - 4x + 3 = 0, we have k = 4h - 3. The circle passes through (2, 3) and (4, 5), so \( (h - 2)^2 + (k - 3)^2 = (h - 4)^2 + (k - 5)^2 \). Expanding and simplifying: h² - 4h + 4 + k² - 6k + 9 = h² - 8h + 16 + k² - 10k + 25. This reduces to -4h - 6k + 13 = -8h - 10k + 41, or 4h + 4k = 28, giving h + k = 7. Substituting k = 4h - 3: h + 4h - 3 = 7, so 5h = 10, thus h = 2. Then k = 5. The radius is \( \sqrt{(2-2)^2 + (5-3)^2} = 2 \). The equation is (x - 2)² + (y - 5)² = 4, or x² + y² - 4x - 10y + 25 = 0.
In simple words: Use the line equation to express one coordinate in terms of the other, then apply the equal-distance condition.

Exam Tip: Substitute the line's constraint early to reduce the number of unknowns before expanding.

 

Question 49. Find the equation of the circle which passes through the points (2, -2), (3, 4) and whose centre is on the line x + y = 2. (NCERT)
Answer: Let the centre be (h, k). From x + y = 2, we get k = 2 - h. Since the circle passes through both (2, -2) and (3, 4), the distances are equal: \( (h - 2)^2 + (k + 2)^2 = (h - 3)^2 + (k - 4)^2 \). Expanding: h² - 4h + 4 + k² + 4k + 4 = h² - 6h + 9 + k² - 8k + 16. Simplifying: -4h + 4k + 8 = -6h - 8k + 25, which gives 2h + 12k = 17. Substituting k = 2 - h: 2h + 12(2 - h) = 17, so 2h + 24 - 12h = 17, giving 10h = 7, thus h = 0.7. Then k = 1.3. The radius is \( \sqrt{(0.7-2)^2 + (1.3+2)^2} = \sqrt{1.69 + 10.89} = \sqrt{12.58} \). Alternatively, multiply through to clear decimals: the equation becomes 5x² + 5y² - 7x - 13y - 52 = 0.
In simple words: Express the centre using the line constraint, then use equal distances to find exact coordinates.

Exam Tip: If decimals arise, multiply the final equation by an appropriate factor to clear them and match the answer key format.

 

Question 50. Find the equation of the circle passing through the three points
(i) (0, 0), (0, 1) and (2, 3)
(ii) (0, 0), (5, 0) and (3, 3)
(iii) (1, 0), (-1, 0) and (0, 1)
(iv) (1, 2), (3, -4) and (5, -6)
(v) (20, 3), (19, 8) and (2, -9)
Answer: For a circle through three non-collinear points, use the general form x² + y² + 2gx + 2fy + c = 0 and substitute each point to get three equations.
(i) Substituting (0, 0), (0, 1), (2, 3): From (0, 0): c = 0. From (0, 1): 1 + 2f + 0 = 0, so f = -1/2. From (2, 3): 4 + 9 + 4g - 6 + 0 = 0, so 4g = -7, giving g = -5/4. Thus c = 0, and the equation is x² + y² - 5x/2 - y = 0, or 2x² + 2y² - 5x - 2y = 0. Rearranging: x² + y² - 5x - y = 0. Centre: (5/2, 1/2), Radius: \( \sqrt{26}/2 \).
(ii) Substituting (0, 0), (5, 0), (3, 3): From (0, 0): c = 0. From (5, 0): 25 + 10g + 0 = 0, so g = -5/2. From (3, 3): 9 + 9 + 6g + 6f + 0 = 0, so 18 - 15 + 6f = 0, giving f = -1/2. Thus x² + y² - 5x - y = 0. Centre: (5/2, 1/2), Radius: \( \sqrt{26}/2 \).
(iii) Substituting (1, 0), (-1, 0), (0, 1): From (1, 0): 1 + 2g + c = 0. From (-1, 0): 1 - 2g + c = 0. Adding: 2 + 2c = 0, so c = -1. Subtracting: 4g = 0, so g = 0. From (0, 1): 1 + 2f - 1 = 0, so f = 0. Thus x² + y² - 1 = 0. Centre: (0, 0), Radius: 1.
(iv) Use the three equations to solve for g, f, c. After computation: x² + y² - 22x - 4y + 25 = 0. Centre: (11, 2), Radius: 10.
(v) After computation: x² + y² - 14x - 6y - 111 = 0. Centre: (7, 3), Radius: 13.
In simple words: Substitute each of the three points into the general circle equation and solve the resulting system for the coefficients.

Exam Tip: Always verify your answer by checking that all three given points satisfy the final equation.

 

Question 51. Find the equation of the circle which is circumscribed about the triangle whose vertices are (-2, 3), (5, 2) and (6, -1).
Answer: The circumscribed circle (circumcircle) passes through all three vertices. Use the general form x² + y² + 2gx + 2fy + c = 0. Substituting (-2, 3): 4 + 9 - 4g + 6f + c = 0, or -4g + 6f + c = -13. Substituting (5, 2): 25 + 4 + 10g + 4f + c = 0, or 10g + 4f + c = -29. Substituting (6, -1): 36 + 1 + 12g - 2f + c = 0, or 12g - 2f + c = -37. Solving this system: From the first two equations, 14g - 2f = -16, or 7g - f = -8. From the second and third, 2g - 6f = -8, or g - 3f = -4. Solving: From 7g - f = -8 and g - 3f = -4, multiply the second by 7: 7g - 21f = -28. Subtract from the first: 20f = 20, so f = 1. Then g = -1, and c = -23. The equation is x² + y² - 2x + 2y - 23 = 0.
In simple words: The circle through three points is found by substituting each point into the general form and solving for the unknown coefficients.

Exam Tip: Keep the arithmetic organised by writing coefficients in a clear system. Double-check arithmetic when solving for g, f, and c.

 

Question 52. Show that the points (7, 1), (-2, 4), (5, 5) and (6, 4) are concyclic. Also find the equation, centre and radius of the circle on which they lie.
Answer: Four points are concyclic if they all satisfy the same circle equation. Use the general form x² + y² + 2gx + 2fy + c = 0. Substitute all four points and solve for g, f, c. From (7, 1): 49 + 1 + 14g + 2f + c = 0, or 14g + 2f + c = -50. From (-2, 4): 4 + 16 - 4g + 8f + c = 0, or -4g + 8f + c = -20. From (5, 5): 25 + 25 + 10g + 10f + c = 0, or 10g + 10f + c = -50. From (6, 4): 36 + 16 + 12g + 8f + c = 0, or 12g + 8f + c = -52. Using the first three equations: From (1) and (2): 18g - 6f = -30, or 3g - f = -5. From (1) and (3): 4g - 8f = 0, or g = 2f. Solving: g = 2f and 3(2f) - f = -5 give 5f = -5, so f = -1 and g = -2. From (1): -28 - 2 + c = -50, so c = -20. Check with equation (4): -24 - 8 - 20 = -52. ✓ The circle equation is x² + y² - 4x - 2y - 20 = 0, or (x - 2)² + (y - 1)² = 25. Centre: (2, 1), Radius: 5. Thus, all four points are concyclic.
In simple words: If all four points satisfy the same circle equation, they lie on a circle. Verify the fourth point after finding the equation.

Exam Tip: To prove concyclicity, either find a single circle equation that all four points satisfy, or show that opposite angles in the quadrilateral sum to 180°.

 

11.3 Symmetry

Reflection of a point in a line

If a point P lies on the line l, then the reflection of P in l is the point P itself. If P does not lie on l, let M be the foot of perpendicular from P on l and extend it to a point P' such that MP = MP', then P' is called the reflection of P in l.

Note that if P' is the reflection of P in l, then P is the reflection of P' in l.

It follows that if P does not lie on l, then P' is the reflection of P in l iff l is the perpendicular bisector of the segment PP'.

Reflection of a point in a point

The reflection of a point P in a fixed point M is the point P' such that M is the mid-point of the segment PP'.

In particular, if P = M then P' = M = P.

Note that if P' is the reflection of P in a point M, then P is the reflection of P' in M.

We leave it for the reader to see that the reflection of a point (α, β) in the

(i) origin is the point (-α, -β)

(ii) x-axis is the point (α, -β)

(iii) y-axis is the point (-α, β).

Symmetry of a curve about a line

A curve C is said to be symmetrical about a line l iff for every point P on C, the reflection of P in l also lies on C. If a curve is symmetrical about a line l, then l is called a line of symmetry of C or an axis of C.

Let F(x, y) = 0 be an equation of a curve C, then C is symmetrical about

(i) x-axis iff F(x, y) = F(x, -y)

(ii) y-axis iff F(x, y) = F(-x, y).

Proof. (i) Let P(α, β) be a point on the curve C, then F(α, β) = 0 …(1)

Now the curve C is symmetrical about x-axis iff the reflection of P in the x-axis i.e. the point P'(α, -β) lies on C i.e. iff F(α, -β) = 0 …(2)

From (1) and (2), it follows that C is symmetrical about x-axis iff F(x, y) = F(x, -y).

We leave the proof of (ii) for the reader.

Symmetry of a curve about a point

A curve C is said to be symmetrical about a point M iff for every point P on C, the reflection of P in M also lies on C. If a curve is symmetrical about a point M, then M is called a centre of symmetry of C or a centre of C.

Let F(x, y) = 0 be an equation of a curve C, then C is symmetrical about the origin iff F(x, y) = F(-x, -y).

Proof. Let P(α, β) be a point on the curve C, then F(α, β) = 0 …(1)

Now the curve C is symmetrical about the origin iff the reflection of P in the origin i.e. the point P'(-α, -β) lies on C i.e. iff F(-α, -β) = 0 …(2)

From (1) and (2), it follows that C is symmetrical about origin iff F(x, y) = F(-x, -y).

Illustrative Example

Example. Which of the following curves are symmetrical about the x-axis? or y-axis? or origin?

(i) 7y² = 5x

(ii) 3x² - 4y² + 7 = 0

Solution. (i) The equation of the given curve is F(x, y) = 7y² - 5x = 0.

Here, F(x, -y) = 7(-y)² - 5x = 7y² - 5x = F(x, y);

F(-x, y) = 7y² - 5(-x) = 7y² + 5x ≠ F(x, y);

F(-x, -y) = 7(-y)² - 5(-x) = 7y² + 5x ≠ F(x, y).

Therefore, the given curve is symmetrical about x-axis but neither about y-axis nor about the origin.

(ii) The equation of the given curve is F(x, y) = 3x² - 4y² + 7 = 0.

Here, F(x, -y) = 3x² - 4(-y)² + 7 = 3x² - 4y² + 7 = F(x, y);

F(-x, y) = 3(-x)² - 4y² + 7 = 3x² - 4y² + 7 = F(x, y);

F(-x, -y) = 3(-x)² - 4(-y)² + 7 = 3x² - 4y² + 7 = F(x, y).

Therefore, the given curve is symmetrical about x-axis, y-axis and about the origin.

 

Exercise 11.2

Which of the following curves are symmetrical about the x-axis? or y-axis? or origin?

1. 5x² + 7y = 0

2. 3y² = 7x

3. x² = 3y² + 7

4. x² + y² = 25

 

11.4 Parabola

A parabola is the set of all points in a plane which are equidistant from a fixed line and a fixed point (not on the line) in the plane.

The fixed line (say l) is called the directrix of the parabola and the fixed point (say F) is called the focus of the parabola.

If P₁, P₂, P₃, … are points on the parabola and M₁P₁, M₂P₂, M₃P₃ are perpendiculars to the directrix l, then FP₁ = M₁P₁, FP₂ = M₂P₂, FP₃ = M₃P₃ etc.

The line passing through the focus and perpendicular to the directrix is called the axis of parabola. The point of intersection of parabola with its axis is called vertex of parabola.

The equation of a parabola is in simplest form if its vertex is at the origin and its axis lies along either x-axis or y-axis.

11.4.1 To find the equation of a parabola in the standard form y² = 4ax, a - 0

Let F be the focus, l be the directrix and Z be the foot of perpendicular from F to the line l.

Take ZF as x-axis with positive direction from Z to F. Let O be the mid-point of ZF, take O as origin, then the line through O and perpendicular to ZF becomes y-axis.

Let ZF = 2a (a - 0, because F does not lie on l), then ZO = OF = a.

Since F lies to the right of O and Z lies to the left of O, coordinates of F, Z are (a, 0), (-a, 0) respectively. Therefore, the equation of the line l i.e. directrix is x = -a i.e. x + a = 0.

Let P(x, y) be any point in the plane of the line l and the point F, and MP be the perpendicular distance from P to the line l then P lies on parabola iff FP = MP

\( \iff \sqrt{(x - a)^2 + y^2} = \frac{|x + a|}{1} \)

\( \iff (x - a)^2 + y^2 = (x + a)^2 \)

\( \iff x^2 + a^2 - 2ax + y^2 = x^2 + a^2 + 2ax \)

\( \iff y^2 = 4ax \)

Hence, the equation of a parabola in the standard form is y² = 4ax, a - 0, with focus F(a, 0) and directrix x + a = 0.

Sometimes it is called a first standard form or a right hand parabola.

11.4.2 Some facts about the parabola y² = 4ax, a - 0.

The equation of the parabola is F(x, y) = y² - 4ax = 0 …(i)

We note the following facts about the given parabola:

1. F(x, -y) = (-y)² - 4ax = y² - 4ax = F(x, y) - the given parabola is symmetrical about x-axis. This line is the axis of the parabola.

2. Focus is F(a, 0) and directrix is x + a = 0.

3. The point O(0, 0) where the axis of parabola meets the parabola is the vertex of the parabola, it is mid-point of ZF where F(a, 0) is the focus and Z is the foot of perpendicular from focus to the directrix.

4. If x - 0, then y² = 4ax has no real solutions in y and so there is no point on the curve with negative x-coordinate i.e. on the left of y-axis. When x = 0, we get y² = 0 - y = 0. Thus, (0, 0) is the only point of the y-axis which lies on it, therefore, the entire curve, except the origin, lies to the right of y-axis.

5. If P(x, y) is a point of the parabola in the first quadrant, then the equation (i) gives \( y = 2\sqrt{ax} \) so that as x increases, y also increases and the curve is unbounded.

6. A chord passing through the focus F and perpendicular to the axis of parabola is called latus-rectum and its length is called length of latus-rectum.

7. Length of latus-rectum.

Let chord L'L be the latus-rectum of the parabola, then L'L passes through focus F(a, 0) and is perpendicular to x-axis.

Let LF = k (k - 0), then the points L and L' are (a, k) and (a, -k) respectively.

As L(a, k) lies on the parabola y² = 4ax, we get k² = 4a × a - k = 2a.

Therefore The points L, L' are (a, 2a), (a, -2a) and length of latus-rectum = L'L = 2k = 4a.

Thus, the length of latus rectum = 4a = 2ZF = 4OF.

The end points of the latus-rectum are L(a, 2a), L'(a, -2a) and the equation of the latus-rectum is x - a = 0.

11.4.3 To find the equation of a parabola in other standard forms

Find the equation of a parabola with

(i) focus F(-a, 0), a - 0 and the line x - a = 0 as directrix.

(ii) focus F(0, a), a - 0 and the line y + a = 0 as directrix.

(iii) focus F(0, -a), a - 0 and the line y - a = 0 as directrix.

Solution. (i) Let P(x, y) be any point in the plane of directrix and focus, and MP be the perpendicular distance from P to the directrix, then P lies on parabola iff FP = MP

\( \iff \sqrt{(x + a)^2 + y^2} = \frac{|x - a|}{1} \)

\( \iff (x + a)^2 + y^2 = (x - a)^2 \)

\( \iff x^2 + a^2 + 2ax + y^2 = x^2 + a^2 - 2ax \)

\( \iff y^2 = -4ax, \) a - 0.

It is called 2nd standard form or left hand parabola.

(ii) Proceeding as above, it will be found that the equation of the parabola is x² = 4ay, a - 0 (Do it!) It is called 3rd standard form or upward parabola.

(iii) Proceeding as in part (i), it will be found that the equation of the parabola is x² = -4ay, a - 0 (Do it!) It is called 4th standard form or downward parabola.

11.4.4 Four standard forms of the parabola

The students are advised to trace the following four standard forms of parabolas and to fix in their memory the positions (figures) of these parabolas with respect to co-ordinate axes.

Main facts about the parabola

Illustrative Examples

 

Example 1. Find the co-ordinates of focus, the axis, the equation of the directrix and the length of latus-rectum of the parabola represented by the equation 3y² = 8x.
Answer: The given equation is 3y² = 8x, which becomes \( y^2 = \frac{8}{3}x \). This is the same as y² = 4ax with \( 4a = \frac{8}{3} \), so \( a = \frac{2}{3} \). This is a right-hand parabola with axis along the x-axis. The focus is at \( (a, 0) = (\frac{2}{3}, 0) \). The directrix equation is \( x + a = 0 \), which gives \( 3x + 2 = 0 \). The length of latus-rectum is \( 4a = \frac{8}{3} \).
In simple words: Rewrite the equation in standard form y² = 4ax. The value a tells you the focus, directrix, and latus-rectum length.

Exam Tip: Always identify which standard form the equation matches first - this immediately gives you the focus and directrix directions.

 

Example 2. Find the co-ordinates of focus, the equation of directrix and the length of latus-rectum of the conic represented by the equation x² = -16y. (NCERT)
Answer: The equation x² = -16y matches the fourth standard form x² = -4ay with 4a = 16, so a = 4. This is a downward parabola. The focus is at \( (0, -a) = (0, -4) \). The directrix is \( y - a = 0 \), which gives y = 4 or y - 4 = 0. The length of latus-rectum is 4a = 16.
In simple words: The negative sign tells you the parabola opens downward. The 4a value gives you both the focus and latus-rectum length.

Exam Tip: Pay attention to the sign of a: negative means the parabola opens left or down; positive means right or up.

 

Example 3. Find the equation of the parabola with focus (6, 0) and directrix x = -6. Also find the length of latus-rectum. (NCERT)
Answer: Let P(x, y) be any point on the parabola. By definition, the distance from P to the focus equals the distance from P to the directrix:
\( \sqrt{(x - 6)^2 + y^2} = |x + 6| \)

Squaring both sides:
\( (x - 6)^2 + y^2 = (x + 6)^2 \)

\( x^2 - 12x + 36 + y^2 = x^2 + 12x + 36 \)

\( y^2 = 24x \)

Comparing with y² = 4ax, we get 4a = 24, so a = 6. Therefore, the length of latus-rectum is 4a = 24. Alternatively, since the focus is at (6, 0) on the x-axis, this is a first standard form parabola with a = 6, giving y² = 24x.
In simple words: Use the definition: a point is on the parabola if its distance to the focus equals its distance to the directrix.

Exam Tip: When deriving the equation, always square carefully and cancel like terms on both sides.

 

Example 4. Find the equation of a parabola with focus at (-1, -2) and directrix x - 2y + 3 = 0. (NCERT Exemplar Problems)
Answer: Let P(x, y) be any point on the parabola. The distance from P to the focus F(-1, -2) equals the distance from P to the directrix x - 2y + 3 = 0:
\( \sqrt{(x + 1)^2 + (y + 2)^2} = \frac{|x - 2y + 3|}{\sqrt{1^2 + (-2)^2}} = \frac{|x - 2y + 3|}{\sqrt{5}} \)

Squaring:
\( (x + 1)^2 + (y + 2)^2 = \frac{(x - 2y + 3)^2}{5} \)

\( 5[(x + 1)^2 + (y + 2)^2] = (x - 2y + 3)^2 \)

\( 5(x^2 + 2x + 1 + y^2 + 4y + 4) = x^2 + 4y^2 + 9 - 4xy + 6x - 12y \)

\( 5x^2 + 5y^2 + 10x + 20y + 25 = x^2 + 4y^2 + 9 - 4xy + 6x - 12y \)

\( 4x^2 + 4xy + y^2 + 4x + 32y + 16 = 0 \)

This is the required equation of the parabola.
In simple words: Set distance to focus equal to distance to directrix. The directrix is a line, so use the point-to-line distance formula.

Exam Tip: When the directrix is not parallel to an axis, use the perpendicular distance formula \( \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} \).

 

Example 5. Find the equation of the parabola with vertex at (0, 0) and focus at (-2, 0). (NCERT)
Answer: Since the focus (-2, 0) lies on the x-axis and the vertex is at the origin, the axis is the x-axis itself. The parabola opens to the left (second standard form). With a = 2, the equation is y² = -4ax = -4(2)x = -8x, or y² = -8x.
In simple words: The focus and vertex determine both which axis the parabola uses and which direction it opens.

Exam Tip: When the focus is at (-a, 0) and vertex at origin, the parabola opens left, giving y² = -4ax.

 

Example 6. Find the equation of the parabola with vertex at origin and directrix the line y + 3 = 0. Also find its focus.
Answer: The vertex is at the origin (0, 0). The directrix is y = -3, which is a horizontal line below the origin at distance 3. This means the parabola opens upward (third standard form) with a = 3. The equation is x² = 4ay = 4(3)y = 12y, or x² = 12y. The focus is at (0, a) = (0, 3).
In simple words: If the directrix is horizontal, the parabola opens up or down. If the directrix is vertical, it opens left or right.

Exam Tip: The vertex is always midway between the focus and the directrix.

 

Example 7. Find the equation of the parabola with vertex at origin, symmetric with respect to y-axis and passing through (2, -3). (NCERT)
Answer: The vertex is at the origin and the parabola is symmetric about the y-axis. This means the axis is the y-axis. Since the point (2, -3) is in the fourth quadrant (below the x-axis), the parabola opens downward (fourth standard form). Let the equation be x² = -4ay. Substituting (2, -3): 4 = -4a(-3) = 12a, so a = 1/3. Thus x² = -4(1/3)y, or 3x² = -4y.
In simple words: Use the given point to find the value of a, then write the full equation.

Exam Tip: Check which quadrant the given point is in to determine the opening direction of the parabola.

 

Example 8. If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola. (NCERT Exemplar Problems)
Answer: The vertex A is at (0, 4) and focus F is at (0, 2). Both lie on the y-axis, so the axis is the y-axis. The distance AF = 2 = a. Since the focus is below the vertex, the parabola opens downward. The vertex form is \( (x - h)^2 = -4a(y - k) \) with (h, k) = (0, 4) and a = 2: \( x^2 = -8(y - 4) \), or \( x^2 = -8y + 32 \), or \( x^2 + 8y = 32 \).
In simple words: Find a from the distance between vertex and focus, then use the vertex form of the parabola equation.

Exam Tip: If the focus is below the vertex on the y-axis, the parabola opens downward; if above, it opens upward.

 

Example 9. Find the equations of the lines joining the vertex of the parabola y² = 6x to the points which have abscissa 24. (NCERT Exemplar Problems)
Answer: The vertex of y² = 6x is at the origin O(0, 0). Let P(24, k) be a point on the parabola. Substituting into the equation: k² = 6(24) = 144, so k = ±12. Thus there are two points: P(24, 12) and Q(24, -12). The slope of OP is \( \frac{12}{24} = \frac{1}{2} \). The slope of OQ is \( \frac{-12}{24} = -\frac{1}{2} \). The equations of lines OP and OQ are y = (1/2)x, or x - 2y = 0, and y = (-1/2)x, or x + 2y = 0, respectively.
In simple words: Find the points on the parabola at the given x-coordinate, then find the slope of each line from the vertex to that point.

Exam Tip: A parabola is symmetric about its axis, so if (24, 12) is on it, then (24, -12) is also on it (for parabolas of form y² = 4ax).