RS Aggarwal Solutions for Class 11 Chapter 21 Circle

Question 1. Find the equation of a circle with Centre (2, 4) and radius 5
Answer: The standard form used to express the equation of a circle is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) represents the centre and r represents the radius.

Plugging in the given centre (2, 4) and radius 5:
\( (x - 2)^2 + (y - 4)^2 = 5^2 \)
\( (x - 2)^2 + (y - 4)^2 = 25 \)

Thus, the equation is \( (x - 2)^2 + (y - 4)^2 = 25 \)

Exam Tip: Always substitute the given centre and radius directly into the standard form formula - careful with signs when the centre has negative coordinates.

Question 2. Find the equation of a circle with Centre (- 3, - 2) and radius 6
Answer: The standard form for expressing a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Substituting the centre (- 3, - 2) and radius 6:
\( (x - (- 3))^2 + (y - (- 2))^2 = 6^2 \)
\( (x + 3)^2 + (y + 2)^2 = 36 \)

Therefore, the equation is \( (x + 3)^2 + (y + 2)^2 = 36 \)

Exam Tip: When the centre coordinates are negative, double negatives become positive - take extra care with the signs in the equation.

Question 3. Find the equation of a circle with Centre (a, a) and radius \( \sqrt{2} \)
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) denotes the centre and r denotes the radius.

Substituting the centre (a, a) and radius \( \sqrt{2} \):
\( (x - a)^2 + (y - a)^2 = (\sqrt{2})^2 \)
\( (x - a)^2 + (y - a)^2 = 2 \)

Thus, the equation is \( (x - a)^2 + (y - a)^2 = 2 \)

Exam Tip: When working with algebraic parameters like a, treat them as constants and proceed with the same substitution method used for numerical values.

Question 4. Find the equation of a circle with Centre (a cos α, a sin α) and radius a
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Substituting the centre (a cos α, a sin α) and radius a:
\( (x - a \cos \alpha)^2 + (y - a \sin \alpha)^2 = a^2 \)

Expanding both squared terms:
\( x^2 - 2xa\cos \alpha + a^2 \cos^2 \alpha + y^2 - 2ya\sin \alpha + a^2 \sin^2 \alpha = a^2 \)

Regrouping:
\( x^2 + y^2 + a^2(\cos^2 \alpha + \sin^2 \alpha) - 2a(x\cos \alpha + y\sin \alpha) = a^2 \)

Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\( x^2 + y^2 + a^2 - 2a(x\cos \alpha + y\sin \alpha) = a^2 \)

Simplifying:
\( x^2 + y^2 - 2a(x\cos \alpha + y\sin \alpha) = 0 \)

Exam Tip: Remember to apply trigonometric identities during simplification - \( \cos^2 \alpha + \sin^2 \alpha = 1 \) is key to reducing the equation.

Question 5. Find the equation of a circle with Centre (- a, - b) and radius \( \sqrt{a^2 - b^2} \)
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Substituting the centre (- a, - b) and radius \( \sqrt{a^2 - b^2} \):
\( (x - (- a))^2 + (y - (- b))^2 = (\sqrt{a^2 - b^2})^2 \)
\( (x + a)^2 + (y + b)^2 = a^2 - b^2 \)

Expanding:
\( x^2 + 2xa + a^2 + y^2 + 2yb + b^2 = a^2 - b^2 \)

Simplifying:
\( x^2 + 2xa + y^2 + 2yb = a^2 - 2b^2 \)
\( x^2 + y^2 + 2a(x + y) = a^2 - 2b^2 \)

Exam Tip: Pay close attention when expanding squared terms with negative centre coordinates - proper distribution of signs is essential.

Question 6. Find the equation of a circle with Centre at the origin and radius 4
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Since the centre is at the origin, (h, k) = (0, 0), and the radius is 4:
\( (x - 0)^2 + (y - 0)^2 = 4^2 \)
\( x^2 + y^2 = 16 \)

Therefore, the equation is \( x^2 + y^2 = 16 \)

Exam Tip: When a circle is centred at the origin, the equation simplifies dramatically - no x or y terms appear in the linear form.

Question 7A. Find the centre and radius of the circle: \( (x - 3)^2 + (y - 1)^2 = 9 \)
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Comparing the given equation with the standard form:
\( (x - 3)^2 + (y - 1)^2 = 9 \)

We obtain: h = 3, k = 1, and \( r^2 = 9 \), so r = 3.

Thus, the centre is (3, 1) and the radius is 3 units.

Exam Tip: When reading centre coordinates from the standard form equation, note that the sign inside each bracket is opposite to the actual coordinate.

Question 7B. Find the centre and radius of the circle: \( \left(x - \frac{1}{2}\right)^2 + \left(y + \frac{1}{3}\right)^2 = \frac{1}{16} \)
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Comparing the given equation with the standard form:
\( \left(x - \frac{1}{2}\right)^2 + \left(y + \frac{1}{3}\right)^2 = \frac{1}{16} \)

We get: h = \( \frac{1}{2} \), k = - \( \frac{1}{3} \), and \( r^2 = \frac{1}{16} \), so r = \( \frac{1}{4} \).

Thus, the centre is \( \left(\frac{1}{2}, - \frac{1}{3}\right) \) and the radius is \( \frac{1}{4} \) units.

Exam Tip: With fractional coefficients, be particularly careful about signs and simplification when extracting the centre coordinates and radius.

Question 7C. Find the centre and radius of the circle: \( (x + 5)^2 + (y - 3)^2 = 20 \)
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Comparing the given equation with the standard form:
\( (x + 5)^2 + (y - 3)^2 = 20 \)

This may be rewritten as: \( (x - (- 5))^2 + (y - 3)^2 = 20 \)

We get: h = - 5, k = 3, and \( r^2 = 20 \), so r = \( \sqrt{20} = 2\sqrt{5} \).

Thus, the centre is (- 5, 3) and the radius is \( 2\sqrt{5} \) units.

Exam Tip: Always simplify the radical for the radius - \( \sqrt{20} \) becomes \( 2\sqrt{5} \) by factoring out perfect squares.

Question 7D. Find the centre and radius of the circle: \( x^2 + (y - 1)^2 = 2 \)
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Comparing the given equation with the standard form:
\( (x - 0)^2 + (y - 1)^2 = 2 \)

We obtain: h = 0, k = 1, and \( r^2 = 2 \), so r = \( \sqrt{2} \).

Thus, the centre is (0, 1) and the radius is \( \sqrt{2} \) units.

Exam Tip: When x appears without a linear transformation, recognize that it is equivalent to \( (x - 0)^2 \), indicating a centre with x-coordinate of 0.

Question 8. Find the equation of the circle whose centre is (2, - 5) and which passes through the point (3, 2).
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

We know the centre (h, k) = (2, - 5), but the radius is not given directly. Since the circle passes through the point (3, 2), we can find the radius using the distance formula.

Substituting the point (3, 2) into the circle equation:
\( (3 - 2)^2 + (2 - (- 5))^2 = r^2 \)
\( 1^2 + 7^2 = r^2 \)
\( 1 + 49 = r^2 \)
\( r^2 = 50 \)

Therefore, the equation of the circle is:
\( (x - 2)^2 + (y + 5)^2 = 50 \)

Exam Tip: When a point on the circle is given but the radius is not, always use the given point to calculate the radius by substituting into the standard form equation.

Question 9. Find the equation of the circle of radius 5 cm, whose centre lies on the y - axis and which passes through the point (3, 2).
Answer: The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where (h, k) is the centre and r is the radius.

Since the centre lies on the y - axis, its x - coordinate is 0. Let the centre be (0, k).

The radius is given as 5. Substituting the point (3, 2), which lies on the circle:
\( (3 - 0)^2 + (2 - k)^2 = 5^2 \)
\( 9 + (2 - k)^2 = 25 \)
\( (2 - k)^2 = 16 \)

Taking the square root of both sides:
\( 2 - k = \pm 4 \)

This gives us: k = 2 - 4 = - 2 or k = 2 + 4 = 6

When k = - 2, the equation is: \( x^2 + (y + 2)^2 = 25 \)

When k = 6, the equation is: \( x^2 + (y - 6)^2 = 25 \)

Exam Tip: When solving for a parameter using a constraint like "centre on y-axis," remember that this restriction may lead to multiple solutions - check both possibilities carefully.

Question 10. Find the equation of the circle whose centre is (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4.
Answer: First, we locate the intersection point of the two given lines.

Solving the system: 3x + 2y = 11 and 2x + 3y = 4

From these equations, the intersection point is (5, - 2).

Now we have a centre (2, - 3) and a point (5, - 2) that the circle passes through. We can find the radius using the distance formula.

The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)

Substituting the point (5, - 2) into the equation with centre (2, - 3):
\( (5 - 2)^2 + (- 2 - (- 3))^2 = r^2 \)
\( 3^2 + 1^2 = r^2 \)
\( 9 + 1 = r^2 \)
\( r^2 = 10 \)

Therefore, the equation of the circle is:
\( (x - 2)^2 + (y + 3)^2 = 10 \)

Exam Tip: When the problem involves finding the intersection of lines first, solve that system carefully before proceeding with the circle equation - accuracy in the intersection point is crucial.

Question 11. Find the equation of the circle passing through the point (- 1, - 3) and having its centre at the point of intersection of the lines x - 2y = 4 and 2x + 5y + 1 = 0.
Answer: First, we determine the intersection point of the two given lines.

Solving the system: x - 2y = 4 and 2x + 5y + 1 = 0

From these equations, the intersection point is (2, - 1).

Now we have a centre (2, - 1) and a point (- 1, - 3) on the circle. We can calculate the radius using the distance formula.

The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)

Substituting the point (- 1, - 3) into the equation with centre (2, - 1):
\( (- 1 - 2)^2 + (- 3 - (- 1))^2 = r^2 \)
\( (- 3)^2 + (- 2)^2 = r^2 \)
\( 9 + 4 = r^2 \)
\( r^2 = 13 \)

Therefore, the equation of the circle is:
\( (x - 2)^2 + (y + 1)^2 = 13 \)

Exam Tip: Always verify your intersection point by substituting back into both original line equations - this prevents errors that would propagate through the rest of the solution.

Question 12. If two diameters of a circle lie along the lines x - y = 9 and x - 2y = 7, and the area of the circle is 38.5 sq cm, find the equation of the circle.
Answer: The centre of the circle is located at the intersection of its two diameters.

Solving the system: x - y = 9 and x - 2y = 7

From these equations, the intersection point is (11, 2), which serves as the centre.

Using the area formula for a circle:
\( \text{Area} = \pi r^2 \)
\( 38.5 = \pi r^2 \)
\( r^2 = \frac{38.5}{\pi} = \frac{38.5}{3.14} = 12.25 \text{ sq cm} \)

The standard form for a circle's equation is:
\( (x - h)^2 + (y - k)^2 = r^2 \)

Substituting the centre (11, 2) and \( r^2 = 12.25 \):
\( (x - 11)^2 + (y - 2)^2 = 12.25 \)

Exam Tip: Remember to convert the area into the radius squared using \( r^2 = \frac{\text{Area}}{\pi} \) - this is the critical step when area is given instead of radius.

Question 13A. Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(3, 2) and B(2, 5)
Answer: When the endpoints of a diameter are given, the equation can be found using the formula:
\( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \)

where (x₁, y₁) and (x₂, y₂) are the endpoints of the diameter.

Substituting the points A(3, 2) and B(2, 5):
\( (x - 3)(x - 2) + (y - 2)(y - 5) = 0 \)

Expanding:
\( x^2 - 2x - 3x + 6 + y^2 - 5y - 2y + 10 = 0 \)
\( x^2 + y^2 - 5x - 7y + 16 = 0 \)

Therefore, the equation is: \( x^2 + y^2 - 5x - 7y + 16 = 0 \)

Exam Tip: The diameter endpoint formula is more efficient than finding the centre and radius separately - use it when both endpoints are provided.

Question 13B. Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(5, - 3) and B(2, - 4)
Answer: When the endpoints of a diameter are provided, the formula is:
\( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \)

where (x₁, y₁) and (x₂, y₂) denote the endpoints.

Substituting the points A(5, - 3) and B(2, - 4):
\( (x - 5)(x - 2) + (y + 3)(y + 4) = 0 \)

Expanding:
\( x^2 - 2x - 5x + 10 + y^2 + 3y + 4y + 12 = 0 \)
\( x^2 + y^2 - 7x + 7y + 22 = 0 \)

Therefore, the equation is: \( x^2 + y^2 - 7x + 7y + 22 = 0 \)

Exam Tip: Carefully manage signs when one or both coordinates are negative - errors in sign expansion are common in this type of problem.

Question 13C. Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(- 2, - 3) and B(- 3, 5)
Answer: The formula for a circle when the endpoints of a diameter are known is:
\( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \)

where (x₁, y₁) and (x₂, y₂) are the diameter endpoints.

Substituting the points A(- 2, - 3) and B(- 3, 5):
\( (x - (- 2))(x - (- 3)) + (y - (- 3))(y - 5) = 0 \)
\( (x + 2)(x + 3) + (y + 3)(y - 5) = 0 \)

Expanding:
\( x^2 + 3x + 2x + 6 + y^2 - 5y + 3y - 15 = 0 \)
\( x^2 + y^2 + 5x - 2y - 9 = 0 \)

Therefore, the equation is: \( x^2 + y^2 + 5x - 2y - 9 = 0 \)

Exam Tip: When both coordinates are negative, double - check the expansion of binomials to ensure correct handling of all sign combinations throughout the calculation.

Question 13. D. Find the equation of the circle, the coordinates of the end points of one of whose diameters are A(p, q) and B(r, s)
Answer: A circle passing through two endpoints of a diameter follows the pattern: (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0. Here, x₁ = p, y₁ = q, x₂ = r, y₂ = s. Substituting these values: (x - p)(x - r) + (y - q)(y - s) = 0. Expanding: x² - rx - px + pr + y² - sy - qy + qs = 0. Rearranging: x² + y² - (r + p)x - (s + q)y + (pr + qs) = 0.
In simple words: When you know the two endpoints of a diameter, you can build the circle's equation by multiplying the differences of x-coordinates and y-coordinates and setting the sum equal to zero. This automatically gives you the correct circle.

Exam Tip: Always recognize the diameter endpoints form and apply the direct formula - this saves time compared to finding the centre and radius separately.

Question 14. The sides of a rectangle are given by the equations x = - 2, x = 4, y = - 2 and y = 5. Find the equation of the circle drawn on the diagonal of this rectangle as its diameter.
Answer: The four vertices of the rectangle, arranged clockwise, are: (- 2, 5), (4, 5), (4, - 2), (- 2, - 2). The circle's equation that passes through two diameter endpoints is: (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0. Taking the diagonal endpoints (- 2, 5) and (4, - 2): (x + 2)(x - 4) + (y - 5)(y + 2) = 0. Expanding: x² - 4x + 2x - 8 + y² + 2y - 5y - 10 = 0. Simplifying: x² + y² - 2x - 3y - 18 = 0.
In simple words: Find the opposite corners of the rectangle (the diagonal endpoints), then plug them into the diameter formula. The resulting equation describes the circle.

Exam Tip: Identify all four rectangle corners first; the diagonal connects two opposite corners, not adjacent ones.

Exercise 21B

Question 1. Show that the equation x² + y² - 4x + 6y - 5 = 0 represents a circle. Find its centre and radius.
Answer: The general form of a conic is ax² + 2hxy + by² + 2gx + 2fy + c = 0, where a, b, c, f, g, h are constants. A circle requires a = b and h = 0, giving the standard form x² + y² + 2gx + 2fy + c = 0. For the equation x² + y² - 4x + 6y - 5 = 0, we have 2g = - 4, so g = - 2; 2f = 6, so f = 3; and c = - 5. This confirms it is a circle. The centre is (- g, - f) = (- (- 2), - 3) = (2, - 3). The radius is \( \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + 3^2 - (-5)} = \sqrt{4 + 9 + 5} = \sqrt{18} = 3\sqrt{2} \).
In simple words: Check that the equation fits the circle pattern (x² and y² have equal coefficients, no xy term). Then extract the constants and apply the centre and radius formulas directly.

Exam Tip: Always verify that a = b and h = 0 first - if not, it's not a circle. Show your comparison step explicitly.

Question 2. Show that the equation x² + y² + x - y = 0 represents a circle. Find its centre and radius.
Answer: The general conic equation is ax² + 2hxy + by² + 2gx + 2fy + c = 0. A circle satisfies a = b and h = 0, so the standard form is x² + y² + 2gx + 2fy + c = 0. For x² + y² + x - y = 0, we identify 2g = 1, giving g = 1/2; 2f = - 1, giving f = - 1/2; and c = 0. Since the coefficients of x² and y² are equal and there is no xy term, this is indeed a circle. The centre is (- g, - f) = (- 1/2, - (- 1/2)) = (- 1/2, 1/2). The radius is \( \sqrt{g^2 + f^2 - c} = \sqrt{(1/2)^2 + (-1/2)^2 - 0} = \sqrt{1/4 + 1/4} = \sqrt{1/2} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \).
In simple words: Match the equation to the circle form by checking that x² and y² have the same coefficient and no xy term exists. Then use the formulas to find the centre and radius.

Exam Tip: When f or g is a fraction, be careful with sign changes - negative values inside the centre formula become positive.

Question 3. Show that the equation 3x² + 3y² + 6x - 4y - 1 = 0 represents a circle. Find its centre and radius.
Answer: Divide the entire equation by 3 to get the standard form: x² + y² + 2x - (4/3)y - 1/3 = 0. The general circle form is x² + y² + 2gx + 2fy + c = 0. Here, 2g = 2, so g = 1; 2f = - 4/3, so f = - 2/3; and c = - 1/3. Since the x² and y² coefficients are equal (both 1 after dividing) and the xy term is absent, this is a circle. The centre is (- g, - f) = (- 1, - (- 2/3)) = (- 1, 2/3). The radius is \( \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + (-2/3)^2 - (-1/3)} = \sqrt{1 + 4/9 + 1/3} = \sqrt{1 + 4/9 + 3/9} = \sqrt{(9 + 4 + 3)/9} = \sqrt{16/9} = 4/3 \).
In simple words: When coefficients of x² and y² are not 1, divide the entire equation first. Then apply the centre and radius formulas to the simplified form.

Exam Tip: Simplifying by dividing first prevents arithmetic errors and makes identifying g, f, c much easier.

Question 4. Show that the equation x² + y² + 2x + 10y + 26 = 0 represents a point circle. Also, find its centre.
Answer: The standard circle equation is x² + y² + 2gx + 2fy + c = 0. Comparing with x² + y² + 2x + 10y + 26 = 0: 2g = 2, so g = 1; 2f = 10, so f = 5; c = 26. The centre is (- g, - f) = (- 1, - 5). The radius is \( \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 5^2 - 26} = \sqrt{1 + 25 - 26} = \sqrt{0} = 0 \). Since the radius equals zero, this is a point circle - a circle that degenerates to a single point. The centre is (- 1, - 5).
In simple words: A point circle happens when the radius equals zero. The equation still fits the circle pattern, but it represents only one location instead of a full circular path.

Exam Tip: Always compute the radius; if it equals zero, mention that it is a point circle and state only the centre as the answer.

Question 5. Show that the equation x² + y² - 3x + 3y + 10 = 0 does not represent a circle.
Answer: The standard circle form is x² + y² + 2gx + 2fy + c = 0. For the equation x² + y² - 3x + 3y + 10 = 0: 2g = - 3, so g = - 3/2; 2f = 3, so f = 3/2; c = 10. The radius would be \( \sqrt{g^2 + f^2 - c} = \sqrt{(-3/2)^2 + (3/2)^2 - 10} = \sqrt{9/4 + 9/4 - 10} = \sqrt{18/4 - 10} = \sqrt{4.5 - 10} = \sqrt{-5.5} \). Since the expression under the square root is negative, the radius is not a real number. Therefore, x² + y² - 3x + 3y + 10 = 0 does not represent a circle.
In simple words: If the radius formula gives a negative value inside the square root, the circle cannot exist in real space. This means the equation does not represent an actual circle.

Exam Tip: Check the discriminant (g² + f² - c) before concluding - if negative, state clearly that no real circle exists.

Question 6. Find the equation of the circle passing through the points (i) (0, 0), (5, 0) and (3, 3); (ii) (1, 2), (3, - 4) and (5, - 6); (iii) (20, 3), (19, 8) and (2, - 9). Also, find the centre and radius in each case.
Answer:
(i) Using the determinant form for a circle through three points:
\[ \begin{vmatrix} x^2 + y^2 & x & y & 1 \\ 0 & 0 & 0 & 1 \\ 25 & 5 & 0 & 1 \\ 9 + 9 & 3 & 3 & 1 \end{vmatrix} = 0 \]
Applying Laplace expansion and simplifying: 15(x² + y²) - 75x - 15y = 0, which gives x² + y² - 5x - y = 0. The centre is (2.5, 0.5) and the radius is \( \sqrt{(-2.5)^2 + (-0.5)^2 - 0} = \sqrt{6.25 + 0.25} = \sqrt{6.5} \approx 2.549 \).
(ii) Using the same determinant approach with points (1, 2), (3, - 4), (5, - 6): The expanded and simplified result is 8(x² + y²) - 176x - 32y - 200 = 0, giving x² + y² - 22x - 4y - 25 = 0. The centre is (11, 2) and the radius is \( \sqrt{(-11)^2 + (-2)^2 - (-25)} = \sqrt{121 + 4 + 25} = \sqrt{150} \approx 12.247 \).
(iii) With points (20, 3), (19, 8), (2, - 9): The determinant yields 102(x² + y²) - 1428x - 612y - 11322 = 0, simplifying to x² + y² - 14x - 6y - 111 = 0. The centre is (7, 3) and the radius is \( \sqrt{(-7)^2 + (-3)^2 - (-111)} = \sqrt{49 + 9 + 111} = \sqrt{169} = 13 \).
In simple words: When you have three points on a circle, set up a determinant with their coordinates and expand it. The result, after simplification, is your circle equation. Then extract the centre and radius using the standard formulas.

Exam Tip: Laplace expansion along the first column minimizes computation. Double-check your arithmetic when simplifying the determinant to avoid sign errors.

Question 7. Find the equation of the circle which is circumscribed about the triangle whose vertices are A(- 2, 3), B(5, 2) and C(6, - 1). Find the centre and radius of this circle.
Answer: A circle circumscribed about a triangle passes through all three vertices. Using the general form (x - h)² + (y - k)² = r², substitute each vertex:
From A(- 2, 3): (- 2 - h)² + (3 - k)² = r² ⟹ h² + k² + 4h - 6k + 13 = r²...(ii)
From B(5, 2): (5 - h)² + (2 - k)² = r² ⟹ h² + k² - 10h - 4k + 29 = r²...(iii)
From C(6, - 1): (6 - h)² + (- 1 - k)² = r² ⟹ h² + k² - 12h + 2k + 37 = r²...(iv)
Subtracting (ii) from (iii): - 14h + 2k + 16 = 0 ⟹ - 7h + k + 8 = 0...(v)
Subtracting (ii) from (iv): - 16h + 8k + 24 = 0 ⟹ - 2h + k + 3 = 0...(vi)
Subtracting (v) from (vi): 5h - 5 = 0 ⟹ h = 1. Substituting into (vi): k = - 1. The centre is (1, - 1). The radius is r = \( \sqrt{1^2 + (-1)^2 - (h^2 + k^2 - 13)} = \sqrt{(1 - 1)^2 + (0 + 1)^2 + 13} = 5 \). The equation is (x - 1)² + (y + 1)² = 25.
In simple words: Substitute each of the three vertices into the circle equation separately. This creates a system of three equations. Solve for h and k (the centre), then find the radius. The three vertices will all be equidistant from the centre.

Exam Tip: Use subtraction to eliminate r² and h² + k² terms quickly, leaving linear equations in h and k.

Question 8. Find the equation of the circle concentric with the circle x² + y² + 4x + 6y + 11 = 0 and passing through the point P(5, 4).
Answer: Two or more circles are concentric when they share the same centre but have different radii. From the given circle x² + y² + 4x + 6y + 11 = 0, we have 2g = 4 (g = 2) and 2f = 6 (f = 3), so the centre is (- 2, - 3). A concentric circle has the same g and f values but a different constant: x² + y² + 4x + 6y + c' = 0. Since this circle passes through P(5, 4): 5² + 4² + 4(5) + 6(4) + c' = 0 ⟹ 25 + 16 + 20 + 24 + c' = 0 ⟹ 85 + c' = 0 ⟹ c' = - 85. The required equation is x² + y² + 4x + 6y - 85 = 0.
In simple words: Extract the centre from the given circle. Keep the same x and y terms, but change the constant by substituting the point. Solve for the new constant to complete the equation.

Exam Tip: Concentric circles always have the same 2g and 2f terms. Only the constant c changes.

Question 9. Show that the points A(1, 0), B(2, - 7), C(8, 1) and D(9, - 6) all lie on the same circle. Find the equation of this circle, its centre and radius.
Answer: Using the general form (x - h)² + (y - k)² = r², substitute each point:
From A(1, 0): (1 - h)² + (0 - k)² = r² ⟹ h² + k² + 1 - 2h = r²...(ii)
From B(2, - 7): (2 - h)² + (- 7 - k)² = r² ⟹ h² + k² + 53 - 4h + 14k = r²
Rearranging: (h² + k² + 1 - 2h) + 52 - 2h + 14k = r² ⟹ h - 7k - 26 = 0...(iii)
From C(8, 1): (8 - h)² + (1 - k)² = r² ⟹ 7h + k - 32 = 0...(iv)
Solving (iii) and (iv): From (iv), k = 32 - 7h. Substituting into (iii): h - 7(32 - 7h) - 26 = 0 ⟹ h - 224 + 49h - 26 = 0 ⟹ 50h = 250 ⟹ h = 5. Then k = 32 - 7(5) = - 3. The centre is (5, - 3). From (ii): r² = 5² + (- 3)² + 1 - 2(5) = 25 + 9 + 1 - 10 = 25 ⟹ r = 5. Checking D(9, - 6): (9 - 5)² + (- 6 + 3)² = 4² + (- 3)² = 16 + 9 = 25 = r². All four points lie on the circle. The equation is (x - 5)² + (y + 3)² = 25 or x² + y² - 10x + 6y - 9 = 0.
In simple words: Set up the circle equation with the first point, then use the remaining points to find two linear equations in h and k. Solve these to get the centre. Finally, verify that all four points satisfy the equation.

Exam Tip: Always verify the fourth point - if it doesn't satisfy the equation, your centre calculation has an error.

Question 10. Find the equation of the circle which passes through the points (1, 3) and (2, - 1), and has its centre on the line 2x + y - 4 = 0.
Answer: The general circle equation is x² + y² + 2gx + 2fy + c = 0. Substitute (1, 3): 1 + 9 + 2g + 6f + c = 0 ⟹ 2g + 6f + c = - 10...(ii). Substitute (2, - 1): 4 + 1 + 4g - 2f + c = 0 ⟹ 4g - 2f + c = - 5...(iii). Since the centre (- g, - f) lies on 2x + y - 4 = 0: 2(- g) + (- f) - 4 = 0 ⟹ - 2g - f - 4 = 0...(iv). From (iv): f = - 2g - 4. Substitute into (ii): 2g + 6(- 2g - 4) + c = - 10 ⟹ 2g - 12g - 24 + c = - 10 ⟹ - 10g + c = 14...(v). Substitute f = - 2g - 4 into (iii): 4g - 2(- 2g - 4) + c = - 5 ⟹ 4g + 4g + 8 + c = - 5 ⟹ 8g + c = - 13...(vi). Subtracting (v) from (vi): 18g = - 27 ⟹ g = - 1.5. From (vi): c = - 13 - 8(- 1.5) = - 13 + 12 = - 1. From (iv): f = - 2(- 1.5) - 4 = 3 - 4 = - 1. The equation is x² + y² - 3x - 2y - 1 = 0.
In simple words: Substitute the two given points into the general circle equation to get two conditions. Then use the constraint that the centre lies on the given line to get a third equation. Solve the system to find g, f, and c.

Exam Tip: The centre condition provides a direct link between g and f - use it to reduce the number of unknowns before solving.

Question 11. Find the equation of the circle concentric with the circle x² + y² - 4x - 6y - 3 = 0 and which touches the y-axis.
Answer: The given circle x² + y² - 4x - 6y - 3 = 0 has 2g = - 4 (g = - 2) and 2f = - 6 (f = - 3), so its centre is (- g, - f) = (2, 3). A concentric circle has the same centre. The radius of the given circle is \( \sqrt{(-2)^2 + (-3)^2 - (-3)} = \sqrt{4 + 9 + 3} = \sqrt{16} = 4 \) units. Since the new circle touches (is tangent to) the y-axis, the distance from the centre (2, 3) to the y-axis equals the radius. This distance is the x-coordinate of the centre, which is 2. Therefore, the new radius is 2. The equation of the new circle is (x - 2)² + (y - 3)² = 2² ⟹ x² - 4x + 4 + y² - 6y + 9 = 4 ⟹ x² + y² - 4x - 6y + 9 = 0.
In simple words: Find the centre of the given circle. If a new circle is tangent to the y-axis and has this same centre, the radius must equal the x-coordinate of the centre. Use this radius to write the new circle equation.

Exam Tip: For a circle tangent to the y-axis, radius = |x-coordinate of centre|. For tangency to the x-axis, radius = |y-coordinate of centre|.

Question 12. Find the equation of the circle concentric with the circle x² + y² - 6x + 12y + 15 = 0 and of double its area.
Answer: From x² + y² - 6x + 12y + 15 = 0, we have 2g = - 6 (g = - 3) and 2f = 12 (f = 6), so the centre is (- g, - f) = (3, - 6). The radius of the given circle is \( r = \sqrt{(-3)^2 + 6^2 - 15} = \sqrt{9 + 36 - 15} = \sqrt{30} \). The area of the given circle is πr² = 30π. If the new circle has double the area, then πr'² = 60π ⟹ r'² = 60 ⟹ r' = \( \sqrt{60} = 2\sqrt{15} \). The new circle has the same centre (3, - 6) and radius \( 2\sqrt{15} \). Its equation is (x - 3)² + (y + 6)² = 60 ⟹ x² - 6x + 9 + y² + 12y + 36 = 60 ⟹ x² + y² - 6x + 12y - 15 = 0.
In simple words: If a circle has double the area, its radius squared is doubled. Find the new radius from the area condition, then write the circle equation with the original centre and new radius.

Exam Tip: Area scales with r². If area doubles, r² doubles. Keep the centre the same for concentric circles.

Question 13. Prove that the centres of the three circles x² + y² - 4x - 6y - 12 = 0, x² + y² + 2x + 4y - 5 = 0 and x² + y² - 10x - 16y + 7 = 0 are collinear.
Answer: For the first circle x² + y² - 4x - 6y - 12 = 0: 2g = - 4, 2f = - 6, so the centre is (2, 3). For the second circle x² + y² + 2x + 4y - 5 = 0: 2g = 2, 2f = 4, so the centre is (- 1, - 2). For the third circle x² + y² - 10x - 16y + 7 = 0: 2g = - 10, 2f = - 16, so the centre is (5, 8). To prove collinearity, use the determinant: \[ \begin{vmatrix} 2 & 3 & 1 \\ -1 & -2 & 1 \\ 5 & 8 & 1 \end{vmatrix} \]. Expanding: 2(- 2 - 8) - 3(- 1 - 5) + 1(- 8 + 10) = 2(- 10) - 3(- 6) + 1(2) = - 20 + 18 + 2 = 0. Since the determinant equals zero, the three centres are collinear.
In simple words: Extract the centre of each circle from its equation. Set up a 3×3 determinant with the three centre coordinates. If the determinant is zero, the points are collinear.

Exam Tip: Collinearity is proven when the determinant of coordinates equals zero. Always expand along the easiest row or column.

Question 14. Find the equation of the circle which passes through the points A(1, 1) and B(2, 2) and whose radius is 1. Show that there are two such circles.
Answer: Let the circle have centre (h, k) and radius 1. Since A(1, 1) lies on the circle: (1 - h)² + (1 - k)² = 1...(i). Since B(2, 2) lies on the circle: (2 - h)² + (2 - k)² = 1...(ii). From (i): 1 - 2h + h² + 1 - 2k + k² = 1 ⟹ h² + k² - 2h - 2k + 1 = 0...(iii). From (ii): 4 - 4h + h² + 4 - 4k + k² = 1 ⟹ h² + k² - 4h - 4k + 7 = 0...(iv). Subtracting (iii) from (iv): - 2h - 2k + 6 = 0 ⟹ h + k = 3 ⟹ k = 3 - h...(v). Substituting (v) into (iii): h² + (3 - h)² - 2h - 2(3 - h) + 1 = 0 ⟹ h² + 9 - 6h + h² - 2h - 6 + 2h + 1 = 0 ⟹ 2h² - 6h + 4 = 0 ⟹ h² - 3h + 2 = 0 ⟹ (h - 1)(h - 2) = 0. Thus h = 1 or h = 2. If h = 1, then k = 2. If h = 2, then k = 1. The two circles are (x - 1)² + (y - 2)² = 1 and (x - 2)² + (y - 1)² = 1, or x² + y² - 2x - 4y + 4 = 0 and x² + y² - 4x - 2y + 4 = 0. There are exactly two circles satisfying the conditions.
In simple words: Set up two equations using the fact that both points lie on the circle. Solve for the centre coordinates. You will get a quadratic equation that yields two distinct solutions, proving two circles exist.

Exam Tip: When a quadratic gives two solutions, always verify both by substitution - both should satisfy the original conditions.

Question 14. Find the equation of a circle passing through the points A(1, 1) and B(2, 2) with radius 1.
Answer: We start with the general form of a circle's equation: \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the centre and \( r \) is the radius.

Substituting point A(1, 1) into the equation:
\( (1 - h)^2 + (1 - k)^2 = 1^2 \)

Expanding:
\( h^2 + k^2 + 2 - 2h - 2k = 1 \)

\( h^2 + k^2 - 2h - 2k = -1 \) ...(ii)

Substituting point B(2, 2) into the equation:
\( (2 - h)^2 + (2 - k)^2 = 1^2 \)

Expanding:
\( h^2 + k^2 + 8 - 4h - 4k = 1 \)

\( h^2 + k^2 - 4h - 4k = -7 \)

\( (h^2 + k^2 - 2h - 2k) - 2h - 2k = -7 \)

\( -1 - 2h - 2k = -7 \) [from (ii)]

\( -2h - 2k = -6 \)

\( h + k = 3 \Rightarrow h = 3 - k \)

When \( k = 2 \), \( h = 3 - 2 = 1 \)

Equation of circle 1: \( (x - 1)^2 + (y - 2)^2 = 1 \)

When \( k = 1 \), \( h = 3 - 1 = 2 \)

Equation of circle 2: \( (x - 2)^2 + (y - 1)^2 = 1 \)

Exam Tip: When a circle passes through two given points with a known radius, substitute each point separately to get two equations, then solve them together to find the centre coordinates.

Question 15. Find the equation of a circle passing through the origin and intercepting lengths a and b on the axes.
Answer: From the given setup, we have points D(0, b), E(a, 0), and A(0, 0) lying on the circle, with C as the centre.

Using the general circle equation: \( (x - h)^2 + (y - k)^2 = r^2 \) ...(i), where \( (h, k) \) is the centre and \( r \) is the radius.

Substituting A(0, 0) into equation (i):
\( (0 - h)^2 + (0 - k)^2 = r^2 \)

\( h^2 + k^2 = r^2 \) ...(ii)

Substituting D(0, b) into equation (i):
\( (0 - h)^2 + (b - k)^2 = r^2 \)

\( h^2 + k^2 + b^2 - 2kb = r^2 \)

\( r^2 + b^2 - 2kb = r^2 \)

\( b^2 - 2kb = 0 \)

\( (b - 2k)b = 0 \)

Either \( b = 0 \) or \( k = \frac{b}{2} \)

Substituting E(a, 0) into equation (i):
\( (a - h)^2 + (0 - k)^2 = r^2 \)

\( h^2 + k^2 + a^2 - 2ha = r^2 \)

\( r^2 + a^2 - 2ha = r^2 \)

\( a^2 - 2ha = 0 \)

\( (a - 2h)a = 0 \)

Either \( a = 0 \) or \( h = \frac{a}{2} \)

Therefore, the centre is \( C\left(\frac{a}{2}, \frac{b}{2}\right) \), and the radius squared is \( r^2 = \frac{a^2 + b^2}{4} \).

Substituting these values into equation (i):
\( \left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4} \)

Expanding:
\( x^2 + y^2 + \frac{a^2}{4} + \frac{b^2}{4} - xa - yb = \frac{a^2 + b^2}{4} \)

\( x^2 + y^2 - xa - yb = 0 \)

This is the required equation.

Exam Tip: For circles passing through the origin and intercepts on axes, always find the centre by substituting the given points and using the intercept conditions to simplify the resulting equations.

Question 16. Find the equation of the circle circumscribing the triangle formed by the lines x + y = 6, 2x + y = 4 and x + 2y = 5.
Answer: Solving the three line equations together, we can find the vertices of the triangle.

The required circle equation can be determined using the determinant form. For a circle passing through three points (or in this case, the vertices formed by three lines), we use:

\[ \begin{vmatrix} x^2 + y^2 & x & y & 1 \\ x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ x_2^2 + y_2^2 & x_2 & y_2 & 1 \\ x_3^2 + y_3^2 & x_3 & y_3 & 1 \end{vmatrix} = 0 \]

Upon solving the system of line equations and applying the determinant method, the equation simplifies to a standard circle form.

Exam Tip: When finding the circumcircle of a triangle formed by three lines, solve the lines pairwise to get the three vertices, then use the determinant form or substitute all three points into the general circle equation.

Question 17. Show that the quadrilateral formed by the straight lines x - y = 0, 3x + 2y = 5, x - y = 10 and 2x + 3y = 0 is cyclic and hence find the equation of the circle.
Answer: Solving the four line equations pairwise, we determine the four vertices of the quadrilateral.

To prove the quadrilateral is cyclic, we verify that the sum of opposite angles equals 180°. By calculating the slopes of the sides:

Slope of AB = 1
Slope of CD = 1
AB is parallel to CD

Slope of BD = -1
Slope of AC = -1
AC is parallel to BD

Since opposite sides are parallel, the quadrilateral forms a rectangle with all angles equal to 90°. Therefore, the sum of opposite angles = 90° + 90° = 180°, confirming the quadrilateral is cyclic.

Since the quadrilateral is cyclic and specifically a rectangle, its circumcircle has the diagonal as the diameter. Using the diameter endpoints A(0, 0) and D(6, -4):

\( (x - 0)(x - 6) + (y - 0)(y + 4) = 0 \)

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

This is the required circle equation.

Exam Tip: To show a quadrilateral is cyclic, prove that opposite angles sum to 180° or use the property that all four vertices lie on a single circle. The diameter method works perfectly when the quadrilateral is a rectangle.

Question 18. If (-1, 3) and (α, β) are the extremities of the diameter of the circle x² + y² - 6x + 5y - 7 = 0, find the coordinates (α, β).
Answer: Given the circle equation: \( x^2 + y^2 - 6x + 5y - 7 = 0 \)

First, we rewrite it in standard form to identify the centre. Comparing with the general form \( x^2 + y^2 + 2gx + 2fy + c = 0 \), we get \( 2g = -6 \) and \( 2f = 5 \).

Therefore, the centre is \( C\left(3, -\frac{5}{2}\right) \).

Since (-1, 3) and (α, β) are the two extremities of a diameter, the centre of the circle must be the midpoint of these two points. Using the midpoint formula:

\( \frac{\alpha - 1}{2} = 3 \)

\( \alpha = 7 \)

And:

\( \frac{\beta + 3}{2} = -\frac{5}{2} \)

\( \beta = -8 \)

Therefore, \( (\alpha, \beta) = (7, -8) \)

Exam Tip: When two points are the extremities of a diameter, their midpoint must equal the circle's centre. Extract the centre from the general circle equation and apply the midpoint formula to find the unknown endpoint.