CBSE Class 10 Mathematics Circles Worksheet Set 06

CBSE Class 10 Circles (6). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

CIRCLES

Key Points

1. Circle: A circle is a collection of all points in a plane which are at a constant distance (radius) from a fixed point (centre).

2. Secant & Tangent to a Circle: In fig. 1 the line PQ and the circle have no common point. Line PQ is called non-intersecting. In fig. 2 line PQ a secant to a circle. In fig. 3, there is only 1 point A, which is common to the line PQ and the circle. The line is called a tangent to the circle.

3. Tangent to a Circle :

It is a line that intersects the circle at only one point. There is only one tangent at a point of the circle. The tangent to a circle is a special case of the secant, when the two end points of its corresponding chord coincide.

4. Theorems :

1. The tangent at any point of a circle is perpendicular to the radius through the point of contact.

2. The length of tangents drawn from an external point to a circle are equal

5. Number of tangents from a point on a circle-

(i)There is no tangent to a circle passing through a point lying inside the circle.

(ii)There is one and only one tangent to a circle passing through a point lying on the circle.

(iii)There are exactly two tangents to a circle through a point lying outside the circle.

LEVEL I

1. In the given fig. O is the centre of the circle and PQ is tangent then POQ + QPO is equal to

2. If PQ is a tangent to a circle of radius 5cm and PQ = 12 cm, Q is point of contact, then OP is

3. In the given fig. PQ and PR are tangents to the circle, QOP = 70°, thenQPR is equal to

P a g e 86 | 118

4. In the given fig. QS is a tangent to the circle, OS = 8 cm, OQ = 6 cm then the length of QS is

5. In the given fig PQ is tangent to outer circle and PR is tangent to inner circle. If PQ = 4 cm, OQ = 3 cm and OR = 2 cm then the length of PR is

6. In the given fig. P, Q and R are the points of contact. If AB = 4 cm, BP = 2 cm then the perimeter of ABC is

7. The distance between two tangent parallel to each other to a circle is 12 cm. The radius of circle is

8. The chord of a circle of radius 10cm subtends a right angle at its centre.Find the length of the chord.

9. How many tangents can a circle have?

Circles

Key Points

• Circle: A circle is a collection of all points in a plane which are at a constant distance (radius) from a fixed point (centre).
• Secant & Tangent to a Circle: In a plane, a line and a circle can have three possible relationships:
  • They can be non-intersecting, having no points in common.
  • The line can cut the circle at two distinct points, which is called a secant.
  • The line can touch the circle at exactly one single point, which is called a tangent.

• Tangent to a Circle: A tangent is a line that intersects the circle at only one point. There is only one tangent possible at any given point of the circle. It can be viewed as a limiting case of a secant when the two endpoints of its corresponding chord coincide.
• Core Theorems:
  • The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  • The lengths of tangents drawn from an external point to a circle are equal.
  • Number of Tangents from a Point:
    • There is no tangent possible from a point lying inside the circle.
    • There is exactly one tangent from a point lying on the circle.
    • There are exactly two tangents from a point lying outside the circle.

LEVEL I

Question 1. In the given fig. O is the centre of the circle and PQ is tangent then \(\angle POQ + \angle QPO\) is equal to
Answer: Because the radius of a circle meets a tangent at a right angle at its point of contact, we know that \(\angle OQP = 90^\circ\). Since the interior angles of a triangle always sum up to \(180^\circ\), the remaining two acute angles in right-angled triangle \(\Delta OQP\) must add up to \(90^\circ\). Therefore, we find that \(\angle POQ + \angle QPO = 180^\circ - 90^\circ = 90^\circ\).
In simple words: The radius of a circle meets its tangent at a \(90^\circ\) angle. Because the angles in any triangle add up to \(180^\circ\), the other two angles must add up to \(90^\circ\).

Exam Tip: Always state that the radius is perpendicular to the tangent at the point of contact to establish the \(90^\circ\) angle in your proof.

Question 2. If PQ is a tangent to a circle of radius 5cm and PQ = 12 cm, Q is point of contact, then OP is
Answer: The radius \(OQ\) is perpendicular to the tangent \(PQ\) at the point of contact \(Q\), forming a right-angled triangle \(\Delta OQP\) with \(\angle OQP = 90^\circ\). Using the Pythagoras theorem, we find the hypotenuse \(OP\): \(OP = \sqrt{OQ^2 + PQ^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}\). *(Note - If the question intended for \(OP = 12\text{ cm}\) to find the tangent length \(PQ\) as seen in similar textbook exercises, the calculation would be \(PQ = \sqrt{12^2 - 5^2} = \sqrt{119}\text{ cm}\).)*
In simple words: The radius and the tangent form a right-angled triangle. We can find the longest side of this triangle using the Pythagoras theorem, which gives a length of \(13\text{ cm}\).

Exam Tip: Be sure to identify which side is the hypotenuse before applying the Pythagoras theorem - it is always the side opposite the right angle (in this case, \(OP\)).

Question 3. In the given fig. PQ and PR are tangents to the circle, \(\angle QOP = 70^\circ\), then \(\angle QPR\) is equal to
Answer: In the right-angled triangle \(\Delta OQP\), we know \(\angle OQP = 90^\circ\) because the radius \(OQ\) is perpendicular to the tangent \(PQ\). The sum of angles in the triangle is \(180^\circ\), so: \(\angle QPO = 180^\circ - (90^\circ + 70^\circ) = 20^\circ\). Since the line joining the center of the circle to the external point bisects the angle between the two tangents, we have: \(\angle QPR = 2 \times \angle QPO = 2 \times 20^\circ = 40^\circ\).
In simple words: The top triangle has angles of \(90^\circ\) and \(70^\circ\), which leaves \(20^\circ\) at the corner point \(P\). Since the line down the middle cuts the corner angle exactly in half, the total angle at \(P\) is \(20^\circ \times 2 = 40^\circ\).

Exam Tip: Mention that the line segment \(OP\) bisects the angle \(\angle QPR\) to justify doubling the angle value.

Question 4. In the given fig. QS is a tangent to the circle, OS = 8 cm, OQ = 6 cm then the length of QS is
Answer: Since \(QS\) is a tangent to the circle at point \(Q\), the radius \(OQ\) is perpendicular to \(QS\). Consequently, \(\Delta OQS\) is a right-angled triangle at \(Q\). Applying the Pythagoras theorem: \(OS^2 = OQ^2 + QS^2 \Rightarrow 8^2 = 6^2 + QS^2\) \(\Rightarrow 64 = 36 + QS^2 \Rightarrow QS^2 = 28\) \(\Rightarrow QS = \sqrt{28} = 2\sqrt{7}\text{ cm}\).
In simple words: The tangent and the radius make a right angle. We use Pythagoras to find the missing side of the triangle, which equals the square root of \(8^2 - 6^2\) or \(2\sqrt{7}\text{ cm}\).

Exam Tip: Leave your final answer in simplified radical form, \(2\sqrt{7}\text{ cm}\), rather than a decimal, unless the question asks for a decimal approximation.

Question 5. In the given fig PQ is tangent to outer circle and PR is tangent to inner circle. If PQ = 4 cm, OQ = 3 cm and OR = 2 cm then the length of PR is
Answer: In right-angled triangle \(\Delta OQP\) (where the radius \(OQ\) of the outer circle is perpendicular to its tangent \(PQ\)): \(OP^2 = OQ^2 + PQ^2 = 3^2 + 4^2 = 9 + 16 = 25\). In right-angled triangle \(\Delta ORP\) (where the radius \(OR\) of the inner circle is perpendicular to its tangent \(PR\)): \(OP^2 = OR^2 + PR^2 \Rightarrow 25 = 2^2 + PR^2\) \(\Rightarrow PR^2 = 25 - 4 = 21 \Rightarrow PR = \sqrt{21}\text{ cm}\).
In simple words: First, find the distance to the center of the circle, which is \(5\text{ cm}\). Then, use that distance with the inner circle's radius to find the inner tangent length, giving \(\sqrt{21}\text{ cm}\).

Exam Tip: Keep \(OP^2 = 25\) directly instead of calculating \(OP = 5\) to save a step, since you need \(OP^2\) for the second equation anyway.

Question 6. In the given fig. P, Q and R are the points of contact. If AB = 4 cm, BP = 2 cm then the perimeter of \(\Delta ABC\) is
Answer: Tangents drawn from an external point to a circle are equal in length. Therefore, at vertex B, we have \(BQ = BP = 2\text{ cm}\). Since \(AB = 4\text{ cm}\), we can calculate the length of \(AQ\): \(AQ = AB - BQ = 4 - 2 = 2\text{ cm}\). At vertex A, we have \(AR = AQ = 2\text{ cm}\). Using symmetry for the isosceles triangle, we have \(AC = AB = 4\text{ cm}\), which gives: \(RC = AC - AR = 4 - 2 = 2\text{ cm}\). At vertex C, we have \(CP = CR = 2\text{ cm}\). Thus, the base of the triangle is \(BC = BP + CP = 2 + 2 = 4\text{ cm}\). The perimeter of \(\Delta ABC = AB + BC + CA = 4 + 4 + 4 = 12\text{ cm}\).
In simple words: Tangents coming from the same corner point are always equal in length. By finding these matching lengths around the triangle, we find that each of the three sides is \(4\text{ cm}\), making the total perimeter \(12\text{ cm}\).

Exam Tip: Be sure to write down the equal tangent pairs clearly (like \(BQ = BP\)) and state the theorem of equal external tangents to get full method marks.

Question 7. The distance between two tangent parallel to each other to a circle is 12 cm. The radius of circle is
Answer: Any two parallel tangents to a circle are located at opposite ends of a diameter line. Thus, the total distance separating the parallel tangents is equal to the circle's diameter. Since the distance is given as \(12\text{ cm}\): \(\text{Diameter} = 12\text{ cm}\) \(\text{Radius} = \frac{\text{Diameter}}{2} = \frac{12}{2} = 6\text{ cm}\).
In simple words: Parallel tangents touch the circle at opposite ends of a straight line through the center. This line is the diameter, so the radius is simply half of that, which is \(6\text{ cm}\).

Exam Tip: Write a brief statement explaining that the distance between parallel tangents is equal to the diameter to demonstrate your conceptual understanding.

Question 8. The chord of a circle of radius 10cm subtends a right angle at its centre.Find the length of the chord.
Answer: Let \(AB\) be the chord subtending a right angle at the center \(O\). This forms a right-angled triangle \(\Delta AOB\) where \(\angle AOB = 90^\circ\). The sides \(OA\) and \(OB\) are radii of the circle, meaning \(OA = OB = 10\text{ cm}\). By the Pythagoras theorem: \(AB = \sqrt{OA^2 + OB^2} = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}\text{ cm}\).
In simple words: The two radii and the chord form a right-angled triangle. Because the two legs are each \(10\text{ cm}\), we find the chord is the hypotenuse, which is \(10\sqrt{2}\text{ cm}\).

Exam Tip: Leave your final answer in radical form (\(10\sqrt{2}\text{ cm}\)) to maintain mathematical precision.

Question 9. How many tangents can a circle have?
Answer: A circle is composed of an infinite number of points along its boundary. Since exactly one tangent line can be drawn at any given point on the circle, a circle can have an infinite number of tangents.
In simple words: A circle's boundary is made of endless points. Because you can draw a tangent at any of these points, a circle can have infinitely many tangents.

Exam Tip: State the relation between the points on the boundary and the number of tangents to provide a complete answer.

Question 10. How many tangents can be drawn from a given point to a circle?
Answer: The number of tangents depends on where the point is located relative to the circle:
• If the point lies inside the circle: No tangent can be drawn.
• If the point lies on the boundary of the circle: Exactly one tangent can be drawn.
• If the point lies outside the circle: Exactly two tangents can be drawn.
In simple words: You can draw exactly two tangents from an outside point, only one tangent from a point on the circle, and no tangents if the point is inside the circle.

Exam Tip: It is best to list all three possible scenarios based on the point's position to show a thorough understanding of the topic.

LEVEL - II

Question 11. Two concentric circles of radii a & b (a>b) are given. Find the length of the chord of the larger circle which touches the smaller circle
Answer: Let \(O\) be the common center. Let \(AB\) be the chord of the larger circle (radius \(a\)) that is tangent to the smaller circle (radius \(b\)) at point \(C\). Because a radius is perpendicular to the tangent at its point of contact, we have \(\angle OCA = 90^\circ\). In the right-angled triangle \(\Delta OCA\): \(OA^2 = OC^2 + AC^2 \Rightarrow a^2 = b^2 + AC^2\) \(\Rightarrow AC = \sqrt{a^2 - b^2}\). A perpendicular line drawn from the center to a chord bisects the chord. Hence, \(C\) is the midpoint of \(AB\). Therefore, the total length of the chord \(AB\) is: \(AB = 2 \times AC = 2\sqrt{a^2 - b^2}\).
In simple words: The small circle's radius meets the large circle's chord at a right angle, making a right-angled triangle. Calculating half of the chord gives \(\sqrt{a^2 - b^2}\), so the full chord is twice that length.

Exam Tip: Always state the theorem about the perpendicular from the center bisecting the chord to justify doubling the value of \(AC\).

Question 12. From a point P outside the circle with centre O, tangents PA and PB are drawn to the circle. Prove that OP is the right bisector of the line segment AB.
Answer: Let \(OP\) intersect the chord \(AB\) at point \(M\). We need to prove that \(OP \perp AB\) and \(AM = BM\). In triangles \(\Delta MAP\) and \(\Delta MBP\):
• \(PA = PB\) (Tangents drawn from an external point are equal in length)
• \(\angle APM = \angle BPM\) (The line segment connecting the center to the external point bisects the angle between the two tangents)
• \(MP = MP\) (Common side) Using the SAS congruence criterion, we have \(\Delta MAP \cong \Delta MBP\). By CPCT (Corresponding Parts of Congruent Triangles):
• \(AM = BM\) (Hence, \(M\) is the midpoint of \(AB\))
• \(\angle AMP = \angle BMP\) Since these two angles form a linear pair: \(\angle AMP + \angle BMP = 180^\circ \Rightarrow 2\angle AMP = 180^\circ \Rightarrow \angle AMP = 90^\circ\). Therefore, \(OP\) is the right bisector (perpendicular bisector) of the segment \(AB\).

In simple words: By proving that the top and bottom triangles are identical, we show that the line segment \(AB\) is divided exactly in half at a \(90^\circ\) angle.

Exam Tip: Clearly state your steps using congruent triangle rules (SAS) and then use the linear pair property to establish the \(90^\circ\) angle.

Question 13. A circle is inscribed in a triangle ABC, touching BC, CA and AB at P,Q and R respectively if AB = 10 cm AQ = 7cm CQ = 5 cm. Find BC
Answer: Tangents drawn from an external point to a circle are equal in length. Thus, starting from vertex A: \(AR = AQ = 7\text{ cm}\). Since \(AB = 10\text{ cm}\), we can calculate the length of \(BR\): \(BR = AB - AR = 10 - 7 = 3\text{ cm}\). Now, from vertex B: \(BP = BR = 3\text{ cm}\). From vertex C: \(CP = CQ = 5\text{ cm}\). The total length of side \(BC\) is: \(BC = BP + CP = 3 + 5 = 8\text{ cm}\).

In simple words: The segments of the triangle touching the circle are equal from each corner. By matching these equal pieces around the triangle, we find that the base is made of a \(3\text{ cm}\) segment and a \(5\text{ cm}\) segment, which add up to \(8\text{ cm}\).

Exam Tip: Label the tangent segments on your sketch (like \(AR\), \(AQ\)) to help keep your calculations clear and prevent simple addition errors.

Question 14. A Quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Prove that AB + CD = AD + BC
Answer: Let the circle touch the sides \(AB\), \(BC\), \(CD\), and \(DA\) of the quadrilateral at points \(P\), \(Q\), \(R\), and \(S\) respectively. Since the lengths of tangents drawn from an external point to a circle are equal:
• At vertex A: \(AP = AS\) — (I)
• At vertex B: \(BP = BQ\) — (II)
• At vertex C: \(CR = CQ\) — (III)
• At vertex D: \(DR = DS\) — (IV) Adding equations (I), (II), (III), and (IV): \((AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)\) Substituting the segments back into their respective sides of the quadrilateral: \(AB + CD = AD + BC\). (Hence Proved)

In simple words: By writing down the equal tangent pairs for all four corners and adding them up, they combine to show that the sum of the opposite sides of the quadrilateral is equal.

Exam Tip: When writing the four equations, make sure that segments belonging to the same side (like \(AP\) and \(BP\)) are kept on the same side of the equations to make adding them easier.

Question 15. Two concentric circles are of radii 7 cm and r cm respectively, where r>7. A chord of the larger circle of length 46 cm, touches the smaller circle. Find the value of r.
Answer: Let \(O\) be the common center. The chord \(AB\) of the larger circle (radius \(r\)) is tangent to the smaller circle (radius \(7\text{ cm}\)) at point \(C\). The perpendicular from the center of the circle bisects the chord: \(AC = \frac{AB}{2} = \frac{46}{2} = 23\text{ cm}\). In right-angled triangle \(\Delta ACO\) (where \(OC = 7\text{ cm}\) is perpendicular to the chord \(AB\)): Applying the Pythagoras theorem: \(OA^2 = OC^2 + AC^2 \Rightarrow r^2 = 7^2 + 23^2 = 49 + 529 = 578\) \(r = \sqrt{578} = 17\sqrt{2}\text{ cm}\).

In simple words: The inner radius and half of the chord make a right-angled triangle with the outer radius. We use Pythagoras with sides \(7\) and \(23\) to get the outer radius, which equals \(17\sqrt{2}\text{ cm}\).

Exam Tip: Always show that you have halved the chord length (\(46 / 2 = 23\)) before using the value in the Pythagoras theorem.

Question 16. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Answer: Let us consider a circle with center \(O\) and a tangent line \(XY\) touching the circle at point \(P\). We need to prove that \(OP \perp XY\). Take any point \(Q\) on the tangent line \(XY\) other than the point of contact \(P\). Join the segment \(OQ\). The point \(Q\) must lie completely outside the circle. (If \(Q\) were inside the circle, the line \(XY\) would intersect the circle at two points, making it a secant instead of a tangent). Since \(Q\) lies outside the circle, the distance from the center to \(Q\) must be greater than the circle's radius: \(OQ > OP\). This inequality remains true for every point on the tangent line \(XY\) except the point \(P\). Therefore, \(OP\) represents the shortest possible distance from the center \(O\) to any point on the line \(XY\). Because the shortest path between a point and a line is always the perpendicular distance, we conclude: \(OP \perp XY\). (Hence Proved)
In simple words: Any point on the tangent line other than the point of contact lies outside the circle, meaning it is further from the center than the radius. Because the point of contact is the closest point to the center, it must be the perpendicular point.

Exam Tip: This is a standard geometric proof. Remember to state clearly that "the perpendicular distance is the shortest distance" to complete your logic.

LEVEL - III

Question 17. Prove that the length of tangents drawn from an external point to a circle are equal.
Answer: Let there be a circle with center \(O\). From an external point \(P\), two tangents \(PA\) and \(PB\) are drawn, touching the circle at \(A\) and \(B\) respectively. We need to prove that \(PA = PB\). Join \(OA\), \(OB\), and \(OP\). In right-angled triangles \(\Delta OAP\) and \(\Delta OBP\):
• \(OA = OB\) (Radii of the same circle)
• \(OP = OP\) (Common hypotenuse)
• \(\angle OAP = \angle OBP = 90^\circ\) (The radius is perpendicular to the tangent at the point of contact) Therefore, by the RHS (Right angle - Hypotenuse - Side) congruence rule: \(\Delta OAP \cong \Delta OBP\). By CPCT (Corresponding Parts of Congruent Triangles): \(PA = PB\). (Hence Proved)

In simple words: By drawing lines from the center of the circle to the corners, we make two right-angled triangles. Since these triangles are mirror images (congruent), the two tangent lines must be equal in length.

Exam Tip: Use the RHS congruence rule for this proof, as \(OP\) is the common hypotenuse and the radii are equal sides.

Question 18. Prove that the tangents at the extremities of any chord of a circle, make equal angle with the chord.
Answer: Let \(AB\) be a chord of a circle with center \(O\). The tangents at \(A\) and \(B\) intersect at an external point \(P\). We need to prove that \(\angle PAB = \angle PBA\). Since tangents drawn from an external point to a circle are equal: \(PA = PB\). In \(\Delta PAB\), because two sides are equal, it is an isosceles triangle. Angles opposite to equal sides of a triangle are also equal: \(\angle PAB = \angle PBA\). (Hence Proved)
In simple words: The two tangents meeting at an outside point are equal. This makes an isosceles triangle with the chord, meaning the base angles on the chord are equal.

Exam Tip: Justify your proof by mentioning the properties of an isosceles triangle (equal sides lead to equal opposite angles).

Question 19. PA and PB are tangents to the circle with the centre O from an external point P,touching the circle at A and B respectively. Show that the quadrilateral AOBP IS cyclic.
Answer: In quadrilateral \(AOBP\): The sum of all four interior angles is \(360^\circ\): \(\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ\). Since the radius is perpendicular to the tangent at the point of contact, we have: \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\). Substituting these values: \(\angle AOB + 90^\circ + \angle APB + 90^\circ = 360^\circ\) \(\angle AOB + \angle APB + 180^\circ = 360^\circ\) \(\angle AOB + \angle APB = 180^\circ\). Since the sum of the opposite angles in quadrilateral \(AOBP\) is \(180^\circ\), the quadrilateral is cyclic. (Hence Proved)

In simple words: The two side angles at \(A\) and \(B\) are both \(90^\circ\) because the radii and tangents meet there. Since those two make \(180^\circ\), the other two opposite angles must also add up to \(180^\circ\), proving the shape is a cyclic quadrilateral.

Exam Tip: Recall that a quadrilateral is cyclic if its opposite angles are supplementary (add up to \(180^\circ\)). Proving this satisfies the theorem requirement.

Question 20. Prove that the parallelogram circumscribing a circle is a rhombus.
Answer: Let \(ABCD\) be a parallelogram circumscribing a circle. Since \(ABCD\) is a parallelogram, opposite sides are equal: \(AB = CD\) and \(AD = BC\). As proven in Question 14, for any quadrilateral circumscribing a circle: \(AB + CD = AD + BC\). Substituting \(CD = AB\) and \(AD = BC\) into this equation: \(AB + AB = BC + BC \Rightarrow 2AB = 2BC \Rightarrow AB = BC\). Since adjacent sides are equal (\(AB = BC\)) and opposite sides are equal, all four sides must be equal: \(AB = BC = CD = AD\). A parallelogram with all sides equal is a rhombus. Therefore, \(ABCD\) is a rhombus. (Hence Proved)
In simple words: In a parallelogram circumscribing a circle, we know opposite sides are equal and the sum of opposite sides matches. This forces all four sides of the shape to be equal, making it a rhombus.

Exam Tip: Combine the tangent sum property (\(AB + CD = AD + BC\)) with the standard property of a parallelogram (\(AB=CD\)) to prove this result efficiently.

Question 21. In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersects XY at A and X’Y’ at B. Prove that \(\angle AOB = 90^\circ\).
Answer: Let us join the center \(O\) with the point of contact \(C\) of the third tangent. In right triangles \(\Delta OPA\) and \(\Delta OCA\):
• \(OP = OC\) (Radii of the same circle)
• \(AP = AC\) (Tangents from external point \(A\))
• \(OA = OA\) (Common hypotenuse) Therefore, \(\Delta OPA \cong \Delta OCA\) (by SSS congruence). By CPCT, we have: \(\angle POA = \angle COA\) — (i). Similarly, we can prove \(\Delta OQB \cong \Delta OCB\), which gives: \(\angle QOB = \angle COB\) — (ii). Since \(PQ\) is a diameter of the circle, it is a straight line. Thus, the sum of angles on the line is \(180^\circ\): \(\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ\). Substituting from equations (i) and (ii): \(2\angle COA + 2\angle COB = 180^\circ\) \(2(\angle COA + \angle COB) = 180^\circ\) \(\angle COA + \angle COB = 90^\circ\). Since \(\angle AOB = \angle COA + \angle COB\), we get: \(\angle AOB = 90^\circ\). (Hence Proved)

In simple words: The central line is split into four angles that add up to \(180^\circ\) on a straight line. Since these angles are equal in pairs, the middle two combined must equal half of the straight line, which is \(90^\circ\).

Exam Tip: Remember to prove the congruence of the helper triangles first to establish that \(\angle POA = \angle COA\).

Question 22. Two roads starting from P are touching a circular path at A and B. Sarita runs from P to A, 20km and A to O, 15km and Reeta runs from P to O directly.
(a) Find the distance covered by Reeta.
(b) Who will win the race?
(c) What value is depicted by Reeta?
Answer:
(a) The radius \(OA = 15\text{ km}\) is perpendicular to the tangent road \(PA = 20\text{ km}\), making \(\Delta OAP\) a right-angled triangle at \(A\). By the Pythagoras theorem: \(OP^2 = OA^2 + AP^2 = 15^2 + 20^2 = 225 + 400 = 625\) \(OP = \sqrt{625} = 25\text{ km}\). Therefore, Reeta covers a distance of \(25\text{ km}\).
(b) Sarita covers a total distance of \(20\text{ km} + 15\text{ km} = 35\text{ km}\). Reeta covers only \(25\text{ km}\). Since Reeta's path is shorter, she will win the race.
(c) Reeta chooses the shortest path to reach the destination, depicting intelligence, smart planning, and efficiency.

In simple words: The road along the tangent and the radius make a right triangle. We find the direct diagonal path \(OP\) is \(25\text{ km}\). Since \(25\text{ km}\) is shorter than going the long way around (\(35\text{ km}\)), Reeta wins by being smart.

Exam Tip: For value-based questions, make sure to answer all sub-parts (a, b, and c) separately and write down the full calculations for full marks.

SELF EVALUATION

Question 1. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer: Let us draw a circle with center \(O\). We draw a given reference line \(L\). Then, parallel to \(L\), we construct:
• Line \(T\), which touches the circle at exactly one point (the tangent).
• Line \(S\), which cuts through the circle at two points (the secant).

In simple words: This diagram shows three parallel lines: a reference line, a tangent touching the circle's top, and a secant slicing through the circle.

Exam Tip: Use a dashed line for the given reference line to make it visually distinct from your final construction lines.

Question 2. Prove that perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer: Let us consider a circle with center \(O\) and a tangent \(AB\) touching at point \(P\). We need to prove that the perpendicular to \(AB\) at \(P\) passes through \(O\). Let us assume by contradiction that the perpendicular to \(AB\) at \(P\) does not pass through \(O\), but through some other point \(O'\). Since \(O'P \perp AB\), we have: \(\angle O'PB = 90^\circ\) — (i). We also know by theorem that the radius joining the center \(O\) to the point of contact is perpendicular to the tangent: \(\angle OPB = 90^\circ\) — (ii). Comparing equations (i) and (ii), we get: \(\angle O'PB = \angle OPB\). This is only possible if \(O'\) and \(O\) coincide, because a part cannot be equal to the whole. Therefore, the perpendicular to the tangent at the point of contact must pass through the center \(O\). (Hence Proved)
In simple words: Since there can only be one line perpendicular to the tangent at the point of contact, and we know the radius is perpendicular, the perpendicular line has no choice but to be the radius line, which goes through the center.

Exam Tip: Contradiction is the standard and most convincing way to prove this theorem in your board exams.

Question 3. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Answer: Let \(PA\) and \(PB\) be two tangents drawn from an external point \(P\) to a circle with center \(O\). We need to prove that \(\angle APB + \angle AOB = 180^\circ\). In quadrilateral \(AOBP\): The sum of all four interior angles is \(360^\circ\): \(\angle APB + \angle OAP + \angle AOB + \angle OBP = 360^\circ\). The radius is perpendicular to the tangent at the point of contact, so: \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\). Substituting these values: \(\angle APB + 90^\circ + \angle AOB + 90^\circ = 360^\circ\) \(\angle APB + \angle AOB + 180^\circ = 360^\circ\) \(\angle APB + \angle AOB = 180^\circ\). (Hence Proved)
In simple words: The two side angles where the tangents meet the radius are \(90^\circ\) each. Because the four angles in the loop add up to \(360^\circ\), the remaining two opposite angles must add up to \(180^\circ\).

Exam Tip: State the angle sum property of quadrilaterals clearly as the basis of your algebraic proof.

Question 4. The length of a tangent from a point A at a distance 5cm from the centre of the circle is 4cm. Find the radius of the circle.
Answer: Let \(O\) be the center of the circle and \(P\) be the point of contact of the tangent from \(A\). This forms a right-angled triangle \(\Delta OPA\) where \(\angle OPA = 90^\circ\) (since radius is perpendicular to tangent). The hypotenuse \(OA = 5\text{ cm}\), and the tangent length \(AP = 4\text{ cm}\). Using the Pythagoras theorem: \(OA^2 = OP^2 + AP^2 \Rightarrow 5^2 = OP^2 + 4^2\) \(\Rightarrow 25 = OP^2 + 16 \Rightarrow OP^2 = 9\) \(\Rightarrow OP = \sqrt{9} = 3\text{ cm}\). Therefore, the radius of the circle is \(3\text{ cm}\).
In simple words: The radius, tangent, and distance to the center form a right-angled triangle. With hypotenuse \(5\text{ cm}\) and leg \(4\text{ cm}\), Pythagoras gives the radius as \(3\text{ cm}\).

Exam Tip: Remember the standard 3-4-5 right triangle triplet to quickly verify your calculation on rough paper.

Question 5. Two concentric circles are of radii 6.5cm and 2.5cm. Find the length of the chord of larger circle which touches the smaller circle.
Answer: Let \(O\) be the common center. Let \(AB\) be the chord of the larger circle (radius \(6.5\text{ cm}\)) that touches the smaller circle (radius \(2.5\text{ cm}\)) at point \(C\). Since \(OC\) is perpendicular to \(AB\), \(\Delta OCA\) is a right-angled triangle at \(C\). By the Pythagoras theorem: \(OA^2 = OC^2 + AC^2 \Rightarrow 6.5^2 = 2.5^2 + AC^2\) \(\Rightarrow 42.25 = 6.25 + AC^2 \Rightarrow AC^2 = 36\) \(\Rightarrow AC = \sqrt{36} = 6\text{ cm}\). Since the perpendicular from the center bisects the chord, the total length of the chord \(AB\) is: \(AB = 2 \times AC = 2 \times 6 = 12\text{ cm}\).
In simple words: The small radius and half the chord make a right-angled triangle with the large radius. Solving with Pythagoras gives half the chord as \(6\text{ cm}\), so the total chord is \(12\text{ cm}\).

Exam Tip: Write down both the squaring steps and the doubling steps clearly to make sure you get full credit for the solution.

Question 6. From a point P, 10cm away from the centre of the circle, a tangent PT of length 8cm is drawn. Find the radius of the circle.
Answer: Let \(O\) be the center of the circle. The tangent \(PT\) touches the circle at \(T\), so \(\angle OTP = 90^\circ\). In right-angled triangle \(\Delta OTP\): The hypotenuse \(OP = 10\text{ cm}\), and the tangent length \(PT = 8\text{ cm}\). Using the Pythagoras theorem: \(OP^2 = OT^2 + PT^2 \Rightarrow 10^2 = OT^2 + 8^2\) \(\Rightarrow 100 = OT^2 + 64 \Rightarrow OT^2 = 36\) \(\Rightarrow OT = \sqrt{36} = 6\text{ cm}\). Therefore, the radius of the circle is \(6\text{ cm}\).
In simple words: The radius and tangent make a right angle. Pythagoras on a triangle with hypotenuse \(10\text{ cm}\) and leg \(8\text{ cm}\) gives the missing radius as \(6\text{ cm}\).

Exam Tip: Be sure to write the unit "cm" next to your final answer to avoid losing half a mark for missing units.