Exercises
Q10.1.1. How many tangents can a circle have?
Answer:
A circle can have infinitely many tangents, as a tangent can be drawn at every point on the circumference of the circle.
Q10.1.2(i). A tangent to a circle intersects it in _________ point(s).
Answer:
A tangent to a circle intersects it in exactly one point (the point of contact).
Q10.1.2(ii). A line intersecting a circle in two points is called a _________ .
Answer:
A line intersecting a circle in two points is called a secant.
Q10.1.2(iii). A circle can have _________ parallel tangents at the most.
Answer:
A circle can have two parallel tangents at the most (one on each side of the circle).
Q10.1.2(iv). The common point of a tangent to a circle and the circle is called _________ .
Answer:
The common point of a tangent to a circle and the circle is called the point of contact.
Q10.1.3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is : (A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 cm.
Answer:
Correct answer: (D) √119 cm.
Explanation: Since OP is the radius and PQ is the tangent at P, ∠OPQ = 90°. So, triangle OPQ is right-angled at P. Using Pythagoras theorem: OQ² = OP² + PQ² ⇒ 12² = 5² + PQ² ⇒ 144 = 25 + PQ² ⇒ PQ² = 119 ⇒ PQ = √119 cm.
Q10.1.4. 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:
Draw a circle with centre O. Draw a straight line L outside the circle (non-intersecting). Draw a line T parallel to L that touches the circle at exactly one point (point of contact) — this is the tangent. Draw another line S parallel to L that cuts the circle at two distinct points — this is the secant.
Q2. Fill in the blanks : (i) A tangent to a circle intersects it in _____ point (s). (ii) A line intersecting a circle in two points is called a _____. (iii) A circle can have _____ parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called _____.
Answer:
(i) exactly one (ii) secant (iii) two (iv) point of contact
Q1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm
Answer:
Correct answer: (A) 7 cm. Explanation: In right triangle with tangent, radius² = distance² - tangent² = 625 - 576 = 49, so radius = 7 cm.
Q2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to (A) 60° (B) 70° (C) 80° (D) 90°
Answer:
Correct answer: (B) 70°. Explanation: In quadrilateral OPTQ, ∠OPT = ∠OQT = 90° (radius perpendicular to tangent). Sum of angles = 360°, so ∠PTQ = 360° - 90° - 90° - 110° = 70°.
Q3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to (A) 50° (B) 60° (C) 70° (D) 80°
Answer:
Correct answer: (A) 50°. Explanation: PA = PB, ∠APB = 80°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°, so ∠AOB = 360° - 90° - 90° - 80° = 100°. In isosceles triangle OAP, ∠POA = (180° - 100°)/2 = 40°? Wait, correction: OP bisects ∠APB and ∠AOB, so ∠OPA = 40°. Since ∠OAP = 90°, ∠POA = 90° - 40° = 50°.
Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer:
Let AB be a diameter of a circle with centre O. Let tangents be drawn at A and B. Radius OA is perpendicular to tangent at A, and radius OB is perpendicular to tangent at B. Since OA and OB are opposite directions on the same line, the two tangents are perpendicular to the same line, hence they are parallel.
Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer:
Let PT be a tangent to a circle at point P, and let a line through P be perpendicular to PT. Assume this line does not pass through the centre O. Then from O, draw a perpendicular to PT, meeting at Q. Since the shortest distance from O to PT is OQ, but OP is also perpendicular to PT (theorem 10.1), so Q and P coincide, meaning the perpendicular passes through O.
Q6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Answer:
Using Pythagoras theorem: radius² = distance² - tangent² = 25 - 16 = 9, so radius = 3 cm.
Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Let AB be chord of larger circle touching smaller circle at P. OP is radius of smaller circle = 3 cm and perpendicular to AB. In right triangle OAP, OA = 5 cm, OP = 3 cm, so AP² = 25 - 9 = 16, AP = 4 cm. Since OP bisects AB, AB = 2 × 4 = 8 cm.
Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Answer:
Let the circle touch sides AB, BC, CD, DA at points P, Q, R, S respectively. Then AP = AS, BP = BQ, CR = CQ, DR = DS (tangents from same external point). Adding: AP + BP + CR + DR = AS + BQ + CQ + DS ⇒ (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) ⇒ AB + CD = AD + BC.
Q9. In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.
Answer:
Let tangent at C meet XY at A and X'Y' at B. Join OC. Since OA and OB are joined to points of tangency, ∠OAC = 90° and ∠OBC = 90°. In quadrilateral AOB, sum of angles = 360°, and triangles OAC and OBC are congruent? Alternatively, using properties of tangents from A and B, we get ∠AOC = ∠COB, and since AC and BC are tangents from A and B, ∠AOB = 90°.
Q10. 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 tangents from external point P to circle with centre O. Join OA, OB. Then ∠OAP = ∠OBP = 90°. In quadrilateral AOBP, ∠AOB + ∠APB = 360° - 90° - 90° = 180°, so ∠APB = 180° - ∠AOB, i.e., they are supplementary.
Q11. Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:
Let ABCD be a parallelogram circumscribing a circle. Since opposite sides of a parallelogram are equal, AB = CD and BC = AD. From the property of tangents (for a circumscribed quadrilateral), AB + CD = AD + BC ⇒ 2AB = 2AD ⇒ AB = AD. Since adjacent sides are equal, the parallelogram is a rhombus.
Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Answer:
Let circle touch sides AB at E, AC at F, BC at D. Then BD = 8 cm, DC = 6 cm. Let AE = AF = x. Then AB = AE + EB = x + 8, AC = AF + FC = x + 6. Semi-perimeter s = (x+8 + x+6 + 14)/2 = (2x + 28)/2 = x + 14. Area = rs = 4(x+14) = 4x + 56. Also area by Heron formula: √[s(s-a)(s-b)(s-c)] = √[(x+14)(6)(8)(x)] = √[48x(x+14)]. Equating: (4x+56)² = 48x(x+14) ⇒ 16(x+14)² = 48x(x+14) ⇒ divide by 16(x+14): x+14 = 3x ⇒ 14 = 2x ⇒ x = 7. So AB = 15 cm, AC = 13 cm.
Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer:
Let quadrilateral ABCD circumscribe a circle with centre O. Let the circle touch sides AB, BC, CD, DA at P, Q, R, S respectively. Join OP, OQ, OR, OS. Then ∠AOP = ∠AOS, ∠BOP = ∠BOQ, ∠COQ = ∠COR, ∠DOR = ∠DOS (since triangles are congruent). Now ∠AOB = ∠AOP + ∠POB, and ∠COD = ∠COR + ∠ROD. Sum of all angles around O is 360°. Using the above, we get ∠AOB + ∠COD = 180°. Similarly, ∠BOC + ∠AOD = 180°. So opposite sides subtend supplementary angles at the centre.