CBSE Class 10 Mathematics Circles (Tangents) Worksheet Set 08

Answer: The radius OQ is perpendicular to the tangent PQ, so ∠OQP = 90°. In triangle OQP the other two angles add to 180° − 90° = 90°.

Teacher's Note:
1. A tangent is perpendicular to the radius at the point of contact.
2. Use the angle sum of the right triangle.

Answer: Triangle OQP is right-angled at Q, so OP² = 5² + 12² = 169 and OP = 13 cm.

Teacher's Note:
1. The radius to the point of contact is perpendicular to the tangent.
2. Apply Pythagoras theorem.

Answer: In right triangle OQP, ∠OPQ = 90° − 70° = 20°. Since OP bisects ∠QPR, ∠QPR = 2 × 20° = 40°.

Teacher's Note:
1. Find ∠OPQ from the right triangle.
2. The line from P to the centre bisects the angle between the tangents.

Answer: OQ ⊥ QS, so QS² = 8² − 6² = 28 and QS = 2√7 cm ≈ 5.29 cm.

Teacher's Note:
1. The radius and the tangent form a right angle at Q.
2. Use Pythagoras theorem with OS as hypotenuse.

Answer: The distance between parallel tangents equals the diameter, so the radius is 6 cm.

Teacher's Note:
1. Parallel tangents touch at the two ends of a diameter.
2. Divide the distance by 2.

Answer: The triangle formed by the two radii and the chord is right-angled, so the chord = √(10² + 10²) = 10√2 cm.

Teacher's Note:
1. The two radii are perpendicular sides.
2. The chord is the hypotenuse.

Answer: Exactly two tangents can be drawn from an external point. A circle has infinitely many tangents, one at each point of the circle.

Teacher's Note:
1. Each point of the circle has a unique tangent.
2. From an external point there are two touching lines.

Answer: The chord touches the smaller circle at its mid-point, and this radius b is perpendicular to the chord. Half-chord = √(a² − b²), so chord = 2√(a² − b²).

Teacher's Note:
1. The radius to the point of contact bisects the chord.
2. Use Pythagoras theorem in the right triangle.

Answer: Half-chord = √(6.5² − 2.5²) = √36 = 6 cm, so the chord = 12 cm.

Teacher's Note:
1. The chord is a tangent to the inner circle.
2. Double the half-chord.

Answer: Half-chord = 23 cm, so r² = 7² + 23² = 49 + 529 = 578 and r = 17√2 cm ≈ 24.04 cm.

Teacher's Note:
1. Half the chord, the inner radius and the outer radius form a right triangle.
2. Solve for r.

Answer: Radius² = 5² − 4² = 9, so the radius is 3 cm.

Teacher's Note:
1. The radius is perpendicular to the tangent at the point of contact.
2. The distance from the centre is the hypotenuse.

Answer: Radius² = 10² − 8² = 36, so the radius is 6 cm.

Teacher's Note:
1. Form the right triangle OTP.
2. OP is the hypotenuse.

Answer: Tangents from a point are equal: AR = AQ = 7, so BR = 10 − 7 = 3 cm. BP = BR = 3 and CP = CQ = 5, so BC = 3 + 5 = 8 cm.

Teacher's Note:
1. Use equal tangents from each vertex.
2. Add the two parts of BC.

Answer: ∠OAP = 90° and OP bisects ∠APB, so ∠OPA = 40°. In triangle OAP, ∠POA = 180° − 90° − 40° = 50°.

Teacher's Note:
1. OP bisects the angle between the tangents.
2. Use the angle sum of triangle OAP.

Answer: Since OA ⊥ PA, OP² = 20² + 15² = 625, so Reeta covers OP = 25 km. Sarita covers 35 km, so Reeta covers the shorter distance and reaches first.

Teacher's Note:
1. Tangent PA is perpendicular to radius OA.
2. Compare 25 km with 20 + 15 km.

Answer: Let the tangent at P touch the circle with centre O. Take any other point Q on the tangent. Q lies outside the circle, so OQ > OP. This holds for every Q on the tangent other than P, so OP is the shortest distance from O to the tangent. The shortest distance is the perpendicular, so OP ⊥ tangent.

Teacher's Note:
1. Every point of a tangent except the point of contact lies outside the circle.
2. The shortest segment from a point to a line is perpendicular.

Answer: Let PA and PB be tangents from P, with centre O. In triangles OAP and OBP, OA = OB (radii), ∠OAP = ∠OBP = 90°, and OP is common. By RHS congruence the triangles are congruent, so PA = PB.

Teacher's Note:
1. Join the centre to the contact points and to P.
2. Use RHS congruence.

Answer: PA = PB and OA = OB, so both P and O lie on the perpendicular bisector of AB. A line through two points is unique, so OP is the perpendicular bisector of AB.

Teacher's Note:
1. Use equal tangents and equal radii.
2. Each of P and O is equidistant from A and B.

Answer: Let the tangents at A and B meet at P. Then PA = PB, so ∠PAB = ∠PBA in triangle PAB. These are the angles the tangents make with the chord.

Teacher's Note:
1. Tangents from the same external point are equal.
2. Angles opposite equal sides are equal.

Answer: ∠OAP = ∠OBP = 90°. The angle sum of the quadrilateral gives ∠AOB + ∠APB = 360° − 180° = 180°. A quadrilateral with opposite angles supplementary is cyclic.

Teacher's Note:
1. Use the right angles at the points of contact.
2. Check that opposite angles add to 180°.

Answer: Let ABCD circumscribe a circle. Then AB + CD = AD + BC. In a parallelogram AB = CD and AD = BC, so 2AB = 2AD and AB = AD. Thus all four sides are equal and ABCD is a rhombus.

Teacher's Note:
1. Use the result AB + CD = AD + BC for a circumscribed quadrilateral.
2. Use equal opposite sides of a parallelogram.

Answer: Let the circle touch AB, BC, CD and DA at P, Q, R and S. Tangents from a vertex are equal: AP = AS, BP = BQ, CR = CQ, DR = DS. Adding: AB + CD = (AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC.

Teacher's Note:
1. Name the four points of contact.
2. Use equal tangents from each vertex.

Answer: OA bisects ∠PAC and OB bisects ∠QBC, where P and Q are the points of contact on the parallel tangents. Since the tangents are parallel, ∠PAC + ∠QBC = 180°, so ∠OAC + ∠OBC = 90°. In triangle AOB, ∠AOB = 180° − 90° = 90°.

Teacher's Note:
1. Use the property that the centre lies on the bisector of the angle between two tangents.
2. Co-interior angles between parallel lines add to 180°.

Answer: In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. So ∠AOB + ∠APB = 360° − 90° − 90° = 180°. Hence the angles are supplementary.

Teacher's Note:
1. Join the centre to both points of contact.
2. Use the angle sum of a quadrilateral.

Answer: Let the tangent touch the circle at P and suppose the perpendicular at P does not pass through the centre O. Then the radius OP would not lie along that perpendicular, so OP would not be perpendicular to the tangent. This contradicts the tangent-radius property, so the perpendicular must pass through O.

Teacher's Note:
1. Use proof by contradiction.
2. The radius at the point of contact is already known to be perpendicular to the tangent.