Question Bank Tangent Properties of a Circle

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1 Question Bank Tangent Properties of a Circle 1. In quadrilateral ABCD, D = 90, BC = 38 cm and DC = 5 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 7 cm. Find the radius of the circle. In the figure, ABCD is the quadrilateral and a circle with centre O is inscribed in it which touches AD, AB, BC and DC at P, Q, R and S respectively. Join OP and OS. OP = OS...(i) [Radii of the same circle] OP AD and OS DC...(ii) [Radius through point of contact is perpendicular to the tangent] Also, DP = DS...(iii) [Tangents from an exterior point are equal in length] From (i), (ii), and (iii), we conclude that OPDS is a square. Also, BQ = BR = 7 cm [Tangents from an exterior point are equal in length] CR = CS = (CB BR) = (38 7) cm = 11 cm DS = DC CS = (5 11) cm = 14 cm Hence, radius of the circle = 14 cm. Math Class X 1 Question Bank

2 . Two parallel tangents of a circle meet a third tangent at P and Q. Prove that PQ subtends a right angle at the centre. Given : Two parallel tangents of a circle and a third tangent which intersects them at P and Q. To prove : POQ = 90. Construction : Join OA, OB, OC, OP and OQ. Proof : OAP = 90 [Radius through point of contact is perpendicular to the tangent] APC = OAP = 90 [Co-interior angles] Similarly, OBA = 90 and BQC = 90 ABQP is a rectangle. Also, OCP = 90 [Radius through point of contact is perpendicular to the tangent] and, OA = OB [Radii of the circle] CP = CQ Also, PC = PA [Tangents from an exterior point are equal] AOCP is a square. CP = CO CPO = COP = 45 [Angle sum property of a Δ] Similarly, CQO = COQ = 45 Hence, POQ = COP + COQ = = 90. Proved. Math Class X Question Bank

3 3. In the figure, a circle touches the side BC of ΔABC at P and AB and AC produced at Q and R respectively. Prove that : AQ = AR = 1.( perimeter of ΔABC) We know that tangents from an exterior point are equal in length. AQ = AR, BP = BQ and CP = CR Perimeter of ΔABC = AB + BC + CA = AB + BP + PC + CA = AB + BQ + CR + CA [ BP = BQ and CP = CR] = (AB + BQ) + (CR + CA) = AQ + AR = AQ = AR [ AQ = AR] AQ = AR = 1 (perimeter of ΔABC). Math Class X 3 Question Bank

4 4. In figure, PQR is an isosceles triangle with PQ = PR. A circle through Q touches PR at its middle point T and intersects side PQ at S. Prove that PQ = 4PS. PT is a tangent and PQ is a secant. PS PQ = PT PR PS PQ = 4 [ T is the mid-point of PR] PR = 4 PQ PS PQ = 4 [ PQ = PR] PS = PR 4 PQ = 4PS. Proved. 5. AB is a diameter and AC is a chord of a circle such that BAC = 30. The tangent at C intersects AB produced in a point D. Prove that BC = BD. In figure, O is the centre of the circle and BAC = 30. Math Class X 4 Question Bank

5 Join BC. BC is a chord of the circle. BAC = BCD = 30...(i) ACB = 90 ABC = 180 BAC ACB = = 60 CBD = 180 ABC CBD = = 10 In ΔBCD, BDC = 180 ( BCD + CBD) BDC = 180 ( ) = 30...(ii) From (i) and (ii), we have BCD = BDC [Angles in the alternate segments] [Angle in a semi-circle] [Linear pair] BC = BD [Sides opposite to equal angles are equal] 6. In the figure, PAT is a tangent at A and BD is a diameter of the circle. If ÐABD = 8 and BDC = 5, find : (i) TAD (ii) BAD (iii) PAB (iv) CBD (i) TAD = ABD [Angles in the alternate segments] TAD = 8. Ans. (ii) BAD = 90 [Angle in a semi-circle] Ans. (iii) In ΔABD, ADB = 180 ( BAD + ABD) ADB = 180 ( ) = 6 PAB = ADB [Angles in the alternate segments] PAB = 6. (iv) BCD = 90 [Angle in a semi-circle] In ΔBCD, Math Class X 5 Question Bank

6 CBD = 180 ( BCD + BDC) = 180 ( ) = In figure, inscribed circle of ΔPQR touches its sides at A, B, C. Find the angles of ΔABC. In ΔPQR, QPR = 180 ( ) = = 64 QA = QC [Tangents from an exterior point are equal] QCA = QAC [Angles opposite to equal sides are equal] QCA = QAC = [Angle sum property of a triangle] QCA = QAC = 54 QCA = CBA [Angles in the alternate segments] CBA = Similarly, RAB = RBA = = and PCB = PBC = = 58 RAB = ACB [Angles in the alternate segments] ACB = 68 and PCB = CAB = 58 CAB = 58 Hence, angles of the triangle ABC are 58, 54 and 68. Math Class X 6 Question Bank

7 8. Two circles touch internally at P. AB is a chord of the outer circle and it cuts the inner circle at C and D. Prove that CPA = DPB. Draw the tangent QPR at P. QPA = ABP... (i) [Angles in the alternate segments] QPC = CDP... (ii) [Angles in the alternate segments] Subtracting (i) from (ii), we get QPC QPA = CDP ABP CPA = CDP ABP... (iii) In ΔPBD, we have CDP = DBP + DPB [Exterior angle is equal to sum of interior opposite angles] CDP DBP = DPB CDP ABP = DPB... (iv) From (iii) and (iv), we get CPA = DPB. 9. Prove that the line joining the centres of two circles divides their transverse common tangent into the ratio of their radii. Math Class X 7 Question Bank

8 P and Q are the centres of two circles and RS is their tansverse common tangent. RS meets PQ at O. In ΔROP and ΔSOQ, we have PRO = QSO [Each = 90 ] POR = QOS [Vertically opposite angles] ΔROP ~ ΔSOQ [AA similarity] RO PR = PR SQ [Sides of similar triangles are proportional] RO PR = SO SQ. Proved. Math Class X 8 Question Bank

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LLT Education Services 8. 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. (a) 4 cm (b) 3 cm (c) 6 cm (d) 5 cm 9. From a point P, 10 cm away from the