# Udaan School Of Mathematics Class X Chapter 10 Circles Maths

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1 Exercise Fill in the blanks (i) The common point of tangent and the circle is called point of contact. (ii) A circle may have two parallel tangents. (iii) A tangent to a circle intersects it in one point. (iv) A line intersecting a circle in two points A called a secant. (v) The angle between tangent at a point P on circle and radius through the point is 90. Point of contact P 90 Tangent O Secant 2. How many tangents can a circle have? Tangent: A line intersecting circle in one point is called a tangent. As there are infinite number of points on the circle a circle has many (infinite) tangents. 3. O is the center of a circle of radius 8cm. The tangent at a point A on the circle cuts a line through O at B such that AB = 15 cm. Find OB Consider a circle with center O and radius OA = 8cm = r, AB = 15 cm.

2 r A Udaan School Of Mathematics O B (AB) tangent is drawn at A (point of contact) At point of contact, we know that radius and tangent are perpendicular. In OAB, OAB = 90, By Pythagoras theorem OB 2 = OA 2 + AB 2 OB = = = 229 = 17 cm OB = 17 cm 4. If the tangent at point P to the circle with center O cuts a line through O at Q such that PQ = 24cm and OQ = 25 cm. Find the radius of circle Given, PQ = 24 cm OQ = 25 cm OP = radius =? P O 25 cm Q P is point of contact, At point of contact, tangent and radius are perpendicular to each other POQ is right angled triangle OPQ = 90 By Pythagoras theorem, PQ 2 + OP 2 = OQ OP 2 = 25 2 OP = = = 49 = 7cm OP = radius = 7cm

3 Exercise If PT is a tangent at T to a circle whose center is O and OP = 17 cm, OT = 8 cm. Find the length of tangent segment PT. OT = radius = 8cm OP = 17cm PT = length of tangent =? 8 cm T O 17 cm P T is point of contact. at point of contact tangent and radius are perpendicular. OTP is right angled triangle OTP = 90, from Pythagoras theorem OT 2 + PT 2 = OP PT 2 = 17 2 PT = = 225 = 15cm PT = length of tangent = 15 cm. 2. Find the length of a tangent drawn to a circle with radius 5cm, from a point 13 cm from the center of the circle. Consider a circle with center O. OP = radius = 5 cm. A tangent is drawn at point P, such that line through O intersects it at Q, OB = 13cm. Length of tangent PQ =? P Q O A + P, we know that tangent and radius are perpendicular. OPQ is right angled triangle, OPQ = 90 By pythagoras theorem, OQ 2 = OP 2 + PQ = PQ 2 PQ 2 = = 144

4 PQ = 144 = 12cm Length of tangent = 12 cm 3. A point P is 26 cm away from O of circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle. Given OP = 26 cm PT = length of tangent = 10cm radius = OT =? P 10 T Udaan School Of Mathematics 26 O At point of contact, radius and tangent are perpendicular OTP = 90, OTP is right angled triangle. By Pythagoras theorem, OP 2 = OT 2 + PT = OT OT k = ( ) k OT = 576 = 24 cm OT = length of tangent = 24 cm 4. If from any point on the common chord of two intersecting circles, tangents be drawn to circles, prove that they are equal. Let the two circles intersect at points X and Y. XY is the common chord. Suppose A is a point on the common chord and AM and AN be the tangents drawn A to the circle We need to show that AM = AN. In order to prove the above relation, following property will be used.

5 Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersects the circle at points A and B, then PT 2 = PA PB" Now AM is the tangent and AXY is a secant AM 2 = AX AY.. (i) AN is a tangent and AXY is a secant AN 2 = AX AY.. (ii) From (i) & (ii), we have AM 2 = AN 2 AM = AN 5. If the quadrilateral sides touch the circle prove that sum of pair of opposite sides is equal to the sum of other pair. Consider a quadrilateral ABCD touching circle with center O at points E, F, G and H as in figure. The tangents drawn from same external points to the circle are equal in length. 1. Consider tangents from point A [AM AE] AH = AE. (i) 2. From point B [EB & BF] BF = EB. (ii) 3. From point C [CF & GC] FC = CG. (iii) 4. From point D [DG & DH] DH = DG. (iv) Adding (i), (ii), (iii), & (iv) (AH + BF + FC + DH) = [(AC + CB) + (CG + DG)] (AH + DH) + (BF + FC) = (AE + EB) + (CG + DG) AD + BC = AB + DC [from fig.] Sum of one pair of opposite sides is equal to other. 6. If AB, AC, PQ are tangents in Fig. and AB = 5cm find the perimeter of APQ. Perimeter of APQ, (P) = AP + AQ + PQ = AP + AQ + (PX + QX)

6 The two tangents drawn from external point to the circle are equal in length from point A, AB = AC = 5 cm From point P, PX = PB From point Q, QX = QC Perimeter (P) = AP + AQ + (PB + QC) = (AP + PB) + (AQ + QC) = AB + AC = = 10 cms. 7. Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at center. Consider circle with center O and has two parallel tangents through A & B at ends of diameter. Let tangents through M intersects the tangents parallel at P and Q required to prove is that POQ = 90. From fig. it is clear that ABQP is a quadrilateral A + B = = 180 [At point of contact tangent & radius are perpendicular] A + B + P + Q = 360 [Angle sum property] P + Q = = 180..(i) At P & Q APO = OPQ = 1 2 P BQO = PQO = 1 2 Q 2 OPQ + 2 PQO = 180 in (i)

7 OPQ + PQO = 90. (ii) In OPQ, OPQ + PQO + POQ = 180 [Angle sum property] 90 + POQ = 180 [from (ii)] POQ = = 90 POQ = In Fig below, PQ is tangent at point R of the circle with center O. If TRQ = 30. Find PRS. Given TRQ = 30. At point R, OR RQ. ORQ = 90 TRQ + ORT = 90 ORT = = 60 ST is diameter, SRT = 90 [ Angle in semicircle = 90 ] ORT + SRO = 90 SRO + PRS = 90 PRS = = If PA and PB are tangents from an outside point P. such that PA = 10 cm and APB = 60. Find the length of chord AB. AP = 10 cm APB = 60 Represented in the figure A line drawn from center to point from where external tangents are drawn divides or bisects the angle made by tangents at that point APO = OPB = 1 60 = 30 2

8 The chord AB will be bisected perpendicularly AB = 2AM In AMP, sin 30 = opp.side = AM hypotenuse AP AM = AP sin 30 = AP = 10 = 5cm 2 2 AP = 2 AM = 10 cm In AMP, AMP = 90, APM = 30 AMP + APM + MAP = MAP = 180 MAP = 180 In PAB, MAP = BAP = 60, APB = 60 We also get, PBA = 60 PAB is equilateral triangle AB = AP = 10 cm Method (i) -----Method (ii) 10. From an external point P, tangents PA and PB are drawn to the circle with centre O. If CD is the tangent to the circle at point E and PA = 14 cm. Find the perimeter of ABCD. PA = 14 cm Perimeter of PCD = PC + PD + CD = PC + PD + CE + ED The two tangents drawn from external point to the circle are equal in length. From point P, PA = PB = 14cm From point C, CE = CA From point D, DB = ED Perimeter = PC + PD + CA +DB = (PC + CA) + (PD + DB) = PA + PB = = 28 cm. 11. In the fig. ABC is right triangle right angled at B such that BC = 6cm and AB = 8cm. Find the radius of its in circle.

9 BC = 6cm AB = 8cm As ABC is right angled triangle By Pythagoras theorem AC 2 = AB 2 + BC 2 = = 100 AC = 10 cm Consider BQOP B = 90, BPO = OQB = 90 [At point of contact, radius is perpendicular to tangent] All the angles = 90 & adjacent sides are equal BQOP is square BP = BQ = r The tangents drawn from any external point are equal in length. AP = AR = AB PB = 8 r QC = RC = BC BQ = 6 r AC = AR + RC 10 = 8 r + 6 r 10 = 14 2r 2r = 4 Radius = 2cm 12. From a point P, two tangents PA and PB are drawn to a circle with center O. If OP = diameter of the circle shows that APB is equilateral. OP = 2r Tangents drawn from external point to the circle are equal in length PA = PB At point of contact, tangent is perpendicular to radius. opp.side In AOP, sin θ = = r = 1 hypotenuse 2r 2 θ = 30

10 APB = 20 = 60, as PA = PB BAP = ABP = x. In PAB, by angle sum property APB + BAP + ABP = 180 2x = 120 x = 60 In this triangle all angles are equal to 60 APB is equilateral. 13. Two tangent segments PA and PB are drawn to a circle with center O such that APB = 120. Prove that OP = 2AP A + P OP bisects APB APO = OPB = 1 APB = = At point A OA AP, OAP = 90 In PDA, cos 60 = AP DP 1 = AP DP = 2AP 2 DP 14. If ABC is isosceles with AB = AC and C (0, 2) is the in circle of the ABC touching BC at L, prove that L, bisects BC. Given ABC is isosceles AB = AC The tangents from external point to circle are equal in length From point A, AP = AQ But AB = AC AP + PB = AQ + QC PB = PC. (i)

11 From B, PB = BL;.(ii) from C, CL = CQ..(iii) From (i), (ii) & (iii) BL = CL L bisects BC. 15. In fig. a circle touches all the four sides of quadrilateral ABCD with AB = 6cm, BC = 7cm, CD = 4cm. Find AD. the tangents drawn from any external point to circle are equal in length. From A AS = AP.(i) From B QB = BP. (ii) From C QC = RC..(iii) From D DS = DR. (iv) Adding (i), (ii), (iii) & (iv) (AS + QB + QC + DS) = (AB + BP + RC + OR) (AS + DS) + (QB + QC) = (AP + BP) + (RC + DR) AD + BC = AB + CD AD + 7 = AD = 3cm AD = 10 7 = 3cm 16. Prove that the perpendicular at the point of contact to a circle passes through the centre of the circle.

12 The at point of contact, the tangent is perpendicular to the radius. Radius is line from center to point on circle. Therefore, perpendicular to tangent will pass through center of circle. 17. In fig.. O is the center of the circle and BCD is tangent to it at C. Prove that BAC + ACD = 90 Given O is center of circle BCD is tangent. Required to prove: BAC + ACD = 90 Proof: OA = OC [radius] In OAC, angles opposite to equal sides are equal. OAC = OCA. (i) OCD = 90 [tangent is radius are perpendicular at point of contact] ACD + OCA = 90 ACD + OAC = 90 [ OAC = BAC] ACD + BAC = 90 Hence proved 18. Two circles touch externally at a point P. from a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and E respectively. Prove that TQ = TR. Let the circles be represented by (i) & (ii) respectively TQ, TP are tangents to (i) TP, TR are tangents to (ii) The tangents drawn from external point to the circle will be equal in length. For circle (i), TQ = TP. (i)

13 For circle (ii), TP = TR. (ii) From (i) & (ii) TQ = TR 19. In the fig. a circle is inscribed in a quadrilateral ABCD in which B = 90 if AD = 23cm, AB = 29cm and DS = 5cm, find the radius of the circle. Given AD = 23 cm AB = 29 cm B = 90 DS = 5cm From fig in quadrilateral POQB OPB = OQB = 90 = B = POQ and PO = OQ. POQB is a square PB = BQ = r Tangents drawn from external point to circle are equal in length. Tangents drawn from external point to circle are equal in length. From A, AR = AQ. (i) From B, PB = QB. (ii) From C, PC = CS. (iii) From D, DR = DS. (iv) (i) + (ii) + (iv) AR + DB + DR = AQ + QB + DS (AR + DR) + r = (AQ + QB) + DS AD + r = AB + DS 23 + r = r = = 11 cm radius = 11 cm 20. In fig. there are two concentric circles with Centre O of radii 5cm and 3cm. From an external point P, tangents PA and PB are drawn to these circles if AP = 12cm, find the tangent length of BP.

14 Given Udaan School Of Mathematics OA = 5 cm OB = 3 cm AP = 12 cm BP =? At the point of contact, radius is perpendicular to tangent. For circle 1, OAP is right triangle By Pythagoras theorem, OP 2 = OA 2 + AP 2 OP 2 = = = 169 OP = 169 = 13 cm For circle 2, OBP is right triangle by Pythagoras theorem, OP 2 = OB 2 + BP = BP 2 BP 2 = = 160 BP = 160 = 4 10 cm 21. In fig. AB is chord of length 16cm of a circle of radius 10cm. The tangents at A and B intersect at a point P. Find the length of PA. Given length of chord AB = 16cm. Radius OB = OA = 10 cm. Let line through Centre to point from where tangents are drawn be intersecting chord AB at M. we know that the line joining Centre to point from where tangents are drawn be intersecting chord AB at M. we know that

15 The line joining Centre to point from where tangents are drawn bisects the chord joining the points on the circle where tangents intersects the circle. AM = MB = 1 (AB) = 1 16 = 8cm 2 2 Consider OAM from fig. AMO = 90 By Pythagoras theorem, OA 2 = AM 2 + OM = OM 2 OM = = 36 = 6cm In AMP, AMP = 90 by Pythagoras theorem AP 2 = AM 2 + PM 2 AP 2 = (OP OM) 2 PA 2 = 64 + (OP 6) 2 (OP 6) 2 = 64 + PA 2.(i) In APO, PAO = 90 [At point of contact, radius is perpendicular to tangent] OP 2 = OA 2 + PA 2 [Pythagoras theorem] PA 2 = OP = OP (ii) 22. In figure PA and PB are tangents from an external point P to the circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN Given O is Centre of circle PA and PB are tangents The tangents drawn from external point to the circle are equal in length. From point P, PA = PB PL + AL = PN + NB. (i) From point L & N, AL = LM and MN = NB }. Substitute in (i) PL + Lm = PN + MN Hence proved. 23. In the fig. PO QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

16 Given PO OQ Consider quadrilateral OQTP. POQ = 90 OPT = OQT = 90 [At point of contact, tangent and radius are perpendicular] PTO = 90 OP = OQ = radius In this quadrilateral, all the angles are equal and pair of adjacent sides are equal. OQTP is a square. 24. In the fig two tangents AB and AC are drawn to a circle O such that BAC = 120. Prove that OA = 2AB. Consider Centre O for given circle BAC = 120 AB and AC are tangents From the fig. In OBA, OBA = 90 [radius perpendicular to tangent at point of contact] OAB = OAC = 1 BAC = = [Line joining Centre to external point from where tangents are drawn bisects angle formed by tangents at that external point1] In OBA, cos 60 = AB OA 1 = AB OA = 2AB 2 OA

17 25. In the fig. BC is a tangent to the circle with Centre O. OE bisects AP. Prove that AEO ~ ABC. Given BC is tangent to circle OE bisects AP, AE = EP Consider AOP 26. The lengths of three consecutive sides of a quadrilateral circumscribing a circle are 4cm, 5cm and 7cm respectively. Determine the length of fourth side. Let us consider a quadrilateral ABCD, AB = 4cm, BC = 5 cm, CD = 7cm, CD as sides circumscribing circle with centre O. and intersecting at points E, F, G, H. as in fig. the tangents drawn from external point to the circle are equal in length. From point A, AE = AH. (i) From point B, BE = BF.. (ii) From point C, GC = CE.(iii) From point D, GD = DH.(iv) (i) + (ii) + (iii) + (iv) (AE + BE + GC + GD) = (AH + BF + CF + DH) (AE + BE) + (GC + GD) = (AH + DH) + (BF + CF) AB + CD = AD + BC = 5 + AD AD = 11 5 = 6 cm Fourth side = 6 cm

18 27. In fig common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS. Consider Two circles namely (i) & (ii) as shown with common tangents as PQ and RS. The tangents from external point to the circle are equal in length. From A to circle (i) AP = AR (i) From A to circle (ii), AQ = AS. (ii) (i) + (ii) AP + AQ = AR + RS PQ = RS 28. Equal circles with centers O and O touch each other at X. OO produced to meet a circle with Centre O at A. AC is tangent to the circle whose Centre is a O D is perpendicular to AC. Find the value of DO /CO Given circles with centers O and O O D AC. Let radius = r O A = O X = OX = r In triangles, AO D and AOC A = A [Common angle] ADO = ACO = 90 [O D AC and at point of contact C, radius tangent] By A A similarity AO D ~ AOC. when two triangles are similar then their corresponding sides will be in proportion By A.A similarity AO D~ AOC When two triangles are similar then their corresponding sides will be in proportion AO AO = DO CO DO CO = r r+r+r = r 3r = 1 3

19 DO CO = 1 3 Udaan School Of Mathematics 29. In figure OQ : PQ = 3 : 4 and perimeter of PDQ = 60cm. determine PQ, QR and OP. Given OQ: PQ = 3 : 4 Let OQ = 3x PQ = 4x OP = y OQP = 90 [since at point of contact, tangent is perpendicular to radius] In OQP, by Pythagoras theorem OP 2 = OQ 2 + QP 2 y 2 = (3x) 2 + (4x) 2 y 2 = 9x x 2 = 25x 2 y 2 = 25x 2 = 5x Perimeter = OQ + PQ + OP = 3x + 4x + 5x = 12x According to problem perimeter = 60 12x = 60 x = = 5cm OQ = 3 5 = 15cm PQ = 4 5 = 20 cm OP = 5 5 = 25cm

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