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Q1. In the figure, AP and AQ are tangents to the circle with centre O, from an external point A. If PAQ = 70^o then APQ is equal to:

• 1) 110^o
• 2) 125^o
• 3) 35^o
• 4) 55^o

Solution

We know that tangents from an external point to the circle are equal. So, AP = AQ AQP = APQ In APQ, APQ + AQP + PAQ = 180^o 2APQ = 180^o - 70^o(PAQ = 70^o) 2APQ = 110^o APQ = 55^o

Q2. In fig., PT and PS are tangents to a circle from a point P such that PT = 5 cm and TPS = 60^o. Find the length of chord TS.

Solution

We know that the lengths of the tangents drawn from an external point are equal. PT = PS PTS = PST Using angle sum property, PTS + PST + 60^o = 180^o 2 PST= 120^o

Q3. In a right triangle ABC, right angled at B, BC = 15 cm, and AB = 8 cm. A circle is inscribed in triangle ABC. The radius of circle is :

• 1) 1 cm
• 2) 4 cm
• 3) 3 cm
• 4) 2 cm

Solution

Let r be the radius of the incircle. AB, BC and CA tangents to the circle at P, N and M. OP = ON = OM = r (radius of the circle) Area of ABC = By Pythagoras theorem, CA² = AB² + BC² = 8² + 15² = 289 CA = 17 cm Area of ABC = Area OAB + Area OBC + Area OCA Thus, the radius of the incircle is 3 cm.

Q4. The figure shows two concentric circles with centre O. AB and APQ are tangents to the inner circle from point A lying on the outer circle. If AB = 7.5 cm then AQ is equal to:

• 1) 18 cm
• 2) 15 cm
• 3) 10 cm
• 4) 12 cm

Solution

AB = AP……Length of tangents from an external point on a circle are equal. AP = AB = 7.5 cm Also, AQ is the chord to the outer circle. Therefore, OP is the perpendicular bisector to AQ. OP AQ AQ = 2AP = 2 X 7.5 cm = 15 cm

Q5. In fig., AB = 12 cm, BC = 8 cm and AC = 10 cm. Find AD, BE and CF.

Solution

The lengths of the tangent drawn from an external point are equal. Therefore, we have: AD = AF = x₁ BD = BE = x₂ CE = CF = x₃ AB = x₁ + x₂ = 12 cm BC = x₂ + x₃ = 8 cm CA = x₁ + x₃ = 10 cm On adding, AB + BC + CA = 2(x₁ + x₂ + x₃) = 12 + 8 + 10 x₁ + x₂ + x₃ = 15 _{Therefore, we get:} x₂ = 15 - 10 = 5 x₁ = 12 - 5 = 7 x₃ = 10 - 7 = 3 AD = x₁ = 7 cm, BE = x₂ = 5 cm, CF = x₃ = 3 cm

Q6. In fig., two concentric circles of radii a and b (a > b) are given. The chord AB of larger circle touches the smaller circle at C. The length of AB is:

• 1)
• 2)
• 3)
• 4)

Solution

_{We know that radius is perpendicular to the tangent at the point of contact. Given, OA = a and OC = b Now, In right triangle AOC, we have: AC2 = OA² - OC² = a² - b² AC = Now, we know that perpendicular drawn through the centre (O) to a chord (AB) bisects the chord.}

Q7.   ∆OPA and ∆OQA are congruent by the ____ test.

• 1) SSS
• 2) ASA
• 3) SAS
• 4) RHS

Solution

In ∆OPA and ∆OQA,  ∠APO = ∠AQO = 90° … (tangent ⊥ radius)  AO is common and is the hypotenuse.  OP = OQ … (radius)  So, ∆OPA and ∆OQA are congruent by the RHS test. 

Q8. How many tangents can be drawn from an external point to a circle?

• 1) One
• 2) Infinite
• 3) Three
• 4) Two

Solution

Two tangents can be drawn from an external point to a circle.

Q9. From the given figure, ∠TPQ and ∠TOQ are 

• 1) None of these
• 2) Supplementary angles
• 3) Complementary angles
• 4) Equal angles

Solution

∠TPQ and ∠TOQ are complementary angles as ∠OTP =∠OQP = 90°. 

Q10. The ___ at any point on the circle is perpendicular to the radius through the point of contact.

• 1) Secant
• 2) Radius
• 3) Tangent
• 4) None of these

Solution

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

Q11. The length of the tangent drawn from a point 8 cm away from the centre of a circle, of radius 6 cm, is :

• 1) 10 cm
• 2) 5 cm
• 3)
• 4)

Solution

We know that the tangent to a circle is perpendicular to its radius at the point of contact. Using Pythagoras theorem, we have: Thus, the length of the tangent is .

Q12. A circle touches all the four sides of a quadrilateral ABCD, prove that AB + CD = AD + BC

Solution

  Lengths of the tangents from an external point are equal. AP = AS, BP = BQ, CR = CQ and DR = DS AB + CD = (AP + BP) + (CR + DR) = AS + BQ + CQ + DS = (AS + DS) + (BQ + CQ) = AD + BC Hence AB + CD = AD + BC.

Q13. A tangent is a

• 1) Line which intersects exactly at one point.
• 2) Line which intersects a circle at two distinct points.
• 3) Line which does not intersect a circle.
• 4) None of these

Solution

A tangent is a line which intersects exactly at one point.

Q14. From the given figure, ∠TPQ and ∠TOQ are 

• 1) Equal angles
• 2) Complementary angles
• 3) Supplementary angles
• 4) None of these

Solution

∠TPQ and ∠TOQ are complementary angles as ∠OTP =∠OQP = 90°. 

Q15. The number of tangents which can be drawn to a circle from a point lying within is 

• 1) Infinite
• 2) Less than two
• 3) Only one
• 4) Zero

Solution

The number of tangents which can be drawn to a circle from a point lying within is zero. 

Q16. ____ drawn from the external point to a circle are equal in length. 

• 1) None of these
• 2) Tangents
• 3) Secants
• 4) Lines

Solution

Tangents drawn from the external point to a circle are equal in length. 

Q17. If PQ = 4 cm and OQ = 5 cm, then find the radius of the tangent. 

• 1) 6 cm
• 2) 8 cm
• 3) 2 cm
• 4) 3 cm

Solution

OP² + PQ² = OQ² OP² + 4² = 5² OP² = 9 OP = r = 3 cm 

Q18. Number of tangents from a point lying inside the circle is

• 1) None
• 2) One
• 3) Two
• 4) Infinitely many

Solution

All the lines through a point inside a circle intersect the circle in two points. So, it is not possible to draw any tangent to a circle through a point inside it.

Q19. O is the centre of the circle. ST and SR are tangents segments. Show that the quadrilateral STOR is cyclic.

Solution

OT TS and OR RS (Tangent at a point to a circle  to the radius at that point) OTS = 90^o and  ORS = 90^o OTS +ORS= 180^o In quadrilateral STOR, OTS +ORS + TOR + TSR = 360^o (OTS +ORS)+ (TOR + TSR) = 360^o (TOR + TSR) + 180^o = 360^o TOR + TSR = 180^o Opposite angles of the quadrilateral STOR are supplementary, hence it is cyclic.

Q20. In the given fig., angle OBC = 30^o, then value of x is:

• 1) 15^o
• 2) 100^o
• 3) 30^o
• 4) 110^o

Solution

OCB = OBC = 30^o (Since, OB = OC) We know that the radius is perpendicular to the tangent at the point of contact. ABO = 90^o Using angle sum property in triangle ABC, x + ABO + OBC + OCB = 180^o x + 90^o + 30^o + 30^o = 180^o x = 180^o - 150^o = 30^o

Q21. In fig., two circles with centres A and B touch each other externally at k. The length of PQ (in cm) is

• 1) 27 cm
• 2) 20 cm
• 3) 18 cm
• 4) 24 cm

Solution

Join points B and T, A and S. We know that radius is perpendicular to the tangent at the point of contact. ASP = BTQ = 90^o Using Pythagoras theorem in triangles ASP and BTQ, we have: AS² = (13)² - (12)² = 169 - 144 = 25 AS = 5 cm BT² = (5)² - (3)² = 25 - 9 = 16 BT = 4 cm Thus, AK = AS = 5 cm and Bk = BT = 4 cm PQ = PA + Ak + Bk + BQ = (13 + 5 + 4 + 5) cm = 27 cm

Q22. In the figure, if from an external point T, TP and TQ are two tangents to a circle with centre O so that POQ = 110^o, then, PTQ is:

• 1) 60^o
• 2) 80^o`
• 3) 70^o
• 4) 90^o

Solution

TP and TQ are tangents to the circle from a point T outside the circle. The radius drawn to these tangents will be perpendicular to the tangents. Thus, OP TP and OQ TQ OPT = 90°and OQT = 90° In quadrilateral POQT, OPT + POQ + OQT + PTQ = 360° ₍ Sum of all interior angles of a quadrilateral = 360°) 90°+ 110° + 90° + PTQ = 360 PTQ = 70°

Q23.   If ∠OMN = 20°, then find ∠MON. 

• 1) 10°
• 2) 20°
• 3) 140°
• 4) 40°

Solution

In ∆MON, MO = ON ∠ONM = ∠NMO = 20°… (angles opposite to equal sides in a triangle) ∠MON + ∠ONM + ∠NMO = 180°… (sum of angles of a triangle) ∠MON + 40° = 180°  ∠MON = 140° 

Q24. The angle between two tangents drawn from an external point to a circle is 110^o. The angle subtended at the centre by the segments joining the points of contact to the centre of circle is:

• 1) 70^o
• 2) 90^o
• 3) 110^o
• 4) 55^o

Solution

TP and TQ are tangents to the circle from a point T outside the circle. The radius drawn to these tangents will be perpendicular to the tangents. Thus, OP TP and OQ TQ OPT = 90^o and OQT = 90^o In quadrilateral POQT, OPT + POQ + OQT + PTQ = 360° (Sum of all interior angles of a quadrilateral = 360°) 90°+ POQ + 90° + 110^o = 360^o POQ = 70 ^o

Q25. A circle is inscribed in triangle ABC. If the tangents of a triangle are 4, 5, 7, respectively, then find the perimeter of the triangle. 

• 1) 4
• 2) 16
• 3) 8
• 4) 32

Solution

Consider, sides of the triangle are x, y and z. x + y + z = 2 × 4 + 2 × 5 + 2 × 7 Tangents drawn from external points to a circle are equal. x + y + z = 8 + 10 + 14 = 32 

Q26. In fig., APB is a tangent to a circle with centre O, at point P. QPB = 50^o, then the measure of POQ is:

• 1) 100^o
• 2) 120^o
• 3) 140^o
• 4) 150^o

Solution

We know that the radius is perpendicular to the tangent at the point of contact. So, OPB = 90^o OPQ = OPB - QPB = 90^o - 50^o = 40^o In OPQ, OPQ = OQP (OQ = OP, radii of same circle. And, in a triangle, angles opposite to equal sides are equal) Using angle sum property, OPQ + OQP + POQ = 180^o 40^o + 40^o + POQ = 180^o POQ = 100^o

Q27. In figure, l and m are two parallel tangents at A and B. The tangent at C makes an intercept DE between l and m. Prove that DE subtends a right angle at the centre O of the circle.

Solution

Join OC. In DAO and DCO, OA = OC = r OAD = OCD = 90^o (Radius is perpendicular to tangent at the point of contact) OD = OD (common) DAO DCO (RHS congruency criterion) We know corresponding parts of congruent triangles are equal. 1 = 2 ... (i) Similarly OBE OCE Then, 3 = 4 ... (ii) As AOB is a line, 1 + 2 + 3 + 4 = 180^o 22 + 23 = 180^o 2 + 3 = 90^o, i.e., DOE = 90^o

Q28. In fig., PQR is the tangent to a circle at Q whose centre is O. AB is a chord parallel to PR and angle BQR = 70^o, then angle AQB is equal to

• 1) 35^o
• 2) 20^o
• 3) 40^o
• 4) 45^o

Solution

Since AB || PR and OQ PR (Radius is perpendicular to tangent), OL AB. OL bisects chord AB. Thus, QAB is isosceles. LQB = LQA But LQB = 90^o - 70^o = 20^o LQA = 20^o AQB = 2 (20^o) = 40^o

Q29. Find the semi-perimeter of a triangle.  

• 1) 72 cm
• 2) 9 cm
• 3) 36 cm
• 4) 18 cm

Solution

PC = PA = 6 cm, QA = QB = 5 cm, RC = RB = 7 cmSemi-perimeter of a triangle =  = 18 cm

Q30. In fig., if PT is a tangent of the circle with centre O and TPO = 25^o, then the measure of x is :

• 1) 125^o
• 2) 120^o
• 3) 115^o
• 4) 110^o

Solution

We know that the radius is perpendicular to the tangent at the point of contact. So, OTP = 90^o We also know that the measure of an exterior angle of a triangle is equal to the sum of its two interior opposite angles. x = OTP + TPO = 90^o + 25^o = 115^o

Q31. If two parallel lines are intersected by a transversal such that the pair of ______ is not equal. 

• 1) Alternate angles
• 2) Interior angles
• 3) Alternate interior angles
• 4) Corresponding angles

Solution

If two parallel lines are intersected by a transversal such that the pair of interior angle is not equal. 

Q32. In fig., PA is a tangent to a circle of radius 6 cm and PA = 8 cm, then length of PB is

• 1) 18 cm
• 2) 10 cm
• 3) 12 cm
• 4) 16 cm

Solution

PB=PO+OB If we join OA, PO²= PA²+OA²      = 64 + 36 = 100 PO = 10 cm PB = 10+6      = 16 cm

Q33. In the figure, O is the centre of a circle and BCD is tangent to it at C. Prove that BAC + ACD = 90^o.

Solution

OA = OC (radii of same circle) OAC = OCA ...(1) Since the tangents at any point of a circle is perpendicular to radius at the point of contact. OCD = 90^o ACD + OCA = 90^o ACD + OAC = 90^o [From (1)] ACD + BAC = 90^o Hence proved.

Q34. In fig., triangle ABC is a right angled triangle with AB = 6 cm, AC = 8 cm and A = 90^o. A circle with centre O is inscribed inside the triangle. Find the radius 'r'.

Solution

ABC is right angled at A. AB² + AC² = BC² 6² + 8² = 36 + 64 = 100 = 10² BC = 10 cm OP = OQ = OR = r, radius of the circle. OP AB, OQ BC. (Radius is to the tangent) In OPAQ, A, P, Q are 90^o each then the 4^th angle will also be 90^o. So, OPAQ is square (adjacent sides are equal and interior angles are 90^o). Since, OP = r, PA = AQ = r (sides of the square) Tangents from an external point are equal, so, AP = AQ, BP = BR, CQ = CR….(i) If AP = AQ = r then BP = 6 - r, QC = 8 - r, So, BR = 6 - r and CR = 8 - r from(i) BC = 10 = (6 - r) + (8 - r) 10 = 14 - 2r -4 = -2r r = 2

Q35. In fig., if OC = 9 cm and OB = 15 cm, then BC + BD is equal to:

• 1) 18 cm
• 2) 24 cm
• 3) 36 cm
• 4) 12 cm

Solution

We know that the radius is perpendicular to the tangent at the point of contact. So, OCB = 90^o Using Pythagoras theorem in OCB, BC² = (15)² - (9)² = 225 - 81 = 144 BC = 12 cm We also know that the lengths of tangents drawn from an external point are equal. So, BC = BD BC + BD = 2 BC = 2 12 cm = 24 cm

Q36. The number of tangents which can be drawn to a circle from a point lying outside it is 

• 1) Two
• 2) Infinite
• 3) Less than two
• 4) Only one

Solution

The number of tangents which can be drawn to a circle from a point lying outside it is two. 

Q37. A secant intersects a circle at _ points. 

• 1) 1
• 2) more than 1
• 3) 2
• 4) many

Solution

A secant intersects a circle at 2 points.

Q38. TA is a tangent to the circle, centre O, at A. OA = 20 cm, TA = 21 cm. and OTA = 44^o Find OT and x

Solution

OA is the radius and TA is the tangent. OAT = 90^o        In right triangle, OAT OT² = OA² + AT² 20² + 21² = 841 Hence OT = 29 cm. In triangle OAT, AOT = 180^o – (44^o + 90^o) (By angle sum) = 46^o COB = 180^o – 46^o = 134^o In triangle COB, CO = OB  x + x + 134^o = 180^o (By angle sum) x =  = 23^o OT = 29 cm and   x = 23^o.

Q39. Three circles with centres at P, Q and R touch each other externally such that PQ = 6 cm, PR = 8 cm and QR = 12 cm. Find the greatest radius. 

• 1) 7 cm
• 2) 5 cm
• 3) 12 cm
• 4) 13 cm

Solution

Let a, b and c be the radii of the circles, respectively. a + b = 6, b + c = 8 and a + c = 12 … (i)  2(a + b + c) = 6 + 8 + 12 a + b + c = 13 … (ii) Solving (i) and (ii), We get a = 5, b = 1 and c = 7  The greatest radius is (a + c) = 7 cm. 

Q40. In the 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 AOB = 90^o.

Solution

OP xy (tangent radius) OC AB In OPA and OCA OPA = OCA = 90^o OP = OC (radii) AP = AC (tangents from an external point) OPA OCA (SAS) 1 = 2 (CPCT) 2 = PAC Similarly, 3 = 4 3 = QBC xy ||x'y' and AB is transversal PAB + QBA = 180^o (interior angles on same side of transversal) Or PAC = QBC = 180^o 2 + 3 = 90^o In OAB, AOB + 2 + 3 = 180^o AOB = 90^o Hence proved.

Q41. ∠POQ = 65°, then find ∠TPO.  

• 1) 100°
• 2) 50°
• 3) 25°
• 4) 130°

Solution

∠POQ = ∠POT = 65°∠TOQ = ∠POQ +∠POT ∠TOQ = 130°  ∠TPQ and ∠TOQ are supplementary. ∠TPQ = 180° − 130°  ∠TPQ = 50°  ×∠TPQ = 25°  ∠TPO = ∠QPO = 25° 

Q42. OA = 17 cm and AB = 8 cm. Find OB.  

• 1) 7 cm
• 2) 15 cm
• 3) 23 cm
• 4) None of these

Solution

In ∆OBA, OA² = OB² + AB² 17² = OB² + 8² OB² = 225 OB = 15 cm OB² = 225 - 64 OB² = 225 OB = 15 cm 

Q43. The length of tangent drawn from an external point P to a circle, with centre O, is 8 cm. If the radius of the circle is 6 cm, then the length of OP (in cm) is:

• 1)
• 2) 10.5 
• 3)
• 4) 10 

Solution

Given, AP = 8 cm and OA = 6 cm We know that radius is perpendicular to the tangent at the point of contact. _{Using Pythagoras theorem in OAP,} OP² = OA² + AP² = 36 + 64 = 100 OP = 10 cm

Q44. By which of the following properties is ∆OPR congruent to ∆OQR?  

• 1) SSS
• 2) ASA
• 3) SAS
• 4) AAA

Solution

By the SSS property of congruence, ∆OQR.

Q45. A circle can have ____ number of tangent/s.

• 1) One
• 2) Two
• 3) More than two
• 4) Infinite

Solution

A circle can have infinite number of tangents.

Q46. The maximum number of common tangents that can be drawn to two circles intersecting at two distinct point is

• 1) 1
• 2) 3
• 3) 2
• 4) 4

Solution

There can be maximum 2 such tangents.

Q47. In the figure, OP is equal to the diameter of the circle. Prove that ABP is an equilateral triangle.

Solution

Say,Radius (OA) = r OP = 2r………given,OP= diameter of the circle OAP = 90^o (Tangent is to the radius through the point of contact) In right OAP sin (OPA) = OPA = 30^o Similarly OPB = 30^o APB = 30^o + 30^o = 60^o Since PA = PB (Lengths of tangents from an external point are equal) PAB = PBA In APB, APB + PAB + PBA = 180^o (angle sum property) 60^o + 2PAB = 180^o PAB = 60^o PBA = 60^o Since all angles are 60^o, ABP is equilateral triangle.

Q48.

Solution

Q49. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

Solution

Let NM be a chord of a circle with centre C. Let tangents at M and N meet at the point O. Since OM is a tangent, OM CM, i.e., OMC = 90° Since ON is a tangent, ON CN, i.e., ONC = 90° In DCMN, CM = CN (Radius of the same circle) CMN = CNM Now, OMC = ONC OMC - CMN = ONC - CNM OML = ONL Thus, tangents make equal angles with the chord.

Q50. A circle is inscribed in triangle ABC. If the tangents of a triangle are 4, 5, 7, respectively, then find the perimeter of the triangle. 

• 1) 8
• 2) 32
• 3) 16
• 4) 4

Solution

Consider, sides of the triangle are x, y and z. x + y + z = 2 × 4 + 2 × 5 + 2 × 7 Tangents drawn from external points to a circle are equal. x + y + z = 8 + 10 + 14 = 32 

Q51. A tangent PA is drawn from an external point P to a circle of radius 3 cm such that the distance of the point P from O is 6 cm as shown figure. The value of APO is

• 1) 60^o
• 2) 30^o
• 3) 75^o
• 4) 45^o

Solution

Q52. Prove that tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution

Given: A circle C (O, r) and a tangent AB at a point P. To Prove: OP is perpendicular to AB. Construction: Take any point Q, other than P, on the tangent AB. Join OQ. Since, Q is a point on the tangent AB, other than the point of contact P, so Q will be outside the circle. Let OQ intersect the circle at R. Then, OQ=OR+RQ OQ>OR OQ>OP (OR=OP=radius) Thus, OP<OQ, i.e., OP is shorter than any other segment joining O to any point of AB. But, among all the line segments, joining the point O to a point on AB, the shortest one is the perpendicular from O on AB. Hence, OP is perpendicular to AB.

Q53. From the figure, ∠P + ∠O is  

• 1) 90°
• 2) 360°
• 3) None of these
• 4) 180°

Solution

∠OTP = ∠OQP = 90°… (tangent ⊥ radius) Hence, ∠P + ∠O = 180° 

Q54. From a point Q, the length of the tangent to a circle is 13 cm and the distance of Q from the centre is 12 cm. So, the radius of the circle is 

• 1) 7.5 cm
• 2) 3 cm
• 3) 6 cm
• 4) 5 cm

Solution

Using the Pythagoras theorem, √(13² - 12²) = √(169 - 144) = √25 Radius = 5 cm 

Q55. By which of the following properties are triangles ∆OPB and ∆OPA congruent?  

• 1) SAS
• 2) SSS
• 3) Hypotenuse side property
• 4) All of the above

Solution

In ∆OPB and ∆OPA, OB = OA radii AP = PB OP is perpendicular bisector of AB OP = OP common∆OPB and ∆OPA are congruent by the SSS test. ? ∆OPB and ∆OPA are congruent by the SAS test. OP = OP  ∵ common side PB = PA  ∵ Radius-tangent property∠OPB =∠OPA OP perpendicular to AB∆OPB ≅∆OPA  ∵ SAS Also, OP = OP  ∵ common side OA = OB  ∵ radii of the same circle∆OPB ≅∆OPA Hypotenuse-side property

Q56.

Solution

Q57. In fig., AP = 2 cm, BQ = 3 cm and RC = 4 cm, then the perimeter of ABC (in cm) is

• 1) 20
• 2) 16
• 3) 18
• 4) 21

Solution

We know that the lengths of tangents drawn from an external point to the circle are the same. Therefore, AR = AP = 2 cm BP = BQ = 3 cm CQ = CR = 4 cm Perimeter of ABC = AB + BC + CA = (AP + PB) + (BQ + CQ) + (AR + CR) = (2 + 3 + 3 + 4 + 4 + 2) cm = 18 cm

Q58. In the given figure, O is the centre of the circle. If PA and PB are tangents from an external point P to the circle, then AQB is equal to

• 1) 80^o
• 2) 100^o
• 3) 70^o
• 4) 50^o

Solution

 

Q59. In fig., triangle ABC is circumscribing a circle. Then the length of BC is:

• 1) 7 cm
• 2) 8 cm
• 3) 10 cm
• 4) 9 cm

Solution

We know that the lengths of tangents drawn from an external point are equal. BL = BN = 4 cm and CL = CM = 6 cm BC = BL + LC = 4 cm + 6 cm = 10 cm

Q60. The radius of a circle is ____ to the tangent. 

• 1) Perpendicular
• 2) Both A and B
• 3) Coincides
• 4) Parallel

Solution

The radius of a circle is perpendicular to the tangent.

Q61. Which of the following is correct? 

• 1) A line intersecting a circle at one point is called the Tangent of a Circle.
• 2) If a line and a circle do not have a point in common, then the line is outside the circle.
• 3) A line intersecting a circle at two points is called the secant of a circle.
• 4) All of these

Solution

A,B and C are correct.

Q62. In the following fig., triangle ABC is isosceles in which AB = AC, circumscribed about a circle. Prove that base is bisected by the point of contact.

Solution

Given: AB = AC To prove: BE = CE _{Proof: We know that the lengths of the tangents drawn from an external point to the circle are equal. Therefore,} AD = AF, BD = BE CE = CF _{We have: AB = AC (given)} AB - AD = AC - AD AB - AD = AC - AF BD = CF BE = EC (Since, BD = BE and CE = CF)

Q63. In the figure, the pair of tangents AP and AQ, drawn from an external point A to a circle with centre O, are perpendicular to each other and length of each tangent is 4 cm, then the radius of the circle is

• 1) 4 cm
• 2) 10 cm
• 3) 7.5 cm
• 4) 2.5 cm

Solution

We know that radius is perpendicular to tangent at the point of contact. OPA= OQA = 90^o Clearly, it can be observed that OPAQ is a square. All the sides of a square are equal. Radius of the circle = OP = PA = 4 cm

Q64. If two parallel tangents PQ and RS touch the circle C at A and B, respectively, then AB is the ___ of the circle.

• 1) Chord
• 2) Secant
• 3) Radius
• 4) Diameter

Solution

Two parallel tangents PQ and RS touch the circle C at A and B, respectively, then AB is the diameter of the circle. 

Q65. Prove that the lengths of tangents drawn from an external point to a circle are equal.

Solution

Consider a circle with centre O. Let PT and PQ be the two tangents drawn from the point P to the circle. We have to prove that PT = PQ. Join OP, OT and OQ. In ΔOTP and ΔOQP, OT = OQ (Radii of the same circle) OTP = OQP = 90° (The tangent is perpendicular to the radius through the point of contact) OP = OP (Common) (By R.H.S. congruence criterion) Since, corresponding parts of congruent triangles are equal, PT = PQ. Hence, proved.

Q66. In the figure if ATB = 40^o, find AOB.

Solution

  Here quadrilateral AOBT is a cyclic quadrilateral and OAAT and OBBT. OAT = OBT = 90^o  ATB + AOB = 180^o  Also given is ATB = 40^o AOB.= 140^o

Q67. In two concentric circles, prove that all chords of the outer circle that touch the inner circle are of equal length

Solution

    Let PQ and RS be the chords of the circle that touch the inner circle at M and N respectively. PQ and RS are the tangents to the inner circle, and OM and ON are the radii of the smaller circle. OM = ON Thus PQ and RS are equidistant from the centre, therefore they are equal. Hence PQ = RS.

Q68. In the following fig., from an external point P, PA and PB are tangents to the circle with centre O. If CD is another tangent at point E to the circle and PA = 12 cm. Find the perimeter of PCD.

Solution

Tangents drawn from an external point to the circle are equal. PA = PB = 12 cm Also, Perimeter of PCD = PC + CD + PD = PC + CE + ED + PD = PC + AC + BD + PD = PA + PB = 24 cm

Q69. Two concentric circles have centre at O, OP = 4 cm and OB = 5 cm. AB is a chord of the other circle and a tangent to the inner circle at P. Find the length of AB.

Solution

OP = 4 cm, OB = 5 cm We know that the radius is perpendicular to the tangent at the point of contact. In right triangle OPB, OB² = OP² + PB² (5)² = (4)² + PB² PB² = 25 - 16 = 9 PB = 3 cm We know that perpendicular from the centre to the chord bisect the chord. AB = 2PB = 6 cm

Q70. In two concentric circles the radius of inner is 5 cm a chord of length 24 m of outer circle becomes a tangent to the inner circle. Find the radius of the larger circle.

Solution

Let O be the centre of circle and AB be the chord OT is radius of smaller circle So OTAB since tangent is to radius at its point of contact. AT = TB = 12 cm (since perpendicular from centre to the chord bisects it) So, In triangle OAT, OA² = OT² + AT² OA² = 5² + 12² So, OA = 13 cm Thus, the radius of the larger circle is 13 cm.  

Q71. Prove that tangents drawn at the end-points of a diameter of a circle are parallel.

Solution

To prove: XY || AB We know that radius is perpendicular to the tangent at the point of contact. 1 = 90^o, 2 = 90^o These are alternate angles made by the transversal PQ on two lines XY and AB. XY || AB

Q72. Quadrilateral ABCD circumscribes a circle as shown in figure. The side of the quadrilateral which is equal to AP + BR is:

• 1) AC
• 2) AD
• 3) AB
• 4) BC

Solution

In the quadrilateral ABCD: AP + BR = AQ + BQ (AQ = AP and BQ = BR…lengths of tangents drawn from an external point to a circle are equal.) =AB

Q73.

Solution

Q74. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that PTQ = 2OPQ.

Solution

We know that tangent to a circle is perpendicular to its radius at the point of contact. _{So, OPT = OQT = 90}^o In quadrilateral POQT, we have: OPT + PTQ + OQT + POQ = 360^o PTQ = 180^o - POQ = 1 + 2 (Using angle sum property in POQ) = 21 (1 = 2 as OP = OQ) = 2OPQ

Q75. In the diagram, the circle touches AB produced at P, AC produced at R and BC produced at Q. AB = 23 cm, BC = 17 cm. and AC = 20 cm. Find the lengths Of AP and CQ

Solution

  Tangents from an external point are equal. AP = AR,      BP = BQ,         CQ = CR BP = BQ = BC – CQ or BC – CR or 17 – CR. Hence BP = 17 – CR. Now AP = AR Or   AB + BP = AC + CR Or   23 + 17 – CR = 20 + CR Or    2 CR = 40 – 20 CR = 10 cm. Now BP = 17 – 10 = 7 cm, and AP = AB + BP  23 + 7 = 30 cm. Hence AP = 30 cm and CR = CQ = 10 cm.

Q76. Which of the following is incorrect? 

• 1) RC = PB
• 2) AP = AS
• 3) CQ = CR
• 4) DR = DS

Solution

RC and PB are not drawn from a common external point. 

Q77. ____ drawn from the external point to a circle are equal in length.

• 1) Secants
• 2) None of these
• 3) Tangents
• 4) Lines

Solution

Tangents drawn from the external point to a circle are equal in length. 

Q78. Prove that the parallelogram circumscribing a circle is a rhombus.

Solution

Since ABCD is a parallelogram, AB = CD (i) BC = AD (ii) Now, we may observe that DR = DS (tangents on circle from point D) CR = CQ (tangents on circle from point C) BP = BQ (tangents on circle from point B) AP = AS (tangents on circle from point A) Adding all these equations, we get, DR + CR + BP + AP = DS + CQ + BQ + AS (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ) CD + AB = AD + BC From equation (i) and (ii), we get, 2AB = 2BC AB = BC Or, AB = BC = CD = DA Hence, ABCD is a rhombus.

Q79.

Solution

Q80. Two concentric circles have centre O. A chord PQ of the larger circle is bisected by the smaller circle at point R, so PQ is ____ to the smaller circle. 

• 1) Secant
• 2) Tangent
• 3) Diameter
• 4) Radius

Solution

Two concentric circles have centre O. A chord PQ of the larger circle is bisected by the smaller circle at point R, so PQ is tangent to the smaller circle. 

Q81. Which of the following is correct from the following figure?  

• 1) OQ = OR + RQ
• 2) All of the above
• 3) OP < OQ
• 4) OP = OR

Solution

From the figure, A, B and C are correct. 

Q82. In the following fig., triangle ABC is a right angled triangle with AB = 6 cm, AC = 8 cm and A = 90^o. A circle with centre O is inscribed inside the triangle. Find the radius 'r'.

Solution

ABC is right angled at A. AB² + AC² = BC² 6² + 8² = 36 + 64 = 100 = 10² BC = 10 cm OP = OQ = OR = r, radius of the circle. OP AB, OQ BC. (Radius is to the tangent) In OPAQ, A, P, Q are 90^o each then the 4^th angle will also be 90^o. So, OPAQ is square (adjacent sides are equal and interior angles are 90^o). Since, OP = r, PA = AQ = r (sides of the square) Tangents from an external point are equal, so, AP = AQ, BP = BR, CQ = CR….(i) If AP = AQ = r then BP = 6 - r, QC = 8 - r, So, BR = 6 - r and CR = 8 - r from(i) BC = 10 = (6 - r) + (8 - r) 10 = 14 - 2r -4 = -2r r = 2

Q83. In fig., ABC is a triangle right angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and L is the centre of incircle of ABC.

Solution

Given, A = 90^o, AB = 6 cm and BC = 10 cm Area of ABC = Let r be the radius of incircle. Join LA, LB and LC. We know that radius is perpendicular to the tangent at the point of contact. Therefore, Area of ABC = Area of ALB + Area of BLC + Area of ALC Thus, the radius of the incircle is 2 cm. Area of shaded region = area of triangle ABC - area of circle

Q84. In the given figure, the diameter AB of the circle with centre O is extended to a point P and PQ is a tangent to the circle at the point T. If BPT = x and ATP = y, then prove that x + 2y = 90^o.

Solution

Join OT. It is known that the tangent at any point of a circle is perpendicular to the radius through the point of contact. OTP = 90^o In a triangle, the measure of an exterior angle is equal to the sum of the measures of its interior opposite angles. Therefore, in PAT, OAT = APT + ATP OAT = x + y In OAT, OA = OT (Radii of the same circle) OAT = OTA x + y = OTP - ATP x + y = 90^o - y x + 2y = 90^o

Q85. In fig., O is the centre of two concentric circles of radii 6 cm and 4 cm. PQ and PR are tangents to the two circles from an external point P. If PQ = 10 cm, find the length of PR.

Solution

We know that the radius is perpendicular to the tangent at the point of contact. PQO = PRO = 90^o In POQ, Given PQ = 10 cm, OQ = 6 cm OP = In POR, OP² = 4² + PR² 136 - 16 = PR² PR² = 120 PR =

Q86. In the figure, AB and CD are two parallel tangents to a circle with centre O. ST is the tangent segment between two parallel tangents to a circle with centre O. ST is the tangent segment between two parallel tangents touching the circle at Q. Show that SOT = 90^o.

Solution

Join OP and OQ. In OPS and OQS: OP=OQ……Radius of circle OPS=OQS=90⁰ (Tangent is to the radius through the point of contact) OS=OS….common Therefore, by Right angle-Hypotenuse-Side criterion of congruence, we have OPS OQS (RHS) The corresponding parts of the congruent triangles are congruent. POS = QOS……cpct, OSP = OSQ ……..cpct In quadrilateralPSQO: _POQ+PSQ=180⁰ as the total sum of interior angles in a quadrilateral=360⁰ _POQ=PSQ=90⁰ So OSP = OSQ = 45^o_{As PSQ=90}⁰ _STC=90⁰ sum of two co interior angles is 180 ^o. SOT is an isosceles triangle having two sides: OS=OT Similarly, QOT = 45^o SOT = 90^o

Q87. Prove that the line segment joining the points of contact of two parallel tangents to circle is a diameter of the circle.

Solution

Construction: Through O draw OR || BA or OR || CD as AB and CD are parallel tangents. Proof: OPA = 90^o (radius is always perpendicular to tangent) Since OR || BA (By construction) OPA + POR = 180^o POR = 180^o - 90^o = 90^O Similarly QOR = 90^o POR + QOR = 180^o PQ is straight line through O. So PQ is diameter.

Q88. In fig., tangent segments PS and PT are drawn to a circle with centre O such that SPT = 120^o. Prove that OP = 2 PS.

Solution

Tangents drawn from an external point to the circle are equal. _{PS = PT Radius is perpendicular to the tangent at the point of contact. In OST and OTP, PS = PT OS = OT (Radii of same circle)} OPS = OPT SPT = 120^o OPS = 60^o In OSP, OP = 2 PS

Q89. In fig., a quadrilateral ABCD circumscribes a circle, Prove that AB + DC = AD + BC.

Solution

We know that the lengths of the tangents drawn from an external point are equal. Therefore, we have: AP = AS, BP = BQ, CR = QC, DR = DS _{Adding the above equations, we have:} AP + PB + RC + DR = AS + BQ + QC + DS AB + DC = AD + BC

Q90. AB and AC are tangents from A to the circle with center O. M is a point on the circle. Prove that, AP + PM = AQ + QM.

Solution

Lengths of the tangents from an external point to a circle are equal. AB = AC, PB = PM, QC = QM AB = AC or AP + PB = AQ + QC AP + PM = AQ + QM    (PB = PM, QC = QM proved above)