RS Aggarwal (2015) Solutions for Class 10 Math Chapter 12 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among class 10 students for Math Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal (2015) Book of class 10 Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal (2015) Solutions. All RS Aggarwal (2015) Solutions for class 10 Math are prepared by experts and are 100% accurate.

Page No 529:

Question 1:

Find the length of the tangent drawn to a circle with radius 8 cm from a point 17 cm away from the centre of the circle.

Answer:


Let O be the centre of the given circle.Let P be a point, such that OP=17 cm.Let OT be the radius, where OT=5 cmJoin TP,where TP is a tangent.Now,tangent drawn from an external point is perpendicular to the radius at the point of contact.OTPTIn the right OTP, we have:OP2=OT2+TP2   [By Pythagoras' theorem:]TP=OP2OT2=17282=28964=225=15 cm.The length of the tangent is 15 cm.

Page No 529:

Question 2:

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

Answer:


Draw a circle and let P be a point such that OP=25 cm.Let TP be the tangent, so that TP=24 cmJoin OT,where OT is radius.Now, tangent drawn from an external point is perpendicular to the radiusat the point of contact.OTPTIn the right OTP, we have:OP2=OT2+TP2   [By Pythagoras' theorem:]OT2=OP2TP2=252242=625576=49=7 cmThe length of the radius is 7 cm.

Page No 529:

Question 3:

In the given figure, PA and PB are the tangent segments to a circle with centre O. Show that the points A, O, B and P are concyclic.

Answer:

Given, PA and PB are tangents to the circle with centre O and OA is a radius.Now, tangents drawn from an external point are perpendicular to the radiusat the point of contact.OAPA and OBPB.OAP=900 and OBP=900OAP+OBP=900+900=1800Thus, the sum of a pair of opposite angles of quadrilateral AOBP is 1800.Now, AOB+APB+PAO+PBO=3600 (Sum of all the angles of a quadrilateral is 3600)=>AOB+APB=1800   (OAP+OBP=1800)AOBP is a cyclic quadrilateral.

Page No 529:

Question 4:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.

Answer:


Given, PA and PB are the tangents to a circle with centre O and CD is a tangent at E and PA=14 cm.Tangents drawn from an external point are equal.PA=PB,CA=CE and DB=DEPerimeter of PCD=PC+CD+PD=(PACA)+(CE+DE)+(PBDB)=(PACE)+(CE+DE)+(PBDE)=(PA+PB)=2PA   (PA=PB)=(2×14) cm=28 cmPerimeter of PCD=28 cm.

Page No 529:

Question 5:

A circle is inscribed in ΔABC, touching AB, BC and AC at P, Q and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.

Answer:

Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P,Q,R.Tangents drawn to a circle from an external point are equal.AP=AR=7 cm, CQ=CR=5 cm.Now, BP=(ABAP)=(107)=3 cmBP=BQ=3 cmBC=(BQ+QC)=>BC=3+5=>BC=8The length of BC is 8 cm.

Page No 529:

Question 6:

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Answer:


Let the circle touch the sides of the quadrilateral  AB,BC,CD and DA at P,Q,R and  S respectively.Given, AB=6 cm, BC=7 cm and CD=4 cm.Tangents drawn from an external point are equal.AP=AS,BP=BQ,CR=CQ and DR=DSNow, AB+CD=(AP+BP)+(CR+DR)=>AB+CD=(AS+BQ)+(CQ+DS)=>AB+CD=(AS+DS)+(BQ+CQ)=>AB+CD=AD+BC=>AD=(AB+CD)BC=>AD=(6+4)7=>AD=3 cm.The length of AD is 3 cm.

Page No 529:

Question 7:

Two tangent segments BC and BD are drawn to a circle with centre O, such that ∠CBD = 120°. Prove that OB = 2BC.

Answer:

Given, BC and BD are the two tangents drawn to the circle with centre O, such that CBD=1200.They are equally inclined to the line segment joining the centre to that point.OBC=OBD=12CBD=600Now, tangent drawn from an external point is perpendicular to the radiusat the point of contact.OCB=ODB=900From ΔOBC, BOC=1800(OCB+OBC)BOC=1800(900+600)BOC=300.Now from right-angled OBC, BCOB=sin300BCOB=12OB=2BC.Hence proved.



Page No 530:

Question 8:

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.

Answer:

Given, O is the centre of two concentric circles of radii OA=6 cm and OB=4 cm.PA and PB are the two tangents to the outer and inner circles respectively and PA=10 cm.Now, tangent drawn from an external point is perpendicular to the radiusat the point of contact.OAP=OBP=900From right-angled OAP, OP2=OA2+PA2=>OP=OA2+PA2=>OP=62+102=>OP=136 cm.From right-angled OAP, OP2=OB2+PB2=>PB=OP2OB2=>PB=13616=>PB=120 cm=>PB=10.9 cm.The length of PB is 10.9 cm.

Page No 530:

Question 9:

In the given figure, ABCD is a quadrilateral in which ∠D = 90°. A circle C(O, r) touches the sides AB, BC, CD and DA at P, Q, R, S, respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, find the value of r.

Answer:


Given, ABCD is a quadrilateral in which D=900. A circle C(O,r) touches the sidesAB,BC,CD and DA at P,Q,R,S respectively and BC=38 cm, CD=25 cm and BP=27 cm.Now,OR=OS(Radii of the same circle)Tangents drawn from an external point are perpendicular to the radiusat the point of contact.ORDR and OSDS.ORDS is a square.Tangents drawn from an external point are equal.BP=BQ,CQ=CR and DR=DSBQ=BP=27 cm=>BCCQ=27=>38CQ=27=>CQ=3827=>CQ=11 cmCR=11 cm=>CDDR=11 cm=>25DR=11=>DR=2511=14 cmNow, DR=DS=OR=OS (Sides of a square are equal)OR=OS=14 cmr=14 cmThe radius is 14 cm.



Page No 531:

Question 1:

In the given figure, PT is a tangent to the circle with centre O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is


(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm

Answer:

(a) 8 cm

PT  is a tangent to the circle with center O. Given that OT=6 cm and OP=10 cm.The tangent at any point of a circle is perpendicular to the radius at the point of contact.OTPTIn right-angled triangle PTO,PT2+OT2=OP2[Using Pythagoras' theorem]PT2=OP2-OT2PT2=100-36PT2=64PT=64PT=8 cm

Page No 531:

Question 2:

In a circle of radius 7 cm, tangent PT is drawn from a point P, such that PT = 24 cm. If O is the centre of the circle, then OP = ?


(a) 30 cm
(b) 28 cm
(c) 25 cm
(d) 18 cm

Answer:

(c) 25 cm
The tangent at any point of a circle is perpendicular to the radius at the point of contact.
OTPTFrom right-angled triangle PTO,OP2=OT2+PT2[Using Pythagoras' theorem]OP2=72+242OP2=49+576OP2=625OP=625OP=25 cm



Page No 532:

Question 3:

A point P is 26 cm away from the centre of a circle and the length of the tangent drawn from P to the circle is 24 cm. The radius of the circle is


(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 14 cm

Answer:

(b) 10 cm
Given, point P is 26 cm away from the centre of a circle and the length of the tangent drawn from P to the circle is 24 cm. The centre of the circle is O.
The tangent at any point of a circle is perpendicular to the radius at the point of contact.
OTPTOT2=OP2-PT2OT2=262-242OT2=676-576OT2=100OT=100OT=10 cmRadius of the circle is 10 cm.

Page No 532:

Question 4:

If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is


(a) 3 cm
(b) 6 cm
(c) 332 cm
(d) 33 cm

Answer:

(d) 33 cm

Given, PA and PB are tangents to circle with centre O and radius 3 cm and APB=600.Tangents drawn from an external point are equal; so, PA=PB.And OP is the bisector of APB, which gives OPB=OPA=300.OAPA. So, from right-angled OPA, we have: OAAP=tan300OAAP=133AP=13AP=33 cm

Page No 532:

Question 5:

If PA and PB are two tangents to a circle with centre O, such that ∠AOB = 110°, find ∠APB.


(a) 90°
(b) 60°
(c) 70°
(d) 55°

Answer:

(c) 70°
Given, PA and PB are tangents to a circle with centre O, with AOB=1100.Now, we know that tangents drawn from an external point are perpendicularto the radius at the point of contact.So, OAP=900 and OBP=900.OAP+OBP=900+900=1800, which shows that OABP is a cyclicquadrilateral.AOB+APB=18001100+APB=1800APB=1800-1100APB=700

Page No 532:

Question 6:

If PA and PB are two tangents to a circle with centre O, such that ∠APB = 80°, then ∠AOP = ?


(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer:

(b) 50°
Given, PA and PB are two tangents to a circle with centre O and APB=800.APO=12APB=400Since they are equally inclined to the line segment joining the centre to that pointand OAP=900 Since tangents drawn from an external point are perpendicular to the radius at the point of contactNow, in triangle AOP: AOP+OAP+APO=1800AOP+900+400=1800AOP=1800-1300AOP=500

Page No 532:

Question 7:

In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR, such that ∠BQR = 70°. Then, AQB = ?


(a) 20°
(b) 35°
(c) 40°
(d) 45°

Answer:

(c) 40°

Since, ABPR, BQ is transversal.BQR=ABQ=700 Alternate anglesOQPQR(Tangents drawn from an external point are perpendicular  to the radius at the point of contact) and ABPQRQLAB; so, OLABOL bisects chord AB Perpendicular drawn from the centre bisects the chordFrom QLA and QLB:QLA=QLB=900LA=LB(OL bisects chord AB )QL is the common side.QLAQLB By SAS congruencyQAL=QBLQAB=QBAAQB is isosceles.LQA=LQRLQP=LQR=900LQB=900-700=200LQA=LQB=200AQB=LQA+LQB=400

Page No 532:

Question 8:

In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord, such that ∠QPT = 50°, then ∠PQT = ?


(a) 100°
(b) 90°
(c) 80°
(d) 75°

Answer:

(a) 100°Given, QPT=500and OPT=900(Tangents drawn from an external point are perpendicularto the radius at the point of contact)OPQ=OPT-QPT=900-500=400OP=OQ Radius of the same circleOQP=OPQ=400In POQ, POQ+OQP+OPQ=1800POQ=1800-400+400=1000

Page No 532:

Question 9:

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent at P, such that ∠QPT = 60°. Then, ∠PRQ = ?


(a) 135°
(b) 150°
(c) 120°
(d) 110°

Answer:


(c) 120°Angle between a chord and a tangent is equal to the angle formed in the corresponding alternate segment.QPA=PRQ QPA+QPT=1800 linear pairBut QPA=1800-600=1200PRQ=1200 



Page No 533:

Question 10:

If the angle between two radii of a circle is 130°, the angle between the tangents at the ends of the radii is


(a) 65°
(b) 50°
(c) 40°
(d) 70°

Answer:

(b) 50°

OA and OB are the two radii of a circle with centre O. Also AT and BT are the tangents to thecircle. Then,TAO=TBO=900TAO+TBO=1800AOB+ATB=18001300+ATB=1800ATB=1800-1300ATB=500

Page No 533:

Question 11:

PA and PB are tangents to the circle with centre O, such that ∠APB = 50°. Then, ∠OAB = ?


(a) 25°
(b) 30°
(c) 40°
(d) 50°

Answer:

(a) 25°In quadrilateral OAPB, we have:OAP=OBP=900(Tangents drawn from an external point are perpendicular to the radius at the point of contact)OAP+OBP=1800AOB+APB=1800 Sum of the angles of a quadrilateral is 3600AOB+500=1800AOB=1800-500AOB=1300In AOB, OAB+OBA=1800-AOB=1800-1300=500Now OA=OB (Radii of the same circle)OAB=OBA=250Hence, OAB=250

Page No 533:

Question 12:

O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quadrilateral PQOR is


(a) 60 cm2
(b) 32.5 cm2
(c) 65 cm2
(d) 30 cm2

Answer:

(a) 60 cm2

Given, OQ=OR=5 cm, OP=13 cm.OQP=ORP=900(Tangents drawn from an external point are perpendicular to the radius at the point of contact)From right-angled POQ:PQ2=OP2-OQ2PQ2= 132-52PQ2=169-25 PQ2=144PQ=144PQ=12 cmarOQP=12×PQ×OQarOQP=12×12×5cm2arOQP=30 cm2Similarly, arORP=30 cm2arquad.PQOR=30+30 cm2=60 cm2

Page No 533:

Question 13:

In the given figure, O is the centre of a circle. AOC is its diameter, such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT = ?


(a) 40°
(b) 50°
(c) 60°
(d) 65°

Answer:

(b) 50°
ABC=900Angle in a semicircleIn ABC, we have: ACB+CAB+ABC=1800500+CAB+900=1800CAB=1800-1400CAB=400Now, CAT=900(Tangents drawn from an external point are perpendicular to the radius at the point of contact)CAB+BAT=900400+BAT=900BAT=900-400BAT=500

Page No 533:

Question 14:

In the given figure, AT is a tangent to the circle with centre O, such that OT = 4 cm and ∠OTA = 30°. Then, AT = ?


(a) 4 cm
(b) 2 cm
(c) 23 cm
(d) 43 cm

Answer:

(c) 23 cm

OAATSo,ATOT=cos 300AT4=32AT=32×4AT=23 

Page No 533:

Question 15:

In the given figure, O is the centre of a circle. BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30°, then ∠PTA = ?


(a) 60°
(b) 30°
(c) 15°
(d) 45°

Answer:

(b) 30°
BPA=900 angle in a semicircleIn PBA, BPA+PBA+BAP=1800900+300+BAP=1800BAP=1800-1200BAP=600BAT is a straight angle.BAP+PAT=1800600+PAT=1800PAT=1800-600PAT=1200OA=OP Radius of the same circleOAP=OPA=600 Since BAP=600 Now, OPT=900 (Tangents drawn from an external point are perpendicular   to the radius at the point of contact)OPA+APT=900600+APT=900APT=900-600APT=300In PAT, we have:PAT+APT+PTA=18001200+300+PTA=1800PTA=1800-1500PTA=300



Page No 534:

Question 16:

In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively, and S is a point on the circle, such that ∠SQL = 50° and ∠SRM = 60°. Find ∠QSR.


(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer:

(d) 70°PQL is a tangent OQ is the radius; so, OQL=900. OQS=900-500=400Now, OQ=OS (Radius of the same circle)OSQ=OQS=400Similarly, ORS=900-600=300,And ,OR=OS (Radius of the same circle)OSR=ORS=300QSR=OSQ+OSRQSR=400+300QSR=700

Page No 534:

Question 17:

In the given figure, O is the centre of two concentric circles of radii 3 cm and 5 cm. AB is a chord of the outer circle, which touches the inner circle. The length of AB is


(a) 4 cm
(b) 7 cm
(c) 8 cm
(d) 34cm

Answer:

(c) 8 cm

Given, OP=3 cm and OA=5 cmNow, OPA=OPB=900(Tangents drawn from an external point are perpendicular to the radius at the point of contact)In OPA,OA2=(OP2+AP2)OA2-OP2=AP252-32=AP2AP2=16AP=16AP=4cmAB=2×APcm=2×4cm=8cm (Perpendicular drawn from the centreto a chord bisects the chord)

Page No 534:

Question 18:

In the given figure, ΔABC is circumscribed, touching the circle at P, Q and R. AP = 4 cm, BP = 6 cm, AC = 12 cm and BC = x cm. Find x.


(a) 10 cm
(b) 14 cm
(c) 18 cm
(d) 12 cm

Answer:

(b) 14 cm
Tangents drawn from an external point to a circle are equal.So, AR=AP=4 cmCR=AC-ARCR=12-4 cmCR=8 cmBQ=BP=6 cmBC=BQ+CQBC=6+8 cm [CQ=CR=8 cm]BC=14 cm

Page No 534:

Question 19:

In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5 cm, BC = 7 cm and CS = 3 cm, AB = ?


(a) 9 cm
(b) 10 cm
(c) 12 cm
(d) 8 cm

Answer:

(a) 9 cm
Tangents drawn from an external point to a circle are equal.So, AQ=AP=5 cmCR=CS=3 cm and BR=BC-CRBR=7-3 cmBR=4 cmBQ=BR=4 cmAB=AQ+BQAB=5+4 cmAB=9 cm

Page No 534:

Question 20:

In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm, then the perimeter of quadrilateral ABCD is


(a) 18 cm
(b) 27 cm
(c) 36 cm
(d) 32 cm

Answer:

(c) 36 cmGiven, AP=6 cm, BP=5 cm, CQ=3  cm and DR=4 cm.Tangents drawn from an external points to a circle are equal.So, AP=AS=6 cm, BP=BQ=5 cm, CQ=CR=3 cm, DR=DS=4 cm.AB=AP+PB=6+5=11 cmBC=BQ+CQ=5+3=8 cmCD=CR+DR=3+4=7 cmAD=AS+DS=6+4=10 cmPerimeter of quadrilateral ABCD=AB+BC+CD+DA=11+8+7+10 cm=36 cm



Page No 535:

Question 21:

In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S such that ∠DAB = 90°, If CS = 27 cm and CB = 38 cm and radius of the circle is 10 cm, then AB = ?


(a) 28 cm
(b) 21 cm
(c) 19 cm
(d) 17 cm

Answer:

(b) 21 cmGiven, CS=27 cm, CB=38 cm and DAB=900.The lengths of tangents drawn from an external point to a circle are equal.So, CR=CS=27 cm BR=BC-CRBR=38-27 cmBR=11 cm BQ=BR=11 cmJoin OQ. A=Q=O=P=90°  and OP=OQ (radius).So, PAQO is a square. AQ=PQ=10 cmHence, AB=AQ+BQAB=10+11 cmAB=21 cm

Page No 535:

Question 22:

In the given figure, ΔABC is right-angled at B, such that BC = 6 cm and AB = 8 cm. A circle with centre O has been inscribed in the triangle. OPAB, OQBC and OR AC.
If OP = OQ = OR = x cm, then x = ?


(a) 2 cm
(b) 2.5 cm
(c) 3 cm
(d) 3.5 cm

Answer:

(a) 2 cm
Given,AB=8 cm,BC=6 cmNow, in ABC:AC2=AB2+BC2AC2=82+62AC2=64+36AC2=100AC=100AC=10 cmPBQO is a square.CR=CQ Since the lengths of tangents drawn from an external point are equalCQ=BC-BQ=6-x cmSimilarly, AR=AP=AB-BP=8-x cm AC=AR+CR=8-x+6-x cm10 =14-2x cm2x=4x=2 cmThe radius of the circle is 2 cm.

Page No 535:

Question 23:

In the given figure, three circles with centres A, B, C, respectively, touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm, the radius of the circle with centre A is


(a) 1.5 cm
(b) 2 cm
(c) 2.5 cm
(d) 3 cm

Answer:

(b) 2 cm

Given, AB=5 cm,BC=7 cm and CA=6 cm.Let, AR=AP=x cm.BQ=BP=y cmCR=CQ=z cmSince the length of tangents drawn from an external point are equalThen, AB=5 cmAP+PB=5 cmx+y=5..iSimilarly, y+z=7iiand z+x=6.iiiAdding i,ii andiii, we get: x+y+ y+z+ z+x=182x+y+z=18x+y+z=9.ivNow, iv-ii:x+y+z- y+z=9-7x=2 The radius of the circle with centre A is 2 cm.

Page No 535:

Question 24:

In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If PA = 12, then PD = ?


(a) 52 cm
(b) 35 cm
(c) 410 cm
(d) 510 cm

Answer:



(c) 410 cm

Given,OP=5 cm,PA=12 cmNow, join O and B.Then, OB=3 cm.Now, OAP=900(Tangents drawn from an external point are perpendicular to the radius at the point of contact) Now, in OAP:  OP2=OA2+PA2OP2=52+122OP2=25+144OP2=169OP=169OP=13Now, in OBP:PB2=OP2-OB2 PB2=132-32 PB2=169-9 PB2=160PB=160PB=410 cm

Page No 535:

Question 25:

In the given figure, PA and PB are tangents to the given circle, such that PA = 5 cm and ∠APB = 60°. The length of chord AB is


(a) 52 cm
(b) 5 cm
(c) 53 cm
(d) 7.5 cm

Answer:

(b) 5 cm
The lengths of tangents drawn from a point to a circle are equal.
So, PA=PB and therefore, PAB=PBA=x say.Then, in PAB:PAB+PBA+APB=1800x+x+600=18002x=1800-6002x=1200x=600Each angle of PAB is 600 and therefore, it is an equilateral triangle.AB=PA=PB=5 cmThe length of the chord AB is 5 cm.

Page No 535:

Question 26:

Which of the following statements is not true?
(a) If a point P lies inside a circle, no tangent can be drawn to the circle passing through P.
(b) If a point P lies on a circle, then one and only one tangent can be drawn to the circle at P.
(c) If a point P lies outside a circle, then only two tangents can be drawn to the circle from P.
(d) A circle can have more than two parallel tangents parallel to a given line.

Answer:

(d) A circle can have more than two parallel tangents, parallel to a given line.
This statement is false because there can only be two parallel tangents to the given line in a circle.



Page No 536:

Question 27:

Which of the following statements is not true?
(a) A tangent to a circle intersects the circle exactly at one point.
(b) The point common to a circle and its tangent is called the point of contact.
(c) The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
(d) A straight line can meet a circle at one point only.

Answer:

 (d)A straight line can meet a circle at one point only.
This statement is not true because a straight line that is not a tangent but a secant cuts the circle at two points.

Page No 536:

Question 28:

Which of the following statements is not true?
(a) A line which intersects a circle at two points, is called a secant of the circle.
(b) A line intersecting a circle at one point only is called a tangent to the circle.
(c) The point at which a line touches the circle is called the point of contact.
(d) A tangent to the circle can be drawn from a point inside the circle.

Answer:

(d) A tangent to the circle can be drawn from a point inside the circle.
This statement is false because tangents are the lines drawn from an external point to the circle that touch the circle at one point.

Page No 536:

Question 29:

Assertion (A)
At point P of a circle with centre O and radius 12 cm, a tangent PQ of length 16 cm is drawn. Then, OQ = 20 cm.

Reason (R)
The tangent at any point of a circle is perpendicular to the radius through the point of contact.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:



(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

In OPQ, OPQ=900.OQ2=OP2+PQ2OQ=OP2+PQ2=122+162=144+256=400=20 cm



Page No 537:

Question 30:

Assertion (A)
If two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.

Reason (R)
A parallelogram circumscribing a circle is a rhombus.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

Assertion :-
We know that if two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.

Reason:-



Given, a parallelogram ABCD circumscribes a circle with centre O.
 AB=BC=CD=AD
We know that the
tangents drawn from an external point to circle are equal .
AP=AS.i                    [tangents from A]BP=BQ.ii                  [tangents from B]CR=CQ.iii                 [tangents from C]DR=DS..iv               [tangents from D]AB+CD=AP+BP+CR+DR=AS+BQ+CQ+DS [from i,ii,iii and iv]=AS+DS+(BQ+CQ)=AD+BCThus,AB+CD=AD+BC2AB=2AD  opposite sides of a parallelogram are equalAB=ADCD=AB=AD=BC
Hence, ABCD is a rhombus.

Page No 537:

Question 31:

Assertion (A)
In the given figyre, a quadrilateral ABCD is drawn to circumscribe a given circle, as shown.
Then, AB + BC = AD + DC.
Figure

Reason (R)
In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:

 (d) Assertion (A) is false and Reason (R)is true.
Assertion (A) is false[ Since the correct result is AB+CD=AD+BC]
Since the length of tangents drawn from an external point are equal, therefore:AP=ASBP=BQCQ=CRRD=SD AB+CD=AP+PB+CR+DR=AS+BQ+CQ+SD=AS+SD+BQ+CQ=AD+BC



Page No 541:

Question 1:

In the given figure, O is the centre of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?


(a) 130°
(b) 100°
(c) 90°
(d) 75°

Answer:

(b) 100°Given,∠QPT=500Now,∠OPT=900 (Since tangents drawn from an external point areperpendicular to the radius at point of contact)∠OPQ=∠OPT-∠QPT=900-500=400OP=OQRadii of the same circle∠OPQ=∠OQP=400InPOQ,POQ+OPQ+∠OQP=1800POQ+400+400=1800POQ=1800-400+400∠POQ=1800-800∠POQ=1000

Page No 541:

Question 2:

If the angle between two radii of a circle is 130°, then the angle between the angles at the ends of the radii is
Figure

(a) 65°
(b) 40°
(c) 50°
(d) 90°

Answer:


(c) 50°OA and OB are the two radii of a circle with centre O. Also, AP and BP are the tangents to the circle.Given, AOB=1300Now, OAB=OBA=900(Since tangents drawn from an external point areperpendicular to the radius at point of contact)In quadrilateral OAPB,AOB+OAB+OBA+APB=36001300+900+900+APB=3600APB=3600-1300+900+900APB=3600-3100APB=500

Page No 541:

Question 3:

If tangents PA and PB from a point P to a circle with centre O are drawn, so that ∠APB = 80°, then ∠POA = ?


(a) 40°
(b) 50°
(c) 80°
(d) 60°

Answer:

(b) 50°

From OPA and OPB,OA=OB Radii of the same circleOP (Common side)PA=PB Since tangents drawn from an external point to a circle are equalOPA OPB SSS ruleAPO=BPOAPO=12APB=400 And OAP=900 (Since tangents drawn from an external point areperpendicular to the radius at point of contact)Now, inOAP, AOP+OAP+APO=1800AOP+900+400=1800AOP=1800-1300=500



      

Page No 541:

Question 4:

In the given figure, AD and AE are the tangents to a circle with centre O and BC touches the circle at F. If AE = 5 cm, then perimeter of ∆ABC is


(a) 15 cm
(b) 10 cm
(c) 22.5 cm
(d) 20 cm

Answer:

(b) 10 cm
Since the tangents from an external point are equal, we have:
AD=AE,CD=CF,BE=BFPerimeter of ABC=AC+AB+CB=AD-CD+CF+BF+(AE-BE)=AD-CF+CF+BF+AE-BF=AD+AE=2AE=2×5=10 cm

Page No 541:

Question 5:

Find the length of the tangent drawn to a circle with radius 7 cm from a point 25 cm away from the centre of the circle.

Answer:


Since, triangle OPT is a right-angled triangle, by Pythagoras' Theorem:OP2=OT2+TP2TP2=OP2-OT2TP=OP2-OT2TP=252-72TP=625-49TP=576TP=24 cm
The length of the tangent to the circle is 24 cm.



Page No 542:

Question 6:

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Answer:

Here, OA=OBAnd OAAP, OABP (Since tangents drawn from an external point areperpendicular to the radius at the point of contact)OAP=900, OBP=900OAP+OBP=900+900=1800AOB+APB=1800 (Since,OAP+OBP+AOB+APB=3600)Sum of opposite angle of a quadrilateral is 180°.Hence, A,O,B and P are concyclic.

Page No 542:

Question 7:

In the given figure, PA and PB are tangents, such that PA = 9 cm and ∠APB = 60°. Find the length of chord AB.

Answer:

The lengths of tangents drawn from a point to a circle are equal.
So, PAPB.
Suppose, PAB=PBA=xThen, in PAB:PAB+PBA+APB=1800x+x+600=18002x=1200x=12002=600The angles of PAB are 600 .It is an equilateral triangle. AB=PA=PB=9 cm
The length of chord AB is 9 cm.

Page No 542:

Question 8:

Two tangents BC and BD are drawn to a circle with centre O, such that ∠CBD = 120°. Prove that OB = 2BC.

Answer:

Here, OB is the bisector of CBD.(Two tangents are equally inclined to the line segment joining the centre to that point)CBO=DBO=12CBD=600From BOD, BOD=300Now, from right-angled BOD, BDOB=sin300BDOB=12OB=2BDOB=2BC Since tangents from an external point are equal, i.e.BC=BDOB=2BC

Page No 542:

Question 9:

Fill in the blanks.
(i) A line intersecting a circle at two distinct points is called a ....... .
(ii) A circle can have ....... parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the ....... .
(iv) A circle can have ...... tangents.

Answer:

(i) A line intersecting a circle at two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinite tangents.

Page No 542:

Question 10:

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

Answer:


Given two tangents AP and AQ are drawn from a point A to a circle with centre O.
To prove: AP=AQJoin OP, OQ and OA.AP is tangent at P and OP is the radius.OPAP(since tangents drawn from an external point areperpendicular to the radius at the point of contact)Similarly, OQAQIn the right OPA and OQA, we have:OP=OQ   [radii of the same circle]OPA=OQA(=900)OA=OA   common sideOPAOQA     By R.H.S-CongruenceHence, AP=AQ

Page No 542:

Question 11:

Prove that the tangents drawn at the ends of the diameter of a circle are parallel.

Answer:

  
Here, PT and QS are the tangents to the circle with centre O and AB is the diameter.
Now, radius of a circle is perpendicular to the tangent at the point of contact.
OAAT and OBBS (since tangents drawn from an external point areperpendicular to the radius at point of contact)OAT=OBQ=900But OAT and OBQ are alternate angles.AT is parallel to BS.

Page No 542:

Question 12:

In the given figure, if AB = AC, prove that BE = CE.
Figure

Answer:

Given,AB=AC
We know that the tangents from an external point are equal.
AD=AF,BD=BE and CF=CE..(i)Now,AB=ACAD+DB=AF+FCAF+DB=AF+FC  [fromi]DB=FCBE=CE      [fromi]Hence proved.

Page No 542:

Question 13:

If two tangents are drawn to a circle from an external point,show that they subtend equal angles at the centre.

Answer:


Given : A circle with centre O and a point A outside it. Also, AP and AQ are the two tangents to the circle.
:To prove: AOP=AOQ .Proof : In AOP andAOQ,we have:AP=AQ   tangents from an external point are equalOP=OQ    radii of the same circleOA=OA   common sideAOPAOQ  by SSS-congruenceHence,AOP=AOQ (c.p.c.t) .

Page No 542:

Question 14:

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Answer:


Let RA and RB be two tangents to the circle with centre O and let AB be a chord of the circle.We have to prove that RAB=RBA. Now, RA=RB (Since tangents drawn from an external point to a circle are equal)In RAB, RAB=RBA (Since opposite sides are equal, their base angles are also equal)



Page No 543:

Question 15:

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:



Given, a parallelogram ABCD circumscribes a circle with centre O.
 AB=BC=CD=AD
We know that the lengths of tangents drawn from an exterior point to a circle 
are equal.
AP=AS.i                    [tangents from A]BP=BQ.ii                  [tangents from B]CR=CQ.iii                 [tangents from C]DR=DS..iv               [tangents from D]AB+CD=AP+BP+CR+DR=AS+BQ+CQ+DS [from i,ii,iii and iv]=AS+DS+(BQ+CQ)=AD+BCThus,AB+CD=AD+BC2AB=2AD  opposite sides of a parallelogram are equalAB=ADCD=AB=AD=BC
Hence, ABCD is a rhombus.

Page No 543:

Question 16:

Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.

Answer:


Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
                  smaller circle at C; also, OA=5 cm and OC=3 cm.  
In OAC, OA2=OC2+AC2AC2=OA2-OC2AC2=52-32AC2=25-9AC2=16AC=4 cmAB=2AC (since perpendicular drawn from the centre of the circlebisects the chord)AB=2×4=8 cm
The length of the chord of the larger circle is 8 cm.

Page No 543:

Question 17:

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

Answer:


We know that the tangents drawn from an external point to a circle are equal.

AP=ASi          tangents from ABP=BQ.ii      tangents from BCR=CQ.iii  tangents from CDR=DS..iv    tangents from DAB+CD=AP+BP+CR+DR=AS+BQ+CQ+DS      [using i,ii,iiiand (iv)]=AS+DS+BQ+CQ=AD+BCHence, AB+CD=AD+BC

Page No 543:

Question 18:

Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:


Given, a quadrilateral ABCD circumscribes a circle with centre O.
 To prove: AOB+COD=1800and AOD+BOC=1800Join OP,OQ,OR and OS.We know that the tangents drawn from an external point of a circlesubtend equal angles at the centre.1=7,2=3,4=5 and 6=8and 1+2+3+4+5+6+7+8=3600     angles at a point(1+7)+(3+2)+(4+5)+(6+8)=360021+22+26+25=36001+2+5+6=1800 AOB+COD=1800 and AOD+BOC=1800

Page No 543:

Question 19:

Prove that the angles between the two tangents drawn form an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.

Answer:


Given, PA and PB are the tangents drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.
To prove: APB+AOB=1800
We know that the tangent to a circle is perpendicular to the radius through the point of contact.
PAOAOAP=900PBOBOBP=900OAP+OBP=900+900=1800              .i

But we know that the sum of all the angles of a quadrilateral is 3600.OAP+OBP+APB+AOB=3600      .ii
From (i) and (ii), we get:
APB+AOB=1800

Page No 543:

Question 20:

A circle touches the side BC of ∆ABC at P and touches AB and AC produced at Q and R, respectively, as shown in the given figure. Show that AQ=12×(perimeter of ABC).
Figure

Answer:

We know that the lengths of tangents drawn from an external point to a circle are equal.
AQ=AR  i         [Tangents from A]BP=BQ..ii           Tangents from BCP=CR..iii       Tangents from CPerimeter of ABC=AB+BC+AC=AB+BP+CP+AC=AB+BQ+CR+AC   [Using ii and iii]=AQ+AR=2AQ   [Using i]AQ=12Perimeter ofABC



View NCERT Solutions for all chapters of Class 10