ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.3

 Exercise 15.3

Question 1

Find the length of the tangent drawn to a circle of radius 3 cm, from a point distant 5 cm from the centre.

Sol :

In a circle with centre O and radius 3 cm

and P is at a distance of 5 cm.

i.e. OT=3 cm, OP=5 cm

∵OT is the radius of the circle

∴OT⊥PT

Now in right ΔOTP, by pythagorus axiom,

$\mathrm{OP}^{2}=\mathrm{OT}^{2}+\mathrm{PT}^{2}$

$\Rightarrow (5)^{2}=(3)^{2}+\mathrm{PT}^{2}$

$\Rightarrow \mathrm{PT}^{2}=(5)^{2}-(3)^{2}=25-9=16=(4)^{2}$

$\therefore  P T=4 \mathrm{~cm}$

Question 2

A point P is at a distance 13 cm from the centre C of a circle and PT is a tangent to the given circle. If PT = 12 cm, find the radius of the circle.

Sol :

CT is the radius

CP = 13 cm and tangent PT = 12 cm

$\because$CT is the radius and TP is the tangent

$\therefore \mathrm{CT} \perp \mathrm{TP}$

Now in right $\Delta \mathrm{CPT}$

$(\mathrm{CP})^{2}=(\mathrm{CT})^{2}+(\mathrm{PT})^{2}$

(Pythagorus axiom)

$\Rightarrow(13)^{2}=(C T)^{2}+(12)^{2}$

$\Rightarrow 169=(\mathrm{CT})^{2}+144$

$\Rightarrow(C T)^{2}=169-144=25=(5)^{2}$

$\therefore  C T=5 \mathrm{~cm}$

Hence radius of the circle =5 cm

Question 3

The tangent to a circle of radius 6 cm from an external point P, is of length 8 cm. Calculate the distance of P from the nearest point of the circle.

Sol :

Radius of the circle = 6 cm

and length of tangent = 8 cm

Let OP be the distance

i.e. OA = 6 cm, AP = 8 cm .

OA is the radius

$\therefore$ OA $\perp$ AP.

Now in right $\Delta$ OAP

$\mathrm{OP}^{2}=\mathrm{OA}^{2}+\mathrm{AP}^{2}$

(by Pythagorus axiom)

$=(6)^{2}+(8)^{2}=36+64=100=(10)^{2}$

∴OP=10 cm

Question 4

Two concentric circles are of the radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

Sol :

Two concentric circles with centre O

OP and OB are the radii of the circles respectively, then

OP = 5 cm, OB = 13 cm.

AB is the chord of the outer circle which touches the inner circle at P

∵OP is the radius and APB is the tangent to the inner circle.

∴In right ΔOPB, by Pythagorus axiom,

$(\mathrm{OB})^{2}=(\mathrm{OP})^{2}+(\mathrm{PB})^{2}$

$\Rightarrow (13)^{2}=(5)^{2}+(\mathrm{PB})^{2}$

$\Rightarrow 169=25+(\mathrm{PB})^{2}$

$\Rightarrow(\mathrm{PB})^{2}=169-25=144$

$\Rightarrow(P B)^{2}=144=(12)^{2}$

∴PB=12 cm

But P is the mid-point of AB

$\therefore A B=2 \times P B=2 \times 12=24 \mathrm{~cm}$

Question 5

Two circles of radii 5 cm and 2-8 cm touch each other. Find the distance between their centres if they touch :

(i) externally

(ii) internally.

Sol :

Radii of the circles are 5 cm and 2.8 cm.

i.e. OP = 5 cm and CP = 2.8 cm.

(i) When the circles touch externally,

then the distance between their centres = OC = 5 + 2.8 = 7.8 cm.

(ii) When the circles touch internally,

then the distance between their centres = OC = 5.0 – 2.8 = 2.2 cm

Question 6

(a) In figure (i) given below, triangle ABC is circumscribed, find x.

(b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x.

Figure to be added

Sol :

(a) From A, AP and AQ are the tangents to the circle

∴ AQ = AP = 4cm

But AC=12 cm

∴CQ=12-4=8 cm

From B,BP and BR are the tangents to the circle

∴BR=BP=6 cm

Similarly, from C,

CQ and CR the tangents

∴CR=CQ=8 cm

∴x=BC=BR+CR

=6cm+8cm=14cm

(b) From C, CR and CS are the tangents to the circle 

∴CS=CR=3cm

But BC=7cm

∴BS=BC-CS

=7-3=4cm

Now from B, BP and BS are the tangents to the circle

∴BP=BS=4cm

and from A, AP and AQ are the tangents to the circle

∴AP=AQ=5cm

x=AB=AP+BP

=5+4=9cm

Question 7

(a) In figure (i) given below, quadrilateral ABCD is circumscribed; find the perimeter of quadrilateral ABCD.

(b) In figure (ii) given below, quadrilateral ABCD is circumscribed and AD ⊥ DC ; find x if radius of incircle is 10 cm.

Sol :

(a) From A, AP and AS are the tangents to the circle

∴AS = AP = 6

From B, BP and BQ are the tangents

∴BQ = BP = 5

From C, CQ and CR are the tangents

∴CR=CQ=3

and from D, DS and DR are the tangents.

∴DS=DR=4

∴Perimeter of the quadrilateral ABCD

=(6+5+5+3+3+4+4+6)cm

=36cm

(b) In the circle with centre O, radius OS

=10cm, ∠ADC=90°

PB=27cm, BC=38cm

∵OS is radius and AD is the tangent

∴OS AD

∴SD=OS=10cm

Figure to be added

Now from D, DR and DS are the tangents to the circle

∴DR=DS=10cm

From B, BP and BQ are tangents to the circle

∴CR=CQ=11cm

∴DC=x=DR+CR

=10+11=21cm

Question 8

(a) In the figure (i) given below, O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, find the radius of the circle.

(b) In the figure (ii) given below, from an external point P, tangents PA and PB are drawn to a circle. CE is a tangent to the circle at D. If AP = 15 cm, find the perimeter of the triangle PEC.

Sol :

(i) Join OB

∠OBA = 90°

(Radius through the point of contact is

perpendicular to the tangent)

$\Rightarrow \mathrm{OB}^{2}=\mathrm{OA}^{2}-\mathrm{AB}^{2}$

$\Rightarrow r^{2}=(r+7.5)^{2}-15^{2}$

$\Rightarrow r^{2}=r^{2}+56.25+15 \mathrm{r}-225$

$\Rightarrow 15 r=168.75$

$\Rightarrow r=11.25$

Hence, radius of the circle =11.25 cm

(ii) In the figure ,PA and PB are the tangents drawn from P to the circle

CE is a tangent at D

AP=15cm

∵PA and PB are the tangents to the circle

∴AP=BP=15cm

Similarly EA and ED are tangents

∴EA=ED

Similarly, BC=CD

Now perimeter of ΔPEC

=PE+EC+PC

=PE+ED+CD+PC

PE+EA+CB+PC

[∵ED=EA and CB=CD]

=AP+PB

=15+15=30 cm

Question 9

(a) If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by $r=\frac{a+b-c}{2}$

(b) In the given figure, PB is a tangent to a circle with centre O at B. AB is a chord of length 24 cm at a distance of 5 cm from the centre. If the length of the tangent is 20 cm, find the length of OP.

Sol :

(a) Let the circle touch the sides BC, CA and AB of the right triangle ABC at points D, E and F respectively,

where BC = a, CA = b

and AB = c (as showing in the given figure).

As the lengths of tangents drawn from an external point to a circle are equal,

AE=AF, BD=BF and CD=DE

OD⟂BC and OE⟂CA

(∵ tangents is ⟂ to radius)

∴ODCE is a square of side r

$\Rightarrow  \mathrm{DC}=\mathrm{CE}=r$

$\Rightarrow A F=A E=A C-E C=b-r$ and

BF=BD=BC-DC=a-r

Now, AB=AF+BF

$\Rightarrow c=(b-r)+(a-r)$

$\Rightarrow 2 r=a+b-c$

$\Rightarrow r=\frac{a+b-c}{2}$

(b) Construction : Join OB

OC=5 cm (given)

and AB=24 cm

$\mathrm{So}, \mathrm{BC}=\frac{24}{2} \mathrm{~cm}=12 \mathrm{~cm}$

(∵ The perpendicular from the centre to the chord bisects the chord)

In ΔBOC, by Pythagoras Theorem

$\mathrm{OB}^{2}=\mathrm{OC}^{2}+\mathrm{BC}^{2}$

$\Rightarrow \mathrm{OB}^{2}=5^{2}+12^{2}$

$\Rightarrow \mathrm{OB}^{2}=25+144$

$\Rightarrow \mathrm{OB}^{2}=169$

$\Rightarrow \mathrm{OB}=13 \mathrm{~cm}$

Now, In ΔPOB

PB=20cm, OB=20cm and ∠PBO=90°

(∵Tangent to a circle at a point is perpendicular to the radius through the point of contact)

So, by Pythagoras Theorem

$\mathrm{OP}^{2}=\mathrm{BP}^{2}+\mathrm{OB}^{2}$

$\Rightarrow \mathrm{OP}^{2}=20^{2}+13^{2}$

$\Rightarrow \mathrm{OP}^{2}=400+169$

$\Rightarrow \mathrm{OP}=\sqrt{569} \mathrm{~cm}$

Question 10

Three circles of radii 2 cm, 3 cm and 4 cm touch each other externally. Find the perimeter of the triangle obtained on joining the centres of these circles.

Sol :

Three circles with centres A, B and C touch each other externally

at P, Q and R respectively and the radii of these circles are

2 cm, 3 cm and 4 cm.

By joining their centres ΔABC is formed in which

∴AB=2+3=5 cm

BC=3+4=7 cm

and CA=4+2=6 cm

∴Perimeter of the ΔABC

=AB+BC+CA

=5cm+7cm+6cm

=18cm

Question 11

(a) In the figure (i) given below, the sides of the quadrilateral touch the circle. Prove that AB + CD = BC + DA.

(b) In the figure (ii) given below, ABC is triangle with AB = 10cm, BC = 8cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with vertices A, B and C as their centres. Find the radii of the three circles

Sol :

(a) Given: Sides of quadrilateral ABCD touch the circle at

P, Q, R and S respectively.

To prove : AB+CD=BC+DA

Proof : Tangents from A, AP and AS are to the circle

∴AP=AS

Similarly PB=BQ

CR=CQ and DR=DS

Adding them, we get

AP+PB+CR+DR

=AS+BQ+CQ+DS

$\Rightarrow \mathrm{AP}+\mathrm{PB}+\mathrm{CR}+\mathrm{DR}$

$=\mathrm{BQ}+\mathrm{CQ}+\mathrm{AS}+\mathrm{SD}$

$\Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{BC}+\mathrm{DA} .$ Q.E.D.

(b) AB=10 cm, BC=8 cm, AC=6 cm

BP+PA=10 cm...(i)

BC=8 cm (given)

BR+RC=8 cm

∴BP+RC=8...(ii)

Again, AC=6 cm

AQ+QC=6 cm

PA+RC=6 cm...(iii)

(i)+(ii)+(iii) we have

$2(\mathrm{BP}+\mathrm{PA}+\mathrm{RC})=24 \mathrm{~cm}$

$\therefore \mathrm{BP}+\mathrm{PA}+\mathrm{RC}=12 \mathrm{~cm}$...(iv)

(iv)-(i) we have

RC=12-10=2 cm

(iv)-(ii) we have

PA=12-8=4 cm

(iv)-(iii) we have

BP=12-6=6 cm

Radii : BP=6 cm, PA=4 cm, RC=2 cm

Question 12

(a) ln the figure (i) PQ = 24 cm, QR = 7 cm and ∠PQR = 90°. Find the radius of the inscribed circle ∆PQR

(b) In the figure (ii) given below, two concentric circles with centre O are of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP = 12cm, find BP.

Sol :

(a) In the figure, a circle is inscribed in the triangle PQR

which touches the sides. O is centre of the circle.

PQ = 24cm, QR = 7 cm ∠PQR = 90°

OM is joined.

∵OL and OM are the radii of the circle

∴OL⊥PQ and OM⊥BC

and ∠PQR=90°

∴QLOM is a square

∴OL=OM=QL=QM

Now in ΔPQR , ∠O=90

$\therefore P Q^{2}=P Q^{2}+Q R^{2}$

(Pythagoras theorem)

$=(24)^{2}+(7)^{2}=576+49=625=(25)^{2}$

∴PR=25cm

Let OB=x, then 

QM=QL=x

∵PL and PN are the tangents to the circle 

PL=PN

⇒PQ-LQ=PR-RN

⇒24-x=25-PM  (∵RN=RM)

⇒24-x=25-(QR-QM)

⇒24-x=25-(7-x)⇒24-x=25-7+x

⇒24-25+7=2x⇒2x=6

⇒$x=\frac{6}{2}=3 \mathrm{~cm}$

∴Radius of the incircle =3 cm

(b) Radius of outer circle=5 cm

and radius of inner circle=3 cm

∴ OA=5 cm

OB=3 cm

and AP=12 cm

∵OA is radius and AP is the tangent 

∴OA⊥AP

∴In right ΔOAP

$\mathrm{OP}^{2}=\mathrm{GA}^{2}+\mathrm{AP}^{2}=(5)^{2}+(12)^{2}$

$=25+144=169=(13)^{2}$

∴OP=13

Similarly in right ∠OBP

$\mathrm{OP}^{2}=\mathrm{OB}^{2}+\mathrm{BP}^{2}$

$\Rightarrow (13)^{2}=(3)^{2}+\mathrm{BP}^{2}$

$\Rightarrow 169=9+\mathrm{BP}^{2}$

$\Rightarrow B P^{2}=169-9=160$

∴ BP=$\sqrt{160}=\sqrt{16 \times 10}$

$=4 \sqrt{10} \mathrm{~cm}$

Question 13

(a) In the figure (i) given below, AB = 8 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all three semi-circles as shown, find its radius.

(b) In the figure (ii) given below, equal circles with centres O and O’ touch each other at X. OO’ is produced to meet a circle O’ at A. AC is tangent to the circle whose centre is O. O’D is perpendicular to AC. Find the value of :

(i) $\frac{A O^{\prime}}{A O}$

(ii) $\begin{tabular}{ccc} area & of &$\Delta A D O^{\prime}$\\\hline area & of &$\Delta A C O\end{tabular}$

Sol :

(a) Let x be the radius of the circle

with centre C and radii of each equal

Semi circles $=\frac{4}{2}=2 \mathrm{~cm}$

CP=x+2 and CM=4-x

∴In right ΔPCM

$(\mathrm{PC})^{2}=(\mathrm{PM})^{2}+(\mathrm{CM})^{2}$

$\Rightarrow (x+2)^{2}=(2)^{2}+(4-x)^{2}$

$\Rightarrow x^{2}+4 x+4=4+16-8 x+x^{2}$

$\Rightarrow 4 x+8 x+4-4-16=0$

$\Rightarrow 12 x=16$

$\Rightarrow x=\frac{16}{12}=\frac{4}{3}=1 \frac{1}{3} \mathrm{~cm}$

∴Radius $=1 \frac{1}{3} \mathrm{~cm}$

(b) Two equal circles with centre O and O' touch each other externally at X. OO' is joined and produced to meet the circle at A. AC is the tangent to the circle with centre O. O'D is ⊥AC. Join OC

∵OC is radius and AC is tangent, then OC⏊AC

Let radius of each equal circle= r

In ΔADO' and ΔACO

∠A=∠A (common)

∠D=∠C (each 90°)

∴ΔADP'∼ΔACO

(AA axiom of similarity)

(i) $\therefore \frac{\mathrm{AO}^{\prime}}{\mathrm{AO}}=\frac{r}{3 r}=\frac{1}{3}$

(ii) $\because \Delta \mathrm{ADO}^{\prime} \sim \Delta \mathrm{ACO}$

$\therefore \frac{\text { area of } \Delta \mathrm{ADO}^{\prime}}{\text { area of } \Delta \mathrm{ACO}}=\frac{\left(\mathrm{AO}^{\prime}\right)^{2}}{(\mathrm{AO})^{2}}$

$=\left(\frac{\mathrm{AO}^{\prime}}{\mathrm{AO}}\right)^{2}=\left(\frac{1}{3}\right)^{2}$ [from (i)]

$=\frac{1}{9}$

Question 14

The length of the direct common tangent to two circles of radii 12 cm and 4 cm is 15 cm. Calculate the distance between their centres.

Sol :

Let R and r be the radii of the circles

with centre A and B respectively

Let TT’ be their common tangent.

$\therefore \left(T T^{\prime}\right)^{2}=(A B)^{2}-(R-r)^{2}$

$\Rightarrow (15)^{2}=(\mathrm{AB})^{2}-(12-4)^{2}$

$\Rightarrow 225=\mathrm{AB}^{2}-(8)^{2}$

$\Rightarrow(\mathrm{AB})^{2}=225+64=289=(17)^{2}$

∴AB=17 cm

Hence distance between their centres = 17 cm Ans.

Question 15

Calculate the length of a direct common tangent to two circles of radii 3 cm and 8 cm with their centres 13 cm apart.

Sol :

Let A and B be the centres of the circles

whose radii are 8 cm and 3 cm and

let TT’ length of their common tangent and AB = 13 cm.

Now $\left(\mathrm{TT}^{\prime}\right)^{2}=(\mathrm{AB})^{2}-(\mathrm{R}-r)^{2}$

$=(13)^{2}-(8-3)^{2}=169-(5)^{2}$

$=169-25=144=(12)^{2}$

$\therefore \mathrm{TT}^{\prime}=12 \mathrm{~cm}$

Hence length of common tangent

=12 cm

Question 16

In the given figure, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively. Given that AB = 8 cm, calculate PQ.

Sol :

AC is a transverse common tangent to the two circles

with centre P and Q and of radii 6 cm and 3 cm respectively

AB = 8 cm. Join AP and CQ.

In right ΔPAB,

$\mathrm{PB}^{2}=\mathrm{PA}^{2}+\mathrm{AB}^{2}=(6)^{2}+(8)^{2}$

=36+64=100=$(10)^2$

∴PB=10 cm

Now in ΔPAB and ΔBCQ,

∴∠A=∠C (each 90°)

∠ABP=∠CBQ

(vertically opposite angles)

∴ΔPAB∼ΔBCQ

(AA axiom of similarity)

$\therefore \frac{\mathrm{AP}}{\mathrm{CQ}}=\frac{\mathrm{PB}}{\mathrm{BQ}}$

$\Rightarrow \frac{6}{3}=\frac{10}{\mathrm{BQ}}$

$\Rightarrow \mathrm{BQ}=\frac{10 \times 3}{6}=5 \mathrm{~cm}$

$\therefore P Q=P B+B Q=10+5=15 \mathrm{~cm}$

Question 17

Two circles with centres A, B are of radii 6 cm and 3 cm respectively. If AB = 15 cm, find the length of a transverse common tangent to these circles.

Sol :

AB = 15 cm.

Radius of the circle with centre A = 6 cm

and radius of second circle with radius B = 3 cm.

Let AP=x , then PB=15-x

In ΔATP and ΔSBP

∠T=∠S (each 90°)

∠APT=∠BPS 

(vertically opposite angles)

∴ΔATP∼ΔSBP

(AA axiom of similarity)

$\frac{\mathrm{AT}}{\mathrm{BS}}=\frac{\mathrm{AP}}{\mathrm{PB}}$

$\Rightarrow \quad \frac{6}{3}=\frac{x}{15+x}$

$\Rightarrow \frac{x}{15-x}=\frac{6}{3}$

$\Rightarrow 3 x=90-6 x$

$\Rightarrow 3 x+6 x=90$

$\Rightarrow 9 x=90$

$\therefore x=\frac{90}{9}=10$

$\therefore A P=10 \mathrm{~cm}, P B=15-10=5 \mathrm{~cm}$

Now in right ΔATP

$A P^{2}=A T^{2}+T P^{2}$

$\Rightarrow(10)^{2}=(6)^{2}+\mathrm{TP}^{2}$

$\Rightarrow 100=36+\mathrm{TP}^{2}$

$\Rightarrow \mathrm{TP}^{2}=100-36=64$

$\Rightarrow \mathrm{TP}^{2}=(8)^{2}$

∴TP=8 cm

Similarity in right ΔPSB

$P B^{2}=B S^{2}+P S^{2}$

$\Rightarrow \quad(5)^{2}=(3)^{2}+\mathrm{PS}^{2}$

$\Rightarrow 25=9+\mathrm{PS}^{2}$

$\Rightarrow \mathrm{PS}^{2}=25-9=16=(4)^{2}$

∴PS=4 cm

Hence TS=TP+PS

=8+4=12 cm

Question 18

(a) In the figure (i) given below, PA and PB are tangents at a points A and B respectively of a circle with centre O. Q and R are points on the circle. If ∠APB = 70°, find (i) ∠AOB (ii) ∠AQB (iii) ∠ARB

(b) In the figure (ii) given below, two circles touch internally at P from an external point Q on the common tangent at P, two tangents QA and QB are drawn to the two circles. Prove that QA = QB.

Sol :

(a) To find : (i) ∠AOB, (ii) ∠AQB, (iii) ∠ARB

Given: PA and PB are tangents at the points A and B respectively

of a circle with centre O and OA and OB are radii on it.

∠APB = 70°

Construction: Join AB

(i) Since, AP and BP are tangents to the circle and OA and OB are radii on it.

$\therefore O A \perp O P$ and $O B \perp B P$

$\therefore \angle A Q B=180^{\circ}-70^{\circ}=110^{\circ}$

(ii) Arc AB subtends $\angle AOB$ at the centre and $\angle AOB$ at the remaining part of the circle.

$\therefore \angle A Q B=2 \angle A O B$

$\Rightarrow \angle \mathrm{AQB}=\frac{1}{2} \angle \mathrm{AOB}$

$\Rightarrow \angle \mathrm{AQB}=\frac{1}{2} \times 110^{\circ}=55^{\circ}$

(iii) Now in ΔOAB

OA+OB (Radii of the same circle)

∴∠OAB=∠OBA

$=\frac{1}{2}\left(180^{\circ}-110^{\circ}\right)=\frac{1}{2} \times 70^{\circ}=35^{\circ}$

We know , ∠AOB=110°

Reflex ∠AOB=360°-110°=250°

∴$\angle \mathrm{ARB}=\frac{1}{2}$ of reflex $\angle \mathrm{AOB}$

(∵ Angle at the centre of a circle=double the angle at the remaining part of the circle)

$=\frac{1}{2} \times 250^{\circ}=120^{\circ}$

(b) In the figure (ii)

Two circles touch each other internally at P. From a point Q, outside , common tangent QP, QA and QB are drawn to the two circles

To prove : QA=AB

Proof : From Q, QA and QP are the tangents to the outer circle

∴QP=QA..(i)

Similarly, from Q, QB and QP are the tangents to the inner circle

∴QP=QB..(ii)

From (i) and (ii)

QA=QB

Hence proved

Question 19

In the given figure, AD is a diameter of a circle with centre O and AB is tangent at A. C is a point on the circle such that DC produced intersects the tangent of B. If ∠ABC = 50°, find ∠AOC.

Sol :

Given AB is tangent to the circle at A and OA is radius, OA ⊥ AB

In ∆ABD

$\angle \mathrm{OAB}+90^{\circ}+\angle \mathrm{ADC}=180^{\circ}$

$\angle \mathrm{OAB}+90^{\circ}+50^{\circ}=180^{\circ}$

$\angle \mathrm{OAB}+140^{\circ}=180^{\circ}$

$\angle \mathrm{OAB}+180^{\circ}-140^{\circ}=40^{\circ}$

(∵Angle at centre of a circle= double the angle at the remaining part of circle)

∴∠ABC=2°×40°=80°

Question 20

In the given figure, tangents PQ and PR are drawn from an external point P to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ, Find ∠RQS

Sol :

In the given figure,

PQ and PR are tangents to the circle with centre O drawn from P

∠RPQ = 30°

Chord RS || PQ is drawn

To find ∠RQS

∴ PQ = PR (tangents to the circle)

$\therefore \angle \mathrm{PRQ}=\angle \mathrm{PQR}$

But $\angle \mathrm{RPQ}=30^{\circ}$

$\therefore \angle \mathrm{PRQ}=\angle \mathrm{PQR}$

$=\frac{180^{\circ}-30^{\circ}}{2}=\frac{150^{\circ}}{2}=75^{\circ}$

$\because$ SR $\|$ QP

$\therefore \angle$ SRQ $=\angle P Q R=75^{\circ}$

and $\angle \mathrm{RSQ}=\angle \mathrm{QRS}$

(∵QR=QS)

=75°

Now in ΔQRS

∠RQS=180°-(∠RSQ+∠QRS)

=180°-(75°+75°)

=180°-150°=30°

Question 21

(a) In the figure (i) given below, PQ is a tangent to the circle at A, DB is a diameter, ∠ADB = 30° and ∠CBD = 60°, calculate (i) ∠QAB (ii) ∠PAD (iii) ∠CDB.

(b) In the figure (ii) given below, ABCD is a cyclic quadrilateral. The tangent to the circle at B meets DC produced at F. If ∠EAB = 85° and ∠BFC = 50°, find ∠CAB.

Figure to be added

Sol :

(a) PQ is tangent and AD is chord

(i) ∴ ∠QAB = ∠BDA = 30°

(Angles in the alternate segment)

(ii) In ∆ADB,

∠DAB = 90° (Angle in a semi-circle)

and $\angle \mathrm{ADB}=30^{\circ}$

$\therefore \quad \angle \mathrm{ABD}=180^{\circ}-\left(90^{\circ}+30^{\circ}\right)$

$=180^{\circ}-120^{\circ}=60^{\circ}$

But $\angle \mathrm{PAD}=\angle \mathrm{ABD}$

(Angles in the alternate segment)

$=60^{\circ}$ [proved]

(iii) In right ΔBCD

∠BCD=90° (Angle is a semi-circle)

∠CBD=60° (given)

∴∠CDB=180°-(90°+60°)

=180°-150°=30°

(b) ABCD is a cyclic quadrilateral

$\therefore$ Ext. $\angle \mathrm{EAB}=\angle \mathrm{BCD}$

$\therefore \angle \mathrm{BCD}=85^{\circ}$

But $\angle \mathrm{BCD}+\angle \mathrm{BCF}=180^{\circ}$

(Linear pair)

$\Rightarrow 85^{\circ}+\angle B C F=180^{\circ}$

$\Rightarrow  \angle B C F=180^{\circ}-85=95^{\circ}$

Now in ΔBCF,

$\angle \mathrm{BCF}+\angle \mathrm{BFC}+\angle \mathrm{CBF}=180^{\circ}$

$\Rightarrow 95^{\circ}+50^{\circ}+\angle \mathrm{CBF}=180^{\circ}$

$\Rightarrow 145^{\circ}+\angle \mathrm{CBF}=180^{\circ}$

$\Rightarrow \angle \mathrm{CBF}=180^{\circ}-145^{\circ}=35^{\circ}$

$\because \mathrm{BF}$ is a tangent and $\mathrm{BC}$ is the chord.

$\therefore \angle \mathrm{CAB}=\angle \mathrm{CBF}$

(Angles in the alternate segment)

⇒ ∠CAB = 35°

Question 22

(a) In the figure (i) given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the value of x, y and z. (2015)

(b) In the figure (ii) given below, O is the centre of the circle. PS and PT are tangents and ∠SPT = 84°. Calculate the sizes of the angles TOS and TQS.

Sol :

Consider the following figure:

TS ⊥ SP,

∠TSR = ∠OSP = 90°

In ∆TSR,

∠TSR + ∠TRS + ∠RTS = 180°

$\Rightarrow 90^{\circ}+65^{\circ}+x=180^{\circ}$

$\Rightarrow x=180^{\circ}-90^{\circ}-65^{\circ}$

$\Rightarrow x=25^{\circ}$

$\Rightarrow y=2 x$

[Angle subtended at the centre is double that of the angle subtended by the arc at the same centre]

$\Rightarrow y=2 \times 25^{\circ} \Rightarrow y=50^{\circ}$

In ΔOSP,

$\angle \mathrm{OSP}+\angle \mathrm{SPO}+\angle \mathrm{POS}=180^{\circ}$

$\Rightarrow 90^{\circ}+z+50^{\circ}=180^{\circ}$

$\Rightarrow z=180^{\circ}-140^{\circ}$

$\therefore z=40^{\circ}$

Hence $x=25^{\circ}, y=50^{\circ}$ and $z=40^{\circ}$

(b) $\because \mathrm{TP}$ and $\mathrm{SP}$ are the tangents and OS are OT are the radii

$\therefore \mathrm{OS} \perp \mathrm{SP} \text { and } \mathrm{OT} \perp \mathrm{TP}$

$\therefore \angle \mathrm{OSP}=\angle \mathrm{OTP}=90^{\circ}$

$\angle \mathrm{SPT}=84^{\circ}$

But $\angle \mathrm{SOT}+\angle \mathrm{OTP}+\angle \mathrm{TH}_{\mathrm{S}}+\angle \mathrm{OSP}=360^{\circ}$

$\Rightarrow \angle \mathrm{SOT}+90^{\circ}+84^{\circ}+90^{\circ}=360^{\circ}$

$\Rightarrow \angle \mathrm{SOT}+264^{\circ}=360^{\circ}$

$\Rightarrow\angle \mathrm{SOT}=360^{\circ}-264^{\circ}$

$\Rightarrow \angle \mathrm{SOT}=46^{\circ}$

and reflex $\angle \mathrm{SOT}=360^{\circ}-96^{\circ}=264^{\circ}$

Now major arc ST subtends/reflex $\angle$ SOT at the

centre and $\angle$ TQS at the remaining part of the circle.

$\therefore \angle \mathrm{TQS}=\frac{1}{2}$ reflex $\angle \mathrm{SOT}=\frac{1}{2} \times 264^{\circ}=132^{\circ}$

Hence $\angle \mathrm{TOS}=96^{\circ}$ and $\angle \mathrm{TQS}=132^{\circ}$

Question 23

In the given figure, O is the centre of the circle. Tangents to the circle at A and B meet at C. If ∠ACO = 30°, find

(i) ∠BCO (ii) ∠AOR (iii) ∠APB

Sol :

(i) ∠BCO = ∠ACO = 30°

(∵ C is the intersecting point of tangent AC and BC)

(ii) ∠OAC = ∠OBC = 90°

∵∠AOC = ∠BOC = 180° – (90° + 30°) = 60°

(∵ sum of the three angles a ∆ is 180°)

$\therefore \angle \mathrm{AOB}=\angle \mathrm{AOC}+\angle \mathrm{BOC}$

$=60^{\circ}+60^{\circ}=120^{\circ}$

(iii) $\angle \mathrm{APB}=\frac{1}{2} \angle \mathrm{AOB}=\frac{120^{\circ}}{2}=60^{\circ}$

$[\because$ angle subtended at the remaining part of the circle is half the subtended at the centre]

Question 24

(a) In the figure (i) given below, O is the centre of the circle. The tangent at B and D meet at P. If AB is parallel to CD and ∠ ABC = 55°. find: (i)∠BOD (ii) ∠BPD

(b) In the figure (ii) given below. O is the centre of the circle. AB is a diameter, TPT’ is a tangent to the circle at P. If ∠BPT’ = 30°, calculate : (i)∠APT (ii) ∠B OP.

Sol :

(a) AB || CD

(i) ∠ABC = ∠BCD (Alternate angles)

⇒ ∠BCD = 55°

Figure to be added

Now arc BD subtends ∠BOD at the centre and ∠BCD at the remaining part of the circle.

$\therefore \angle \mathrm{BOD}=2 \angle \mathrm{BCD}=2 \times 55^{\circ}=110^{\circ}$

(Angles of quadrilateral)

$\Rightarrow 110^{\circ}+90^{\circ}+90^{\circ}+\angle \mathrm{BPD}=360^{\circ}$

$\Rightarrow 290^{\circ}+\angle \mathrm{BPD}=360^{\circ}$

$\Rightarrow \angle \mathrm{BPD}=360^{\circ}-290^{\circ}=70^{\circ}$

(b) TPT' is the tangent

(i) AB is diameter

$\therefore \angle \mathrm{APB}=90^{\circ}$

(Angle in a semi-circle)

and ∠BPT'=30°

∴∠APT=180°-(90°+30°)

=180°-120°=60°

Figure to be added

(ii) TPT' is tangent and PB is the chord

∴∠BAP=∠BPT'

(Angles in the alternate segment)

=30°

Now arc BP subtends ∠BOP at the centre and ∠BAP at the remaining part of the circle

∴∠BOP=2∠BAP

=2×30°=60°

Question 25

In the adjoining figure, ABCD is a cyclic quadrilateral.

The line PQ is the tangent to the circle at A. If ∠CAQ : ∠CAP = 1 : 2, AB bisects ∠CAQ and AD bisects ∠CAP, then find the measure of the angles of the cyclic quadrilateral. Also prove that BD is a diameter of the circle.

Sol :

ABCD is a cyclic quadrilateral.

PAQ is the tangent to the circle at A.

∠CAD : ∠CAP = 1 : 2.

AB and AD are the bisectors of ∠CAQ and ∠CAP respectively

To prove :

(i) BD is the diameter of the circle

(ii) Find the measure of the angles of the cyclic quadrilateral

Proof : 

(i) ∵AB and AD are the bisectors of 

∠CAQ and ∠CAP respectively and 

∠CAQ+∠CAB=180° (linear pair)

⇒∠BAD=90°

Hence , BD is at diameter of the circle

(ii) ∴∠CAQ : ∠CAP=1 : 2

Let ∠CAQ=x and ∠CAP=2x

∴x+2x=180°

$\Rightarrow 3 x=180^{\circ}$

$\Rightarrow x=\frac{180^{\circ}}{3}=60^{\circ}$

∴∠CAQ=60° and  ∠CAP=120°

∵PAQ is the tangent and AC is the chord of the circle

∴∠ADC=∠CAQ=60°

Similarly ∠ABC=∠CAP=120°

Hence ∠A=90°, ∠B=120°

∠C=90° and ∠D=60°

Question 26

In a triangle ABC, the incircle (centre O) touches BC, CA and AB at P, Q and R respectively. Calculate (i) ∠QOR (ii) ∠QPR given that ∠A = 60°.

Sol :

OQ and OR are the radii and AC and AB are tangents.

OQ ⊥ AC and OR ⊥ AB

Now in the quad. AROQ

$\angle \mathrm{A}=60^{\circ}, \angle \mathrm{ORA}=90^{\circ}$ and $\angle \mathrm{OQA}=90^{\circ}$

$\therefore \angle \mathrm{ROQ}=360^{\circ}-(\angle \mathrm{A}+\angle \mathrm{ORA}+\angle \text{OQA})$

$=360^{\circ}-\left(60^{\circ}+90^{\circ}+90^{\circ}\right)$

$=360^{\circ}-240^{\circ}=120^{\circ}$

Now arc RC subtends ∠ROQ at the centre and ∠RPQ at the remaining part of the circle.

$\therefore  \angle \mathrm{ROQ}=2 \angle \mathrm{RPQ}$

$\Rightarrow \angle \mathrm{RPQ}=\frac{1}{2} \angle \mathrm{ROQ}$

$\Rightarrow  \angle \mathrm{RPQ}=\frac{1}{2} \times 120^{\circ}=60^{\circ}$

Question 27

(a) In the figure (0 given below, AB is a diameter. The tangent at C meets AB produced at Q, ∠CAB = 34°. Find

(i) ∠CBA (ii) ∠CQA (2006)

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. Given ∠APB = 60°, calculate.

(i) ∠AOB (ii) ∠OAB (iii) ∠ACB.

Sol :

(a) AB is the diameter.

$\therefore \quad \angle \mathrm{ACB}=90^{\circ}$ and $\angle \mathrm{CAB}=34^{\circ}$

$\therefore \quad$ In $\Delta \mathrm{ABC}$

$\angle \mathrm{ACB}+\angle \mathrm{CAB}+\angle \mathrm{CBA}=180^{\circ}$

$\Rightarrow  90^{\circ}+34^{\circ}+\angle \mathrm{CBA}=180^{\circ}$

(b) (i) AP and BP are the tangents to the circle and OA and OB are radii on it.

∴OA⊥Ap and OB⊥BP

∴∠AOB=180°-60°=120°

(ii) Now in ΔOAB

OA=OB (radii of the same circle)

$\therefore \angle \mathrm{OAB}=\angle \mathrm{OBA}$

$=\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=\frac{1}{2} \times 60^{\circ}=30^{\circ}$

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

$\therefore \angle \mathrm{AOB}=2 \angle \mathrm{ACB}$

$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}$

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle

∴∠AOB=2∠ACB

$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}$

$\Rightarrow  \angle \mathrm{ACB}=\frac{1}{2} \times 120^{\circ}=60^{\circ}$ [From (i)]

Question 28

(a) In the figure (i) given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY. (1994)

(b) In the figure (ii) given below, O is the centre of the circle and PT is the tangent to the circle at P. Given ∠QPT = 30°, calculate (i) ∠PRQ (ii) ∠POQ.

Sol :

(a) Join OY, OX and OY are the radii of the circle

and XT and YT are the tangents to the circle.

$\therefore O X \perp X T$ and OY $\perp Y T$

and $\angle \mathrm{XTY}=80^{\circ}$

$\therefore \angle \mathrm{XOY}=180^{\circ}-\angle \mathrm{XTY}$

$=180^{\circ}-80^{\circ}=100^{\circ}$

Now $\angle \mathrm{ZOY}=360^{\circ}-(\angle \mathrm{XOZ}+\angle \mathrm{XOY})$

$=360^{\circ}-\left(140^{\circ}+100^{\circ}\right)$

$=360^{\circ}-240^{\circ}=120^{\circ}$

But arc ZY subtends $\angle \mathrm{ZOY}$ at the centree and $\angle \mathrm{ZXY}$ at the remaining part of the circle.

$\therefore \angle Z X Y=\frac{1}{2} \angle Z O Y$

$=\frac{1}{2} \times 120^{\circ}=60^{\circ}$

(b) (i) PT is tangent and OP is radius then OP⏊PT

$\therefore \angle \mathrm{OPT}=90^{\circ}$

But $\angle \mathrm{QPT}=30^{\circ}$

$\therefore  \angle \mathrm{OPQ}=90^{\circ}-30^{\circ}=60^{\circ}$

$\because  \mathrm{OP}=\mathrm{OQ}$

(radii of the same circle)

∴∠OQP=∠OPQ=60°

(ii) Take a point A on the circumference and join AP and AQ

Now arc $\mathrm{PQ}$ subtends $\angle \mathrm{POQ}$ at the centre and $\angle \mathrm{PAQ}$, on the circumference of the circle.

$\therefore \angle \mathrm{PAQ}=\frac{1}{2} \angle \mathrm{POQ}=\frac{1}{2} \times 60^{\circ}=30^{\circ}$

Now APRQ is a cyclic quadrilateral

$\therefore \angle \mathrm{PAQ}+\angle \mathrm{PRQ}=180^{\circ}$

$\Rightarrow \quad 30^{\circ}+\angle P R Q=180^{\circ}$

$\Rightarrow \angle P R Q=180^{\circ}-30^{\circ}=150^{\circ}$

Question 29

Two chords AB, CD of a circle intersect internally at a point P. If

(i) AP = cm, PB = 4 cm and PD = 3 cm, find PC.

(ii) AB = 12 cm, AP = 2 cm, PC = 5 cm, find PD.

(iii) AP = 5 cm, PB = 6 cm and CD = 13 cm, find CP.

Sol :

In a circle, two chords AB and CD intersect

each other at P internally.

$\therefore \mathrm{AP.PB}=\mathrm{CP.PD}$

(i) When AP=6 cm , PB=4 cm, PD=3 cm then

$\Rightarrow 6 \times 4=\mathrm{CP} \times 3$

$\Rightarrow \mathrm{CP}=\frac{6 \times 4}{3}=8 \mathrm{~cm}$

Hence PC=8 cm

(ii) When AB=12 cm, AP=2 cm, PC=5 cm

PB=AB-AP

=12-2=10

$\Rightarrow \quad 2 \times 10=5 \times \mathrm{PD}$

$\Rightarrow \quad P D=\frac{2 \times 10}{5}=4 \mathrm{~cm}$

(iii) When AP=5 cm, PB=6 cm, CD=13 cm

Let CP=x, then PD=CD-CP

or PD=13-x

$\therefore A P \times P B=C P \times P D$

$\Rightarrow 5 \times 6=x(13-x)$

$\Rightarrow 30=13 x-x^{2}$

$\Rightarrow x^{2}-13 x+30=0$

$\Rightarrow x^{2}-10 x-3 x+30=0$

$\Rightarrow x(x-10)-3(x-10)=0$

$\Rightarrow(x-10)(x-3)=0$

Either x-10=0, then x=10

or x-3=0, then x=3

$\therefore  \mathrm{CP}=10 \mathrm{~cm}$ or $3 \mathrm{~cm}$

Question 30

(a) In the figure (i) given below, PT is a tangent to the circle. Find TP if AT = 16 cm and AB = 12 cm.

(b) In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find (i) AB. (ii)the length of tangent PT.

Sol :

(a) PT is the tangent to the circle and AT is a secant.

PT² = TA × TB

Now TA = 16 cm, AB = 12 cm

TB = AT – AB = 16 – 12 = 4 cm

∴ PT² = 16 + 4 = 64 = (8)²

⇒ PT = 8 cm or TP = 8 cm

(b) PT is tangent and PDC is secant out to the circle

∴ PT² = PC × PD

$P T^{2}=(5+7.8) \times 5=12.8 \times 5$

$P T^{2}=64 \Rightarrow P T=8 \mathrm{~cm}$

In $\Delta$ OTP

$P T^{2}+O T^{2}=O P^{2}$

$8^{2}+x^{2}=(x+4)^{2}$

$\Rightarrow 64+x^{2}=x^{2}+16+8 x$

64-16=8 x 

$\Rightarrow 48=8$

$\Rightarrow x=\frac{48}{8}=6 \mathrm{~cm}$$

$\therefore$ Radius $=6 \mathrm{~cm}$

$\mathrm{AB}=2 \times 6=12 \mathrm{~cm}$

Question 31

PAB is secant and PT is tangent to a circle

(i) PT = 8 cm and PA = 5 cm, find the length of AB.

(ii) PA = 4.5 cm and AB = 13.5 cm, find the length of PT.

Sol :

∵ PT is the tangent and PAB is the secant of the circle.

$\therefore \mathrm{PT}^{2}=\mathrm{PA} . \mathrm{PB}$

(i) $\mathrm{PT}=8 \mathrm{~cm}, \mathrm{PA}=5 \mathrm{~cm}$

$\therefore(8)^{2}=5 \times \mathrm{PB} \Rightarrow 64=5 \mathrm{~PB}$

$\Rightarrow P B=\frac{64}{5}=12 \cdot 8 \mathrm{~cm}$

$\therefore A B=P B-P A=12 \cdot 8-5 \cdot 0$

=7.8 cm

(ii) $\mathrm{PT}^{2}=\mathrm{PA} \times \mathrm{PB}$

But $\mathrm{PA}=4 \cdot 5 \mathrm{~cm} , \mathrm{AB}=13 \cdot 5 \mathrm{~cm}$

$\therefore  P B=P A+A B=4 \cdot 5+13 \cdot 5=18 \mathrm{~cm}$

Now $P T^{2}=4 \cdot 5 \times 18=81=(9)^{2}$

∴PT=9 cm

Question 32

In the adjoining figure, CBA is a secant and CD is tangent to the circle. If AB = 7 cm and BC = 9 cm, then

(i) Prove that ∆ACD ~ ∆DCB.

(ii) Find the length of CD.

Sol :

In ∆ACD and ∆DCB

∠C = ∠C (common)

∠CAD = ∠CDB

[Angle between chord and tangent is equal io angle made by chord in alternate segment.

$\therefore \Delta \mathrm{ACD} \sim \Delta \mathrm{DCB}$

$\therefore \frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{DC}}{\mathrm{BC}}$

$\Rightarrow \mathrm{DC}^{2}=\mathrm{AC} \times \mathrm{BC}$

$=16 \times 9=144$

$\Rightarrow \mathrm{DC}=12 \mathrm{~cm}$

Question 33

(a) In the figure (i) given below, PAB is secant and PT is tangent to a circle. If PA : AB = 1:3 and PT = 6 cm, find the length of PB.

(b) In the figure (ii) given below, ABC is an isosceles triangle in which AB = AC and Q is mid-point of AC. If APB is a secant, and AC is tangent to the circle at Q, prove that AB = 4 AP.

Sol :

(a) In the figure (i),

PAB is secant and PT is the tangent to the circle.

PT² = PA × PB

But $\mathrm{PT}=6 \mathrm{~cm},$ and $\mathrm{PA}: \mathrm{AB}=1: 3$

Let $\mathrm{PA}=x,$ then $\mathrm{AB}=3 \mathrm{x}$

$\therefore \mathrm{PB}=\mathrm{PA}+\mathrm{AB}=x+3 x \doteq 4 x$

Now $\mathrm{PT}^{2}=\mathrm{PA} \times \mathrm{PB}$

$\Rightarrow  (6)^{2}=x(4 x)$

$\Rightarrow  36=4 x^{2}$

$\Rightarrow x^{2}=\frac{36}{4}=9=(3)^{2}$

$\Rightarrow  x=3$

$\therefore \mathrm{PB}=4 x=4 \times 3=12 \mathrm{~cm}$

(b) In the figure (ii)

$\Delta \mathrm{ABC}$ is an isosceles triangles in which

AB=AC

$\mathrm{Q}$ is mid-point of $\mathrm{AC}$

$\mathrm{APB}$ is the secant and $\mathrm{AC}$ is the tangent to the circle at $\mathrm{Q}$

To prove : $\mathrm{AB}=4$ AP

Proof $: \because \mathrm{AQ}$ is the tangent and $\mathrm{APB}$ is the secant

$\therefore \quad A Q^{2}=A P \times A B$

$\left(\frac{1}{2} \mathrm{AC}\right)^{2}=\mathrm{AP} \times \mathrm{AB}$

$\Rightarrow\left(\frac{1}{2} \mathrm{AB}\right)^{2}=\mathrm{AP} \times \mathrm{AB}$

$(\because A B=A C)$

$\Rightarrow  \frac{\mathrm{AB}^{2}}{4}=\mathrm{AP} \times \mathrm{AB}$

$\Rightarrow \mathrm{AB}^{2}=4 \mathrm{AP} \times \mathrm{AB}$

$\Rightarrow \frac{\mathrm{AB}^{2}}{\mathrm{AB}}=4 \mathrm{AP}$

$\Rightarrow \mathrm{AB}=4 \mathrm{AP}$

Hence proved

Question 34

Two chords AB, CD of a circle intersect externally at a point P. If PA = PC, Prove that AB = CD.

Sol :

Given: Two chords AB and CD intersect

each other at P outside the circle. PA = PC.

To Prove : AB=CD

Proof : Chords AB and CD intersect each other at P outside the circle

$\therefore \mathrm{PA} \times \mathrm{PB}=\mathrm{PC} \times \mathrm{PD}$

But $\mathrm{PA}=\mathrm{PC}$ (given)...(i)

$\therefore  P B=P D$...(ii)

Subtracting (ii) from (i)

PA-PB=PC-PD

⇒AB=CD Q.E.D

Question 35

(a) In the figure (i) given below, AT is tangent to a circle at A. If ∠BAT = 45° and ∠BAC = 65°, find ∠ABC.

(b) In the figure (ii) given below, A, B and C are three points on a circle. The tangent at C meets BA produced at T. Given that ∠ATC = 36° and ∠ACT = 48°, calculate the angle subtended by AB at the centre of the circle. (2001)

Sol :

(a) AT is the tangent to the circle at A

and AB is the chord of the circle.

Figure to be added

$\therefore \angle \mathrm{ACB}=\angle \mathrm{BAT}$

(Angles in the alternate segment)

=45°

Now in ΔABC

$\angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180^{\circ}$

(Angles of a triangle)

$\Rightarrow \angle \mathrm{ABC}+65^{\circ}+45^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{ABC}+110^{\circ}=180^{\circ}$

$\therefore \angle \mathrm{ABC}=180^{\circ}-110^{\circ}=70^{\circ}$

(b) Join OA, OB and CB. 

In ΔATC

Ext.∠CAB=∠ATC+∠TCA

$=36^{\circ}+48^{\circ}=84^{\circ}$

But $\angle \mathrm{TCA}=\angle \mathrm{ABC}$

(Angles in the alternate segment)

$\therefore \angle \mathrm{ABC}=48^{\circ}$

But in $\triangle \mathrm{ABC}$

$\angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180^{\circ}$

$\Rightarrow 48^{\circ}+84^{\circ}+\angle \mathrm{ACB}=180^{\circ}$

$\Rightarrow 132^{\circ}+\angle \mathrm{ACB}=180^{\circ}$

$\Rightarrow  \angle \mathrm{ACB}=180^{\circ}-132^{\circ}$

$\Rightarrow \angle \mathrm{ACB}=48^{\circ}$

$\because$ Arc AB subtends $\angle A O B$ at the centre and $\angle \mathrm{ACB}$ at the remaining part of the circle

$\therefore \angle A O B=2 \angle A C B=2 \times 48^{\circ}=96^{\circ}$

Question 36

In the adjoining figure ∆ABC is isosceles with AB = AC. Prove that the tangent at A to the circumcircle of ∆ABC is parallel to BC.

Sol :

Given: ∆ABC is an isosceles triangle with AB = AC.

AT is the tangent to the circumcircle at A.

T prove : AT || BC

AB=AC (given)

∴∠C=∠B

(Angles opposite to equal sides)

But AT is the tangent and $\mathrm{AC}$ is the chord.

$\therefore \angle \mathrm{TAC}=\angle \mathrm{B}$

(Angles in the alternate segment)

But $\angle \mathrm{B}=\angle \mathrm{C} \quad$ (proved)

$\therefore \angle \mathrm{TAC}=\angle \mathrm{C}$

But these are alternate angles

$\therefore  \mathrm{AT} \| \mathrm{BC}$  Q.E.D

Question 37

If the sides of a rectangle touch a circle, prove that the rectangle is a square.

Sol :

Given: A circle touches the sides AB, BC, CD and DA

of a rectangle ABCD at P, Q, R and S respectively.

To Prove : ABCD is a square.

Proof : Tangents from a point to the circle are equal

$\therefore \mathrm{AP}=\mathrm{AS}$

Similarly $\mathrm{BP}=\mathrm{BQ}$
$\mathrm{CR}=\mathrm{CQ}$ and $\mathrm{DR}=\mathrm{DS}$

Adding, we get $\mathrm{AP}+\mathrm{BP}+\mathrm{CR}+\mathrm{DR}=\mathrm{AS}+\mathrm{BQ}+\mathrm{CQ}+\mathrm{DS}$

$\Rightarrow \mathrm{AP}+\mathrm{BP}+\mathrm{CR}+\mathrm{DR}=\mathrm{AS}+\mathrm{DS}+\mathrm{BQ}+\mathrm{CQ}$

$\Rightarrow A B+C D=A D+B C$

But $\mathrm{AB}=\mathrm{CD}$ and $\mathrm{AD}=\mathrm{BC}$

(opposite sides of a rectangle)

$\therefore A B+A B=B C+B C$

$\Rightarrow 2 \mathrm{AB}=2 \mathrm{BC} \Rightarrow \mathrm{AB}=\mathrm{BC}$

$\therefore A B=B C=C D=D A$

Hence ABCD is a square

Question 38

(a) In the figure (i) given below, two circles intersect at A, B. From a point P on one of these circles, two line segments PAC and PBD are drawn, intersecting the other circle at C and D respectively. Prove that CD is parallel to the tangent at P.

(b) In the figure (ii) given below, two circles with centres C, C’ intersect at A, B and the point C lies on the circle with centre C’. PQ is a tangent to the circle with centre C’ at A. Prove that AC bisects ∠PAB.

Sol :

Given : Two circles intersect each other at A and B.

From a point P on one circle, PAC and PBD are drawn.

From P, PT is a tangent drawn. CD is joined.

To Prove : PT || CD.

Construction : Join AB.

Proof: $\mathrm{PT}$ is tangent and $\mathrm{PA}$ is chord.

$\therefore \angle \mathrm{APT}=\angle \mathrm{ABP}$

(Angles in the alternate segments)...(i)

But BDCA is a cyclic quadrilateral

$\therefore$ Ext. $\angle \mathrm{ABP}=\angle \mathrm{ACD}$...(ii)

From (i) and (ii)

$\angle \mathrm{APT}=\angle \mathrm{ACD}$

But these are alternate angles

$\therefore \mathrm{CD} \| \mathrm{PT}$  Q.E.D

Given : Two circles with centres $C$ and $C^{\prime}$ intersect each other at $\mathrm{A}$ and $\mathrm{B}$. $\mathrm{PQ}$ is a tangent to the circle with centre $\mathrm{C}^{\prime}$ at $\mathrm{A}$.

To Prove : AC bisects $\angle \mathrm{PAB}$

Construction : Join $\mathrm{AB}$ and $\mathrm{CP}$

Proof : In $\triangle \mathrm{ACB}$

AC=BC (radii of the same circle) 

$\therefore \angle \mathrm{BAC}=\angle \mathrm{ABC}$...(i)

PAQ is tangent and AC is the chord of the circle

$\therefore \angle \mathrm{PAC}=\angle \mathrm{ABC}$...(ii)

From (i) and (ii)

$\angle B A C=\angle P A C$

Hence $\mathrm{AC}$ is the bisector of $\angle \mathrm{PAB}$

Q.E.D

Question 39

(a) In the figure (i) given below, AB is a chord of the circle with centre O, BT is tangent to the circle. If ∠OAB = 32°, find the values of x and y.

(b) In the figure (ii) given below, O and O’ are centres of two circles touching each other externally at the point P. The common tangent at P meets a direct common tangent AB at M. Prove that:

(i) M bisects AB (ii) ∠APB = 90°.

Sol :

AB is a chord of a circle with centre O.

BT is a tangent to the circle and ∠OAB = 32°.

∴ In ∆OAB,

OA = OB (radii of the same circle)

Figure to be added

$\therefore \angle \mathrm{OAB}=\angle \mathrm{OBA}$

$\therefore  x=32^{\circ}$

and $\angle \mathrm{AOB}=180^{\circ}-\left(x+32^{\circ}\right)$

$=180^{\circ}-64^{0^{\circ}}=116^{\circ}$

Now arc $\mathrm{AB}$ subtends $\angle \mathrm{AOB}$ at the centre and $\angle \mathrm{ACB}$ at the remaining part of the circle.

$\therefore \angle \mathrm{AOB}=2 \angle \mathrm{ACB}$

$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}$

$=\frac{1}{2} \times 116^{\circ}=58^{\circ}$

Given : Two circles with centre $\mathrm{O}$ and $\mathrm{O}^{\prime}$ touch each other at P externally. From P, a common tangent is drawn and meets the common direct tangent $\mathrm{AB}$ at $\mathrm{M}$.

To Prove : $(i)$ M bisects AB i.e. AM = MB.

(ii) $\angle \mathrm{APB}=90^{\circ}$

Proof: $(i)$ From $M,$ MA and $M P$ are the tangents.

$\therefore \mathrm{MA}=\mathrm{MP}$...(i)

Similarly $\mathrm{MB}=\mathrm{MP}$...(ii)

From (i) and (ii)

MA=MP

or M is the mid-point of AB

(ii) $\because \mathrm{MA}=\mathrm{MP}$

$\therefore  \angle \mathrm{MAP}=\angle \mathrm{MPA}$...(i)

Similarly $\mathrm{MP}=\mathrm{MB}$

$\therefore  \angle \mathrm{MPB}=\angle \mathrm{MBP}$...(ii)

Adding (i) and (ii)

Ext. $\angle \mathrm{TCQ}=\angle \mathrm{TAC}$...(ii)

∴From (i) and (ii)

$\angle \mathrm{TCQ}=\angle \mathrm{ATP}$...(iii)

(ii) $\angle \mathrm{ATP}=\angle \mathrm{ACT}$...(iv)

(Angles in the alternate segments)

From (iii) and (iv)

$\therefore \quad \angle \mathrm{TCQ}=\angle \mathrm{ACT}$

Hence CT is the bisector of $\angle \mathrm{ACQ}$

(iii) Arc AT subtends ∠AOT at the centre and ∠ACT at the remaining part of the circle

$\therefore \angle \mathrm{AOT}=2 \angle \mathrm{ACT}$

$=\angle \mathrm{ACT}+\angle \mathrm{ACT}$

$=\angle \mathrm{ACT}+\angle \mathrm{TCQ}$

$(\because \angle \mathrm{TCQ}=\angle \mathrm{ACT})$

=∠ACQ  (Q.E.D)

$\angle \mathrm{MAP}+\angle \mathrm{MPB}=\angle \mathrm{MPA}+\angle \mathrm{MBP}$

$\Rightarrow \angle \mathrm{APB}=\angle \mathrm{MAP}+\angle \mathrm{MBP}$

But $\angle \mathrm{APB}+\angle \mathrm{MAP}+\angle \mathrm{MBP}=180^{\circ}$

(Angles of a triangle)

$\Rightarrow \angle \mathrm{APB}+\angle \mathrm{APB}=180^{\circ}$

$\Rightarrow 2 \angle \mathrm{APB}=180^{\circ}$

$\Rightarrow \quad \angle \mathrm{APB}=\frac{180^{\circ}}{2}=90^{\circ}$

Q.E.D