ML Aggarwal Solution Class 10 Chapter 15 Circles Test

 Test

Question 1

(a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC.

(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD

Sol :

(a) ∆ABC is an equilateral triangle.

$\therefore$ Each angle $=60^{\circ}$

$\therefore \angle A=60^{\circ}$

But $\angle \mathrm{A}=\angle \mathrm{D}$

(Angles in the same segment)

∴∠D=60°

Now ABEC is a cyclic quadrilateral

$\therefore \angle A+\angle E=180^{\circ}$

$\Rightarrow 60^{\circ}+\angle \mathrm{E}=180^{\circ}$

$\Rightarrow \angle \mathrm{E}=180^{\circ}-60^{\circ}$

$\therefore \angle E=120^{\circ}$

Hence $\angle \mathrm{BDC}=60^{\circ}$ and $\angle \mathrm{BEC}=120^{\circ}$

(b) AB is diameter of circle with centre O

$\mathrm{OD} \perp \mathrm{AB}$ and $\mathrm{C}$ is a point on arc $\mathrm{DB}$.

(i) $\operatorname{In} \Delta \mathrm{AOD}, \angle \mathrm{AOD}=90^{\circ}$

and OA=OD (radii of the semi-circle)

$\therefore  \angle \mathrm{OAD}=\angle \mathrm{ODA}$

But $\angle \mathrm{OAD}+\angle \mathrm{ODA}=90^{\circ}$

$\Rightarrow \angle \mathrm{OAD}+\angle \mathrm{OAD}=90^{\circ}$

$\Rightarrow 2 \angle \mathrm{OAD}=90^{\circ}$

$\therefore  \angle \mathrm{OAD}=\frac{90^{\circ}}{2}=45^{\circ}$

or $\angle \mathrm{BAD}=45^{\circ}$

(ii) Arc AD subtends $\angle \mathrm{AOD}$ at the centre and $\angle \mathrm{ACD}$ at the remaining part of the circle.

$\therefore \angle \mathrm{AOD}=2 \angle \mathrm{ACD}$

$\Rightarrow \quad 90^{\circ}=2 \angle \mathrm{ACD}$ $(\because O D \perp A B)$

$\Rightarrow \angle \mathrm{ACD}=\frac{90^{\circ}}{2}=45^{\circ}$

Question 2

(a) In the figure given below, AB is a diameter of the circle. If AE = BE and ∠ADC = 118°, find

(i) ∠BDC (ii) ∠CAE

(b) In the figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.

Sol :

(a) Join DB, CA and CB.

∠ADC = 118° (given)

and ∠ADB = 90°

(Angles in a semi-circle)

$\therefore \angle \mathrm{BDC}=\angle \mathrm{ADC}-\angle \mathrm{ADB}$

$=118^{\circ}-90^{\circ}=28^{\circ}$

ABCD is a cyclic quadrilateral

Figure to be added

$\therefore \angle \mathrm{ADC}+\angle \mathrm{ABC}=180^{\circ}$

$\Rightarrow 118^{\circ}+\angle \mathrm{ABC}=180^{\circ}$

$\therefore \angle \mathrm{ABC}=180^{\circ}-118^{\circ}=62^{\circ}$

But in $\Delta \mathrm{AEB}$

$\angle A E B=90^{\circ}$

(Angle in a semi-circle)

and $\angle \mathrm{EAB}=\angle \mathrm{ABE} \quad(\because \mathrm{AE}=\mathrm{BE})$

But $\angle \mathrm{EAB}+\angle \mathrm{ABE}=90^{\circ}$

$\therefore \quad \angle \mathrm{EAB}=90^{\circ} \times \frac{1}{2}=45^{\circ}$

$\therefore \angle \mathrm{CBE}=\angle \mathrm{ABC}+\angle \mathrm{ABE}$

$=62^{\circ}+45^{\circ}=107^{\circ}$

But AEBD is a cyclic quadrilateral

$\therefore \angle \mathrm{CAE}+\angle \mathrm{CBE}=180^{\circ}$

$\Rightarrow  \angle \mathrm{CAE}+107^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{CAE}=180^{\circ}-107^{\circ}=73^{\circ}$

(b) $\mathrm{AB}$ is the diameter of semi-circle $\mathrm{ABCDE}$ with centre $\mathrm{O} \cdot \mathrm{AE}=\mathrm{ED}$ and $\angle \mathrm{BCD}=140^{\circ}$

In cyclic quadrilateral EBCD

Figure to be added

(i)

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

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

$\Rightarrow \angle \mathrm{BED}=180^{\circ}-140^{\circ}=40 \mathrm{P}$

But $\angle \mathrm{AEB}=90^{\circ}$

(Angles in semi-circle)

$\therefore \angle \mathrm{AED}=\angle \mathrm{AEB}+\angle \mathrm{BED}$

$=90^{\circ}+40^{\circ}=130^{\circ}$

(ii)

Now in cyclic quadrilateral AEDB $\angle \mathrm{AED}+\angle \mathrm{DBA}=180^{\circ}$

$\Rightarrow 130^{\circ}+\angle \mathrm{DBA}=180^{\circ}$

$\Rightarrow \angle \mathrm{DBA}=180^{\circ}-130^{\circ}=50^{\circ}$

$\because$ Chord AE=ED  (given)

$\therefore \quad \angle \mathrm{DBE}=\angle \mathrm{EBA}$

But $\angle \mathrm{DBE}+\angle \mathrm{EBA}=50^{\circ}$

$\Rightarrow \angle \mathrm{DBE}+\angle \mathrm{DBE}=50^{\circ}$

$\Rightarrow 2 \angle \mathrm{DBE}=50^{\circ}$

$\therefore \angle \mathrm{DBE}=25^{\circ}$ or $\angle \mathrm{EBD}=25^{\circ}$

In $\Delta \mathrm{OEB}, \mathrm{OE}=\mathrm{OB}$

(radii of the same circle)

$\therefore \angle \mathrm{OEB}=\angle \mathrm{EBO}=\angle \mathrm{DBE}$

But these are alternate angles.

$\therefore \mathrm{OE}|| \mathrm{BD}$ Q.E.D

Question 3

(a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC).

(b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z.

Sol :

(a) Given : O is the centre of the circle.

To Prove : ∠AOC = 2 (∠ACB + ∠BAC).

Proof : In ∆ABC,

∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle)

$\Rightarrow \angle \mathrm{ABC}=180^{\circ}-(\angle \mathrm{ACB}+\angle \mathrm{BAC}) \ldots(i)$

In the circle, arc $A C$ subtends $\angle A O C$ at the centre and $\angle \mathrm{ABC}$ at the remaining part of the circle.

Reflex $\angle \mathrm{AOC}=2 \angle \mathrm{ABC}$...(ii)

Reflex $\angle \mathrm{AOC}=2\left[180^{\circ}-(\mathrm{ACB}+\mathrm{BAC})\right]$

Reflex $\angle \mathrm{AOC}=360^{\circ}-2(\angle \mathrm{ACB}+\angle \mathrm{BAC})$

But $\angle \mathrm{AOC}=360^{\circ}-$ reflex $\angle \mathrm{AOC}$

Figure to be added

$=360-\left[360^{\circ}-2(\angle \mathrm{ACB}+\angle \mathrm{BAC})\right.$

$=360^{\circ}-360^{\circ}+2(\angle \mathrm{ACB}+\angle \mathrm{BAC})$

$=2(\angle \mathrm{ACB}+\angle \mathrm{BAC})$

Hence $\angle \mathrm{AOC}=2(\angle \mathrm{ACB}+\angle \mathrm{BAC})$

(Q.E.D)

(b) Given : In the figure , O is the centre of the circle

To prove : x+y+z

Proof : Arc BC subtends $\angle \mathrm{AOB}$ at the centre and $\angle \mathrm{BEC}$ at the remaining part of the circle.

$\therefore \angle \mathrm{BOC}=2 \angle \mathrm{BEC}$

But $\angle \mathrm{BEC}=\angle \mathrm{BDC}$

(Angles in the same segment) $\therefore \quad \angle B O C=\angle B E C+\angle B D C$

In $\Delta \mathrm{ABD},$ Ext. $\angle y=(\angle \mathrm{EBD}+\angle \mathrm{BEC})$

$\angle \mathrm{BEC}=y-\angle \mathrm{EBD}$

Similarly in $\Delta \mathrm{ABD}$

Ext. $\angle \mathrm{BDC}=x+\angle \mathrm{ABD}$

$=x+\angle E B D$

Substituting the value of $(i i)$ and (iii) in (i) $\angle \mathrm{BOC}=y-\angle \mathrm{EBD}+x+\angle \mathrm{EBD}=x+y$

$\Rightarrow \quad z=x+y$

Q.E.D

Question 4

(a) In the figure (i) given below, AB is the diameter of a circle. If DC is parallel to AB and ∠CAB = 25°, find :

(i)∠ADC (ii) ∠DAC.

(b) In the figure (ii) given below, the centre O of the smaller circle lies on the circumference of the bigger circle. If ∠APB = 70° and ∠BCD = 60°, find :

(i) ∠AOB (ii) ∠ACB

(iii) ∠ABD (iv) ∠ADB.

Sol :

(a) AB is diameter and DC || AB,

∠CAB = 25°, join AD,BD

$\angle \mathrm{BAC}=\angle \mathrm{BDC}$

(Angles in the same segment)

But $\angle \mathrm{ADB}=90^{\circ}$

(Angles in a semi-circle) $\begin{aligned} \therefore \angle \mathrm{ADC}=\angle \mathrm{ADB}+\angle \mathrm{BAC} &=90^{\circ}+25^{\circ} \\ &=115^{\circ} \end{aligned}$

$\because \quad$ DC $\| \mathrm{AB} \quad$ (given)

$\therefore \angle \mathrm{CAB}=\angle \mathrm{ACD} \quad$ (alternate angles)

$\therefore \angle \mathrm{ACD}=25^{\circ}$

Now in $\Delta \mathrm{ACD}$

$\angle \mathrm{DAC}+\angle \mathrm{ADC}+\angle \mathrm{ACD}=180^{\circ}$

(Angles of a triangle)

$\Rightarrow \angle \mathrm{DAC}+115^{\circ}+25^{\circ}=180^{\circ}$

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

$\Rightarrow \angle \mathrm{DAC}=180^{\circ}-140^{\circ}=40^{\circ}$

(b)

(i) Arc AB subtends $\angle \mathrm{AOB}$ at the centre and $\angle \mathrm{APB}$ at the remaining part of the circle.

$\therefore \angle \mathrm{AOB}=2 \angle \mathrm{APB}=2 \times 70^{\circ}=140^{\circ}$

(ii)

$\mathrm{AOBC}$ is a cyclic quadrilateral.

$\therefore \angle \mathrm{AOB}+\angle \mathrm{ACB}=180^{\circ}$

$\Rightarrow \quad 140^{\circ}+\angle \mathrm{ACB}=180^{\circ}$

$\Rightarrow \angle \mathrm{ACB}=180^{\circ}-140^{\circ}=40^{\circ}$

(iii)In cyclic quadrilateral ABDC

$\angle \mathrm{ABD}+\angle \mathrm{ACD}=180^{\circ}$

$\Rightarrow \angle A B D+\angle A C B+\angle B C D=180^{\circ}$

$\Rightarrow \angle A B D+40^{\circ}+60^{\circ}=180^{\circ}$

$\Rightarrow \angle A B D=180^{\circ}-100^{\circ}=80^{\circ}$

(iv)

$\angle \mathrm{ADB}=\angle \mathrm{ACB}$

(Angles in the same segement)

$\therefore \angle \mathrm{ADB}=40^{\circ}$

Question 5

(a) In the figure (i) given below, ABCD is a cyclic quadrilateral. If AB = CD, Prove that AD = BC.

(b) In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ∠ABC = 50°, find ∠BDC and ∠BEC.

Sol :

(a) Given : ABDC is a cyclic quadrilateral AB = CD.

To Prove: AD = BC.

Figure to be added

Construction : Join AD and BC.

Proof: $\ln \Delta \mathrm{ABD}$ and $\Delta \mathrm{CBD}$

AB=CD (given)

BD=BD (common)

$\angle \mathrm{BAD}=\angle \mathrm{BCD}$

(Angles in the same segment) $\therefore \Delta \mathrm{ABC} \cong \Delta \mathrm{CBD}$

(SSA axiom of congruency) $\therefore B C=A D$ Q.E.D.

(b) Given : $\triangle \mathrm{ABC}$ is an isosceles triangle and

$\angle \mathrm{ABC}=50^{\circ}$

In $\triangle \mathrm{ABC},$ an isosceles triangle

$\Rightarrow \angle \mathrm{ACB}=\angle \mathrm{ABC}$

$(\because$ opp. $\angle s$ of an isosceles $\Delta s)$

$\Rightarrow \angle A C B=50^{\circ}$

Also, in $\triangle \mathrm{ABC}$

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

(Sum of an isosceles triangle is $180^{\circ}$ )

$\therefore 50^{\circ}+50^{\circ}+\angle \mathrm{BAC}=180^{\circ}$

$\Rightarrow \angle \mathrm{BAC}=180^{\circ}-100^{\circ}$

$\Rightarrow \angle \mathrm{BAC}=80^{\circ}$

Also $\angle B D C=\angle B A C$

$\angle B D C=80^{\circ}$

(Angles in the same segment)

Now ABEC is a cyclic quadrilateral

$\therefore \angle A+E=180^{\circ}$

$\Rightarrow 80^{\circ}+\angle E=180^{\circ}$

$\Rightarrow \angle E=180^{\circ}-80^{\circ}$

$\therefore \angle E=100^{\circ}$

Hence $\angle \mathrm{BDC}=80^{\circ}$ and $\angle \mathrm{BEC}=100^{\circ}$

Question 6

A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.

Sol :

Join OT, OP = 13 cm and TP = 12 cm

$\because \mathrm{OT}$ is the radius

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

Now in right  ΔOTP

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

$\Rightarrow  (13)^{2}=0 \mathrm{~T}^{2}+(12)^{2}$

$\Rightarrow   169=\mathrm{OT}^{2}+144$

$\Rightarrow  \mathrm{OT}^{2}=169-144=25=(5)^{2}$

$\therefore   \mathrm{OT}=5 \mathrm{~cm}$

The nearest point is A from P to cut circle over OA= radius of the circle =5 cm.

$\therefore A P=O P-O A=13-5=8 \mathrm{~cm}$

Question 7

Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.

Sol :

Given: Two circles with centre O and O’

touch each other internally at P.

PT is any point on the common tangent of circles at P. From T, TA and TB are the tangents drawn to two circles.

To prove : TA=TB

Proof : From T, TA and TP are the tangents to the first circle

∴TA=TP...(i)

Similarly, from T, TB and TP are the tangents to the second circle

∴TB=TP...(ii)

From (i) and (ii)

TA=TB Q.E.D

Question 8

From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that

(i) ∠AOP = ∠BOP.

(ii) OP is the perpendicular bisector of the chord AB.

Sol :

Given: From a point P, outside the circle with centre O.

PA and PB are the tangents to the circle,

OA, OB and OP are joined.

To Prove :

(i) $\angle \mathrm{AOP}=\angle \mathrm{BOP}$

(ii) OP is the perpendicular bisector of the chord $\mathrm{AB}$

Construction : Join AB which intersects OP at M.

Proof : In right $\Delta \mathrm{OAP}$ and $\Delta \mathrm{OBP}$

Hyp. OP=OP (common) 

Side OA=OB  (radii of the same circle)

$\therefore \Delta \mathrm{OAP} \cong \Delta \mathrm{OBP}$

(R.H.S. Axiom of congruency)

$\therefore \angle \mathrm{AOP}=\angle \mathrm{BOP}$ (C.P.C.T)

and $\angle \mathrm{APO}=\angle \mathrm{BPO}$ (C.P.C.T)

Now in ΔAPM and ΔBPM

PM=PM (common)

∠APM=∠BPM (proved)

AP=BP (tangents from P to the circle)

$\therefore \Delta \mathrm{APM} \cong \Delta \mathrm{BPM}$

(SAS axiom of congruency)

$\therefore  \mathrm{AM}=\mathrm{BM}$ (C.P.C.T)

and $\angle \mathrm{AMP}=\angle \mathrm{BMP}$

But $\angle \mathrm{AMP}+\angle \mathrm{BMP}=180^{\circ}$

$\therefore \angle \mathrm{AMP}=\angle \mathrm{BMP}=90^{\circ}$

∴OP is perpendicular bisector of AB at M

Q.E.D

Question 9

(a) The figure given below shows two circles with centres A, B and a transverse common tangent to these circles meet the straight line AB in C. Prove that:

AP : BQ = PC : CQ.

(b) In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.

Sol :

(a) Given : Two circles with centres A and B

and a transverse common tangent to these circles meets AB at C.

To Prove : $A P: B Q=P C: C Q$

Proof: In $\Delta \mathrm{APC}$ and $\Delta \mathrm{BQC}$

$\angle \mathrm{PCA}=\angle \mathrm{QCB}$

(vertically opposite angles)

$\angle \mathrm{APC}=\angle \mathrm{BQC}$ (each 90°)

$\therefore \Delta \mathrm{APC} \sim \Delta \mathrm{BQC}$

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

$\Rightarrow \mathrm{AP}: \mathrm{BQ}=\mathrm{PC}: \mathrm{CQ}$

Q.E.D

(b) Given : In the figure, O is the centre of the circle. AB is diameter. PQ is the tangent and QA || PO

To prove : PB is tangent to the circle

Construction : Join OQ

Proof : In ΔOAQ

OQ=OA (Radii of the same circle)

∴∠OQA=∠OAQ

∵QA||PQ

∴∠OAQ=∠POB (corresponding angles)

and ∠OQA=∠QOP (alternate angles)

But ∠QAO=∠OQA (proved)

∴∠POB=∠QOB

Now , in ΔOPQ and ΔOBP

OP=OP (common)

OQ=OB (Radii of the same circle)

∠BOP=∠POB (Proved)

$\therefore \quad \Delta \mathrm{OPQ} \cong \Delta \mathrm{OBP}$  (SAS axiom)

$\therefore \angle \mathrm{OQP}=\angle \mathrm{OBP}$ (c.p.c.t)

But $\angle \mathrm{OQP}=90^{\circ}$

$(\because \mathrm{PQ}$ is tangent and $\mathrm{OQ}$ is the radius)

∴∠OBP=90°

∴PB is the tangent of the circle

Question 10

In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.

Sol :

In the given figure, two chords with centre A and B touch externally.

PM is a tangent to the circle with centre A

and QN is tangent to the circle with centre B.

PM = 15 cm, QN = 12 cm, PA = 17 cm, QB = 13 cm.

We have to find AB.

AM is radius and PM is tangent

$\therefore A M \perp P M$

Similarly, BN⊥NQ 

Now in right $\Delta$ APM, 

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

$\Rightarrow 17^{2}=A M^{2}+15^{2}$

$\Rightarrow A M^{2}=17^{2}-15^{2}$

$=289-225=64=(8)^{2}$

$\therefore A M=8 \mathrm{~cm}$

Similarly in right $\Delta \mathrm{BNQ}$ $\mathrm{BQ}^{2}=\mathrm{BN}^{2}+\mathrm{NQ}^{2}$

$13^{2}=\mathrm{BN}^{2}+12^{2}$

$\Rightarrow 169=\mathrm{BN}^{2}+144$

$\mathrm{BN}^{2}=169-144=25=(5)^{2}$

$\therefore \mathrm{BN}=5 \mathrm{~cm}$

Now AB=AM+BN

(AR=AM and BR=BN)

=8+5=13 cm

Question 11

Two chords AB, CD of a circle intersect externally at a point P. If PB = 7 cm, AB = 9 cm and PD = 6 cm, find CD.

Sol :

∵ AB and CD are two chords of a circle

which intersect each other at P, outside the circle.

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

$\mathrm{PB}=7 \mathrm{~cm}, \mathrm{AB}=9 \mathrm{~cm}, \mathrm{PD}=6 \mathrm{~cm}$

$\mathrm{AP}=\mathrm{AB}+\mathrm{BP}=9+7=16 \mathrm{~cm}$

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

$\Rightarrow 16 \times 7=\mathrm{PC} \times 6$

$\mathrm{PC}=\frac{16 \times 7}{6}=\frac{56}{3} \mathrm{~cm}$

$\therefore  \mathrm{CD}=\mathrm{PC}-\mathrm{PD}$

$=\frac{56}{3}-6=\frac{38}{3}=12 \frac{2}{3} \mathrm{~cm}$

Question 12

(a) In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle

(b) In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also, find the length of the tangent drawn from P to the circle. 

Sol :

Given : (a) AB is a chord of a circle with centre O

and PT is tangent and CD is the diameter of the circle

which meet at P.

AP = 16 cm, AB = 12 cm, OP = 2 cm

∴PB = PA – AB = 16 – 12 = 4 cm

∵ABP is a secant and PT is tangent.

∴PT² = PA × PB.

$=16 \times 4=64=(8)^{2}$

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

Again $P T^{2}=P D \times P C$

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

$\Rightarrow \mathrm{PC}=\frac{8 \times 8}{2}$

$\Rightarrow \mathrm{PC}=32 \mathrm{~cm}$

$\therefore C D=P C-P D=32-2=30 \mathrm{~cm}$

Radius of the circle $=\frac{30}{2}=15 \mathrm{~cm}$

(b) Chord $\mathrm{AB}$ and diameter $\mathrm{CD}$ intersect each other at Poutside the circle. $A B=8 \mathrm{~cm}$ $\mathrm{BP}=6 \mathrm{~cm}, \mathrm{PD}=4 \mathrm{~cm}$

PT is the tangent to the circle drawn from P

∵ Two chords AB and CD intersect each other at P outside the circle.

PA=AB+PB

=8+6=14 cm

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

$\Rightarrow  14 \times 6=P C \times 4$

$\Rightarrow P C=\frac{14 \times 6}{4}=\frac{84}{4}=21 \mathrm{~cm}$

$\therefore \mathrm{CD}=\mathrm{PC}-\mathrm{PD}=21-4=17 \mathrm{~cm}$

$\therefore$ Radius of the circle $=\frac{17}{2}=8 \cdot 5 \mathrm{~cm}$

(ii) Now PT is the tangent and ABP is secant

$\therefore \mathrm{PT}^{2}=\mathrm{PA} \times \mathrm{PB}=14 \times 6=84$

$\mathrm{PT}=\sqrt{84}=\sqrt{4 \times 21}=2 \sqrt{21} \mathrm{~cm}$

Question 13

In the figure given below, chord AB and diameter PQ of a circle with centre O meet at X. If BX = 5 cm, OX = 10 cm and.the radius of the circle is 6 cm, compute the length of AB. Also, find the length of tangent drawn from X to the circle.

Sol :

Chord AB and diameter PQ meet at X

on producing outside the circle

$\mathrm{BX}=5 \mathrm{~cm}, \mathrm{OX}=10 \mathrm{~cm}$ and radius of the

circle $=6 \mathrm{~cm}$

$\therefore \mathrm{XP}=\mathrm{XO}+\mathrm{OP}=10+6=16 \mathrm{~cm}$

$\therefore  X B . X A=X P . X Q$

$\Rightarrow 5.X A=16 \times 4$

$\Rightarrow \quad X A=\frac{16 \times 4}{5}=\frac{64}{5} \mathrm{~cm}$

$X A=12 \frac{4}{5} c m$

$\therefore A B=X A-X B=12 \frac{4}{5}-5=7 \frac{4}{5} \mathrm{~cm}$

=7.8 cm

Now ∵XT is the tangent of the circle

$\therefore X T^{2}=X P \cdot X Q=16 \times 4=64=(8)^{2}$

XT=8 cm

Question 14

(a) In the figure (i) given below, ∠CBP = 40°, ∠CPB = q° and ∠DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that ∠AQD = 2 q° and the points C, P, B and Q are concyclic, find the values of p and q.

(b) In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, ∠AOB = 130° and ∠ACE = 20°, find:

(i)∠BEC (ii) ∠ACB

(iii) ∠BCD (iv) ∠CED.

Sol :

(a) (i) Given : ABCD is a cyclic quadrilateral.

Ext. ∠PBC = ∠ADC

⇒ 40° = ∠ADC

In $\Delta \mathrm{ADP}$

$p+q+\angle \mathrm{ADP}=180^{\circ}$

$p+q=180^{\circ}-\angle \mathrm{ADP}=180^{\circ}-\angle \mathrm{ADC}$

$=180^{\circ}-40^{\circ}=140^{\circ}$

$\therefore p+q=140^{\circ}$

(ii) $\because \mathrm{C}, \mathrm{P}, \mathrm{B}, \mathrm{Q}$ are concyclic

$\therefore \angle C P B+\angle C Q B=180^{\circ}$

$\Rightarrow q+2 q=180^{\circ}$ $(\because \angle \mathrm{CQB}=\angle \mathrm{DQA})$

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

$\therefore q=\frac{180^{\circ}}{3}=60^{\circ}$

But p+q=140°

$\therefore p+60^{\circ}=140^{\circ}$

$\Rightarrow p=140^{\circ}-60^{\circ}=80^{\circ}$

Hence $p=80^{\circ}, q=60^{\circ}$

(b) $\angle \mathrm{AOB}=130^{\circ}$

But $\angle \mathrm{AOB}+\angle \mathrm{BOC}=180^{\circ}$

(Linear pair)

$\Rightarrow 130^{\circ}+\angle \mathrm{BOC}=180^{\circ}$

$\Rightarrow \angle \mathrm{BOC}=180^{\circ}-130^{\circ}=50^{\circ}$

(i) Now arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle

$\therefore \angle \mathrm{BEC}=\frac{1}{2} \angle \mathrm{BOC}$

$=\frac{1}{2} \times 50^{\circ}=25^{\circ}$

(ii) Similarly arc AB subtends ∠AOB at the centre and ∠ACE at the remaining part of the circle

$\therefore \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 130^{\circ}=65^{\circ}$

(iii) ∵CD||EB

∴∠ECD=∠CEB (alternate angles)

=25°

∴∠BCD=∠ACB+∠ACE+∠ECD

=65°+20°+25°=110°

(iv) ∵EBCD is a cyclic quadrilateral 

∴∠CED+∠BCD=180°

$\Rightarrow \angle C E D+\angle B E C+\angle B C D=180^{\circ}$

$\Rightarrow \angle C E D+25^{\circ}+110^{\circ}=180^{\circ}$

$\Rightarrow \angle C E D+135^{\circ}=180^{\circ}$

$\therefore \angle C E D=180^{\circ}-135^{\circ}=45^{\circ}$

Question 15

(a) In the figure (i) given below, APC, AQB and BPD are straight lines.

(i) Prove that ∠ADB + ∠ACB = 180°.

(ii) If a circle can be drawn through A, B, C and D, Prove that it has AB as a diameter

(b) In the figure (ii) given below, AQB is a straight line. Sides AC and BC of ∆ABC cut the circles at E and D respectively. Prove that the points C, E, P and D are concyclic.

Sol :

(a) Given: In the figure, APC,

AQB and BPD are straight lines.

To Prove : $(i) \angle \mathrm{ADB}+\angle \mathrm{ACB}=180^{\circ}$

(ii) If a circle is drawn through $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D, then AB is a diameter.

Construction : Join PQ.

Proof: In cyclic quad. AQPD.

$\angle \mathrm{ADP}+\angle \mathrm{AQB}=180^{\circ}$

But $\angle \mathrm{AQB}=\angle \mathrm{PCB}$

(Ext. angle of a cyclic quad. is equal to its interior opposite angle)

$\therefore \angle \mathrm{ADP}+\angle \mathrm{PCB}=180^{\circ}$

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

(ii) If A,B,C and D are concyclic then

$\angle \mathrm{ADB}=\angle \mathrm{ACB}$

(Angles in the same segment)

But $\angle \mathrm{ADB}+\angle \mathrm{ACB}=180^{\circ}$

[proved in (i)]

$\therefore \angle \mathrm{ADB}=\angle \mathrm{ACB}=90^{\circ}$

But these are angles on one side of AB 

∴ AB is the diameter of the circle.

Q.E.D

(b) Given : AQB is a straight line. Sides AC and BC of ΔABC cut the circles at E and D respectively.

To prove : C, E, P, D are concyclic

Construction : Join PE, PD and PQ

Proof : In cyclic quad. AQPE,

∠A+∠EPQ=180°

$\Rightarrow \angle E P Q=180^{\circ}-\angle A$...(i)

Similarly PQBD is cyclic quadrilateral 

$\therefore \angle Q P D=180^{\circ}-\angle B$

But $\angle \mathrm{EPD}+\angle \mathrm{EPQ}+\angle \mathrm{QPD}=360^{\circ}$

(angles at a point)

$\Rightarrow \angle E P D+180^{\circ}-\angle A+180^{\circ}-\angle B=360^{\circ}$

$\Rightarrow \angle E P D=\angle A+\angle B$

Adding ∠C both sides

∠EPD+∠C=∠A+∠B+∠C=180°

Hence E,P,D and C are concyclic

Q.E.D

Question 16

(a) In the figure (i) given below, chords AB, BC and CD of a circle with centre O are equal. If ∠BCD = 120°, find

(i) ∠BDC (ii) ∠BEC

(iii) ∠AEC (iv) ∠AOB.

Hence Prove that AOAB is equilateral.

(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. The chord BC of the circle is parallel to the radius OD and the lines OC and BD meet at E. Prove that

(i) ∠CED = 3 ∠CBD (ii) CD = DA.

Sol :

(a) In ∆BCD, BC = CD

∠CBD = ∠CDB

But ∠BCD + ∠CBD + ∠CDB = 180°

(∵ Angles of a triangle)

$\Rightarrow 120^{\circ}+\angle \mathrm{CBD}+\angle \mathrm{CBD}=180^{\circ}$

$\Rightarrow 120^{\circ}+2 \angle \mathrm{CBD}=180^{\circ}$

$\Rightarrow 2 \angle \mathrm{CBD}=180^{\circ}-120^{\circ}=60^{\circ}$

$\therefore \angle \mathrm{CBD}=30^{\circ}$ and $\angle \mathrm{CDB}=30^{\circ}$

$\angle \mathrm{BEC}=\angle \mathrm{CDB}$

(Angles in the same segment)

$\therefore \angle \mathrm{BEC}=30^{\circ}$

$\because \mathrm{CB}=\mathrm{AB}$

$\therefore \angle \mathrm{BEC}=\angle \mathrm{AEB}$

(equal chords subtend equal angles)

$\therefore \angle \mathrm{AEB}=30^{\circ}$

Arc AB subtends $\angle \mathrm{AOB}$ at the centre and $\angle \mathrm{AEB}$ at the remaining part of the circle.

$\therefore \angle \mathrm{AOB}=2 \angle \mathrm{AEB}=2 \times 30^{\circ}=60^{\circ}$

Now in $\Delta \mathrm{AOB}$

$\mathrm{OA}=\mathrm{OB}$ (radii of the same circle)

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

But $\angle \mathrm{OAB}+\angle \mathrm{OBA}+\angle \mathrm{AOB}=180^{\circ}$

$\Rightarrow \angle \mathrm{OAB}+\angle \mathrm{OAB}+60^{\circ}=180^{\circ}$

$\Rightarrow 2 \angle \mathrm{OAB}=180^{\circ}-60^{\circ}=120^{\circ}$

$\therefore \angle \mathrm{OAB}=60^{\circ}$

$\therefore \angle \mathrm{OAB}=\angle \mathrm{OBA}=\angle \mathrm{AOB}=60^{\circ}$

Hence OAB is an equilateral triangle.

Q.E.D

(b) Given : In a circle with centre O, AB is diameter and chord BC || radius OD, OC and BD intersect each other at E.

To Prove : $(i) \angle \mathrm{CED}=3 \angle \mathrm{CBD}$

(ii) CD=DA

Construction : CD, DA are joined

Proof : Arc CD subtends ∠COD at the centre and ∠CBD at the remaining part of the circle

$\therefore \angle \mathrm{COD}=2 \angle \mathrm{CBD}$...(i)

∵BC||OD 

∴∠CBD=∠BDO ..(ii)

In ΔDOE,

Ext. $\angle \mathrm{BEO}=\angle \mathrm{EDO}+\angle \mathrm{EOD}$

$=\angle B D O+\angle C O D$

$=\angle C B D+2 \angle C B D$

From (i) and (ii)

=3∠CBD

 But ∠CED=∠BEO

(vertically opposite angles)

∴∠CED=3∠CBD

(ii) In ΔDBO

OD=OB (radii of the same circle)

∴∠OBD=∠BDO

=∠CBD  [from (ii)]

$\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{CBD}$

$\therefore  \mathrm{AD}=\mathrm{CD}$

$(\because$ Equal chords subtend equal angles)

Q.E.D

Question 17

(a) In the adjoining figure, (i) given below AB and XY are diameters of a circle with centre O. If ∠APX = 30°, find

(i) ∠AOX (ii) ∠APY (iii) ∠BPY (iv) ∠OAX.

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If ∠CBP = 25° and ∠CAP = 40°, find :

(i) ∠ADB (ii) ∠AOB (iii) ∠ACB (iv) ∠APB.

Sol :

(a) AB and XY are diameters of a circle with centre O.

∠APX = 30°.

To find :

(i) ∠AOX (ii) ∠APY

(iii) ∠BPY (iv) ∠OAX

(i) $\because$ arc AX subtends $\angle$ AOX at the centre and $\angle \mathrm{APX}$ at the remaining point of the circle.

$\therefore \angle \mathrm{AOX}=2 \angle \mathrm{APX}$

$=2 \times 30^{\circ}=60^{\circ}$

(ii) $\because \mathrm{XY}$ is the diameter

$\therefore \angle X P Y=90^{\circ}$

(Angle in a semi-circle)

$\therefore \angle \mathrm{APY}=\angle \mathrm{XPY}-\angle \mathrm{APX}$

$=90^{\circ}-30^{\circ}=60^{\circ}$

(iii) $\angle \mathrm{APB}=90^{\circ}$ (Angle in a semi-circle)

$\therefore \angle \mathrm{BPY}=\angle \mathrm{APB}-\angle \mathrm{APY}$

$=90^{\circ}-60^{\circ}=30^{\circ}$

(iv) In ΔAOX

OA=OX (radii of the same circle)

$\therefore \angle \mathrm{OAX}=\angle \mathrm{OXA}$

But $\angle \mathrm{AOX}+\angle \mathrm{OAX}+\angle \mathrm{OXA}=180^{\circ}$

(Angles of a triangle)

$\Rightarrow 60^{\circ}+\angle \mathrm{OAX}+\angle \mathrm{OAX}=180^{\circ}$

$\Rightarrow 2 \angle \mathrm{OAX}=180^{\circ}-60^{\circ}=120^{\circ}$

$\therefore \quad \angle \mathrm{OAX}=\frac{120^{\circ}}{2}=60^{\circ}$

(b) Join CD

(i) $\angle \mathrm{CDB}=\angle \mathrm{CBP}$

(Angles in the alternate segments)

$\therefore \angle \mathrm{CDB}=25^{\circ}$..(i)

Similarly, $\angle \mathrm{CDA}=\angle \mathrm{CAP}=40^{\circ}$

(Angles in the alternate segments)

$\therefore \angle \mathrm{ADB}=\angle \mathrm{CDA}+\angle \mathrm{CDB}$

$=40^{\circ}+25^{\circ}=65^{\circ}$

(ii) $\operatorname{arcACB~subtends~} \angle A O B$ at the centre and

$\angle \mathrm{ADB}$ at the remaining part of circle.

$\therefore \angle \mathrm{AOB}=2 \angle \mathrm{ADB}=2 \times 65^{\circ}=130^{\circ}$

(iii) ACBD is a cyclic quadrilateral

$\therefore \quad \angle \mathrm{ACB}+\angle \mathrm{ADB}=180^{\circ}$

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

$\Rightarrow \angle \mathrm{ACB}=180^{\circ}-65^{\circ}=115^{\circ}$

(iv) $\angle \mathrm{AOB}+\angle \mathrm{APB}=180^{\circ}$

$\Rightarrow 130^{\circ}+\angle \mathrm{APB}=180^{\circ}$

$\Rightarrow \angle \mathrm{APB}=180^{\circ}-130^{\circ}=50^{\circ}$