# Chapter 18 : Tangents and Intersecting Chords - Selina Solutions for Class 10 Maths ICSE

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## Chapter 18 - Tangents and Intersecting Chords Excercise Ex. 18(A)

The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre?

OP = 10 cm; radius OT = 8 cm

Length of tangent = 6 cm.

In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

AB = 15 cm, AC = 7.5 cm

Let 'r' be the radius of the circle.

OC = OB = r

AO = AC + OC = 7.5 + r

In ∆AOB,

AO^{2}
= AB^{2} + OB^{2}

Therefore, r = 11.25 cm

Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

From Q, QA and QP are two tangents to the circle with centre O

Therefore, QA = QP .....(i)

Similarly, from Q, QB and QP are two tangents to the circle with centre O'

Therefore, QB = QP ......(ii)

From (i) and (ii)

QA = QB

Therefore, tangents QA and QB are equal.

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

From Q, QA and QP are two tangents to the circle with centre O

Therefore, QA = QP .......(i)

Similarly, from Q, QB and QP are two tangents to the circle with centre O'

Therefore, QB = QP .......(ii)

From (i) and (ii)

QA = QB

Therefore, tangents QA and QB are equal.

Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.

OS = 5 cm

OT = 3 cm

In Rt. Triangle OST

By Pythagoras Theorem,

Since OT is perpendicular to SP and OT bisects chord SP

So, SP = 8 cm

Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.

AB = 6 cm, AC = 8 cm and BC = 9 cm

Let radii of the circles having centers A, B and C be r_{1}, r_{2} and r_{3} respectively.

r_{1} + r_{3} = 8

r_{3} + r_{2} = 9

r_{2} + r_{1} = 6

Adding

r_{1} + r_{3} + r_{3} + r_{2} + r_{2} + r_{1} = 8+9+6

2(r_{1} + r_{2} + r_{3}) = 23

r_{1} + r_{2} + r_{3 }= 11.5 cm

r_{1} + 9 = 11.5 (Since r_{2} + r_{3 }= 9)

r_{1} = 2.5 cm

r_{2} + 6 = 11.5 (Since r_{1} + r_{3 }= 6)

r_{2} = 5.5 cm

r_{3} + 8 = 11.5 (Since r_{2} + r_{1 }= 8)

r_{3 }= 3.5 cm

Hence, r_{1} = 2.5 cm, r_{2} = 5.5 cm and r_{3 }= 3.5 cm

If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.

Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.

Since AP and AS are tangents to the circle from external point A

AP = AS .......(i)

Similarly, we can prove that:

BP = BQ .......(ii)

CR = CQ .......(iii)

DR = DS ........(iv)

Adding,

AP + BP + CR + DR = AS + DS + BQ + CQ

AB + CD = AD + BC

Hence, AB + CD = AD + BC

If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.

From A, AP and AS are tangents to the circle.

Therefore, AP = AS.......(i)

Similarly, we can prove that:

BP = BQ .........(ii)

CR = CQ .........(iii)

DR = DS .........(iv)

Adding,

AP + BP + CR + DR = AS + DS + BQ + CQ

AB + CD = AD + BC

Hence, AB + CD = AD + BC

But AB = CD and BC = AD.......(v) Opposite sides of a ||gm

Therefore, AB + AB = BC + BC

2AB = 2 BC

AB = BC ........(vi)

From (v) and (vi)

AB = BC = CD = DA

Hence, ABCD is a rhombus.

From the given figure prove that:

AP + BQ + CR = BP + CQ + AR.

Also, show that AP + BQ + CR = x perimeter of triangle ABC.

Since from B, BQ and BP are the tangents to the circle

Therefore, BQ = BP ………..(i)

Similarly, we can prove that

AP = AR …………..(ii)

and CR = CQ ………(iii)

Adding,

AP + BQ + CR = BP + CQ + AR ………(iv)

Adding AP + BQ + CR to both sides

2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR

2(AP + BQ + CR) = AB + BC + CA

Therefore, AP + BQ + CR = x (AB + BC + CA)

AP + BQ + CR = x perimeter of triangle ABC

In the figure, if AB = AC then prove that BQ = CQ.

Since, from A, AP and AR are the tangents to the circle

Therefore, AP = AR

Similarly, we can prove that

BP = BQ and CR = CQ

Adding,

AP + BP + CQ = AR + BQ + CR

(AP + BP) + CQ = (AR + CR) + BQ

AB + CQ = AC + BQ

But AB = AC

Therefore, CQ = BQ or BQ = CQ

Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if -

i) they touch each other externally.

ii) they touch each other internally.

Radius of bigger circle = 6.3 cm

and of smaller circle = 3.6 cm

i)

Two circles are touching each other at P externally. O and O’ are the centers of the circles. Join OP and O’P

OP = 6.3 cm, O’P = 3.6 cm

Adding,

OP + O’P = 6.3 + 3.6 = 9.9 cm

ii)

Two circles are touching each other at P internally. O and O’ are the centers of the circles. Join OP and O’P

OP = 6.3 cm, O’P = 3.6 cm

OO’ = OP - O’P = 6.3 - 3.6 = 2.7 cm

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

i)

ii) OP is the perpendicular bisector of chord AB.

i) In

AP = BP (Tangents from P to the circle)

OP = OP (Common)

OA = OB (Radii of the same circle)

ii) In

OA = OB (Radii of the same circle)

(Proved )

OM = OM (Common)

Hence, OM or OP is the perpendicular bisector of chord AB.

In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that:

i) tangent at point P bisects AB.

ii) Angle APB = 90°

Draw TPT' as common tangent to the circles.

i) TA and TP are the tangents to the circle with centre O.

Therefore, TA = TP ………(i)

Similarly, TP = TB ………..(ii)

From (i) and (ii)

TA = TB

Therefore, TPT' is the bisector of AB.

ii) Now in

Similarly in

Adding,

Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:

In quadrilateral OPAQ,

In triangle OPQ,

OP = OQ (Radii of the same circle)

From (i) and (ii)

Two parallel tangents of a circle meet a third tangent at point P and Q. Prove that PQ subtends a right angle at the centre.

Join OP, OQ, OA, OB and OC.

In

OA = OC (Radii of the same circle)

OP = OP (Common)

PA = PC (Tangents from P)

Similarly, we can prove that

(Sum of interior angles of a transversal)

Now in

ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.

Calculate the value of x, the radius of the inscribed circle.

In

LBNO is a square.

LB = BN = OL = OM = ON = x

Since ABC is a right triangle

In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:

i)

ii)

given that

The incircle touches the sides of the triangle ABC and

i) In quadrilateral AROQ,

ii) Now arc RQ subtends at the centre and at the remaining part of the circle.

In the following figure, PQ and PR are tangents to the circle, with centre O. If , calculate:

i)

ii)

iii)

Join QR.

i) In quadrilateral ORPQ,

ii) In

OQ = QR (Radii of the same circle)

iii) Now arc RQ subtends at the centre and at the remaining part of the circle.

In the given figure, AB is a diameter of the circle, with centre O, and AT is a tangent. Calculate the numerical value of x.

In

OB = OC (Radii of the same circle)

Now in

In quadrilateral ABCD, angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle.

BQ and BR are the tangents from B to the circle.

Therefore, BR =BQ = 27 cm.

Also RC = (38 -; 27) = 11cm

Since CR and CS are the tangents from C to the circle

Therefore, CS = CR = 11 cm

So, DS = (25 - 11) = 14 cm

Now DS and DP are the tangents to the circle

Therefore, DS = DP

Now, (given)

and

therefore, radius = DS = 14 cm

In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.

Given and

Prove that -;

i)

ii) write an expression connecting x and y

(angles in alternate segment)

But OS = OR (Radii of the same circle)

But in

From (i) and (ii)

PT is a tangent to the circle at T. If ; calculate:

i)

ii)

iii)

Join AT and BT.

i) TC is the diameter of the circle

(Angle in a semi-circle)

ii)

(Angles in the same segment of the

circle)

(Angles in the same segment of the circle)

iii) (Angles in the same segment)

Now in

In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.

Join OC.

Therefore, PA and PA are the tangents

In quadrilateral APCO,

Now, arc BC subtends at the centre and at the remaining part of the circle

In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.

∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ)

∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)

∠CAQ + ∠PAC = 180°……(angles in linear pair)

∴∠PAC = 120°

∠PAC = 2∠CAD……(AD is angle bisector of ∠PAC)

∠CAD = 60°

Now,

∠CAD + ∠CAB = 60 + 30 = 90°

∠DAB = 90°

Thus, BD subtends 90° on the circle

So, BD is the diameter of circle

## Chapter 18 - Tangents and Intersecting Chords Excercise Ex. 18(B)

i) In the given figure, 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.

ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4cm. Find CD.

iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.

i) Since two chords AB and CD intersect each other at P.

ii) Since two chords AB and CD intersect each other at P.

iii) Since PAB is the secant and PT is the tangent

In the given figure, 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.

In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ; calculate:

i)

ii)

iii)

i) PAQ is a tangent and AB is the chord.

(angles in the alternate segment)

ii) OA = OD (radii of the same circle)

iii) BD is the diameter.

(angle in a semi-circle)

Now in

If PQ is a tangent to the circle at R; calculate:

i)

ii)

Given: O is the centre of the circle and

PQ is a tangent and OR is the radius.

But in

OT = OR (Radii of the same circle)

In

AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.

Join OC.

(angles in alternate segment)

Arc BC subtends at the centre of the circle and at the remaining part of the circle.

Now in

Now in

Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.

DE is the tangent to the circle at P.

DE||QR (Given)

Since the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment

(DE is tangent and PQ is chord)

from (i) and (ii)

Hence, triangle PQR is an isosceles triangle.

Two circles with centers O and O' are drawn to intersect each other at points A and B.

Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O' at A. Prove that OA bisects angle BAC.

Join OA, OB, O'A, O'B and O'O.

CD is the tangent and AO is the chord.

(angles in alternate segment)

In

OA = OB (Radii of the same circle)

From (i) and (ii)

Therefore, OA is bisector of BAC

Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that:

Draw a tangent TS at P to the circles given.

Since TPS is the tangent, PD is the chord.

Subtracting (i) from (ii)

But in

In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.

TAS is a tangent and AB is a chord

But these are alternate angles

Therefore, TS||BD.

In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If angle BCG=108 and O is the centre of the circle, find:

i) angle BCT

ii) angle DOC

Join OC, OD and AC.

i)

ii)

PCT is a tangent and CA is a chord.

But arc DC subtends at the centre and at the

remaining part of the circle.

Two circles intersect each other at point A and B. A straight line PAQ cuts the circle at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T are concyclic.

Join AB, PB and BQ

TP is the tangent and PA is a chord

(angles in alternate segment)

Similarly,

Adding (i) and (ii)

But they are the opposite angles of the quadrilateral

Therefore, PBQT are cyclic.

Hence, P, B, Q and T are concyclic.

In the figure, PA is a tangent to the circle. PBC is a secant and AD bisects angle BAC.

Show that the triangle PAD is an isosceles triangle. Also show that:

i) PA is the tangent and AB is a chord

( angles in the alternate segment)

AD is the bisector of

In

Therefore, is an isosceles triangle.

ii) In

Two circles intersect each other at point A and B. Their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.

Join AB.

PQ is the tangent and AB is a chord

(angles in alternate segment)

Similarly,

Adding (i) and (ii)

From (iii) and (iv)

Hence, and are supplementary.

In the figure, chords AE and BC intersect each other at point D.

i) if , AB = 5 cm, BD = 4 cm and CD = 9 cm; find DE

ii) If AD = BD, Show that AE = BC.

Join AB.

i) In Rt.

Chords AE and CB intersect each other at D inside the circle

AD x DE = BD x DC

3 x DE = 4 x 9

DE = 12 cm

ii) If AD = BD .......(i)

We know that:

AD x DE = BD x DC

But AD = BD

Therefore, DE = DC .......(ii)

Adding (i) and (ii)

AD + DE = BD + DC

Therefore, AE = BC

Circles with centers P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles are congruent; show that CE = BD.

Join AB and AD

EBM is a tangent and BD is a chord.

(angles in alternate segments)

(Vertically opposite angles)

Since in the same circle or congruent circles, if angles are equal, then chords opposite to them are also equal.

Therefore, CE = BD

In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B. Find

AB is a straight line.

AB i.e. DB is tangent to the circle at point B and BC is the diameter.

Now, OE = OC (radii of the same circle)

(vertically opposite angles)

In

## Chapter 18 - Tangents and Intersecting Chords Excercise Ex. 18(C)

Prove that of any two chord of a circle, the greater chord is nearer to the centre.

Given: A circle with centre O and radius r. . Also AB > CD

To prove: OM < ON

Proof: Join OA and OC.

In Rt.

Again in Rt.

From (i) and (ii)

Hence, AB is nearer to the centre than CD.

OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.

i)If the radius of the circle is 10 cm, find the area of the rhombus.

ii) If the area of the rhombus is , find the radius of the circle.

i) Radius = 10 cm

In rhombus OABC,

OC = 10 cm

In Rt.

Area of rhombus =

ii) Area of rhombus =

But area of rhombus OABC = 2 x area of

Where r is the side of the equilateral triangle OAB.

Therefore, radius of the circle = 8 cm

Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.

If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, AB = (5 - 3) cm = 2 cm.

Also, the common chord PQ is the perpendicular bisector of AB. Therefore, AC = CB = AB = 1 cm

In right ACP, we have AP^{2} = AC^{2} + CP^{2}

5^{2} = 1^{2} + CP^{2}

CP^{2} = 25 -; 1 = 24

CP =

Now, PQ = 2 CP

= 2 x cm

Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of the angle BAC.

Given: AB and AC are two equal chords of C (O, *r*).

To prove: Centre, O lies on the bisector of BAC.

Construction: Join BC. Let the bisector of BAC intersects BC in P.

Proof:

In APB and APC,

AB = AC (Given)

BAP = CAP (Given)

AP = AP (Common)

(SAS congruence criterion)

BP = CP and APB = APC (CPCT)

APB + APC = 180 (Linear pair)

=> 2APB = 180 (APB = APC)

APB = 90

Now, BP = CP and APB = 90

AP is the perpendicular bisector of chord BC.

AP passes through the centre, O of the circle.

The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle?

AB is the diameter and AC is the chord.

Draw

Since and hence it bisects AC, O is the centre of the circle.

Therefore, OA = 10 cm and AL = 6 cm

Now, in Rt.

Therefore, chord is at a distance of 8 cm from the centre of the circle.

ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = and angle BAC = . Find angle DAC and angle DCA.

ABCD is a cyclic quadrilateral in which AD||BC

(Sum of opposite angles of a quadrilateral)

Now in

Now in

In the given figure, C and D are points on the semicircle described on AB as diameter.

Given angle BAD = 70and angle DBC = 30, calculate angle BDC

Since ABCD is a cyclic quadrilateral, therefore, BCD + BAD = 180

(since opposite angles of a cyclic quadrilateral are supplementary)

BCD + 70 = 180

BCD = 180 - 70 = 110

In BCD, we have,

CBD + BCD + BDC = 180

30 + 110 + BDC = 180

BDC = 180 - 140

BDC = 40

In cyclic quadrilateral ABCD, A = 3C and D = 5B. Find the measure of each angle of the quadrilateral.

ABCD is a cyclic quadrilateral.

Similarly,

Hence,

Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Join AD.

AB is the diameter.

ADB = 90º (Angle in a semi-circle)

But, ADB + ADC = 180º (linear pair)

ADC = 90º

In ABD and ACD,

ADB = ADC (each 90º)

AB = AC (Given)

AD = AD (Common)

ABD ACD (RHS congruence criterion)

BD = DC (C.P.C.T)

Hence, the circle bisects base BC at D.

Bisectors of vertex A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF =

Join ED, EF and DF. Also join BF, FA, AE and EC.

In cyclic quadrilateral AFBE,

(Sum of opposite angles)

Similarly in cyclic quadrilateral CEAF,

Adding (ii) and (iii)

In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If , find angle AOD.

Join OB.

In

In

Since DOC is a straight line

Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.

Join OL, OM and ON.

Let D and d be the diameter of the circumcircle and incircle.

and let R and r be the radius of the circumcircle and incircle.

In circumcircle of

Therefore, AC is the diameter of the circumcircle i.e. AC = D

Let radius of the incircle = r

Now, from B, BL, BM are the tangents to the incircle.

(Tangents from the point outside the circle)

Now,

AB+BC+CA = AM+BM+BL+CL+CA

= AN+r+r+CN+CA

= AN+CN+2r+CA

= AC+AC+2r

= 2AC+2r

= 2D+d

P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.

Join AP and BP.

Since TPS is a tangent and PA is the chord of the circle.

(angles in alternate segments)

But

But these are alternate angles

In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent.

Prove that the line NM produced bisects AB at P.

From P, AP is the tangent and PMN is the secant for first circle.

Again from P, PB is the tangent and PMN is the secant for second circle.

From (i) and (ii)

Therefore, P is the midpoint of AB.

In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If and , find:

i)

ii)

iii)

i) PQ is tangent and CD is a chord

(angles in the alternate segment)

ii)

iii) In

The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that:

Join OC.

BCD is the tangent and OC is the radius.

Substituting in (i)

ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D.

Prove that -

i) AC x AD = AB^{2}

ii) BD^{2} = AD x DC.

i) In

and BC is the diameter of the circle.

Therefore, AB is the tangent to the circle at B.

Now, AB is tangent and ADC is the secant

ii) In

From (i) and (ii)

Now in

In the given figure AC = AE.

Show that:

i) CP = EP

ii) BP = DP

In

(angles in the same segment)

AC = AE (Given)

(Common)

(ASA Postulate)

AB = AD

but AC = AE

In

(angles in the same segment)

BC = DE

(angles in the same segment)

(ASA Postulate)

BP = DP and CP = PE (cpct)

ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC=120

Calculate:

i)

ii)

i) Join OC and OB.

AB = BC = CD and

OB and OC are the bisectors of and respectively.

In

Arc BC subtends at the centre and at the remaining part of the circle.

ii) In cyclic quadrilateral BCDE,

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

(i) angle BCO

(ii) angle AOB

(iii) angle APB

In the given fig, O is the centre of the circle and CA and CB are the tangents to the circle from C. Also, ACO = 30

P is any point on the circle. P and PB are joined.

To find: (i)

(ii)

(iii)

Proof:

ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles.

Given: ABC is a triangle with AB = 10 cm, BC= 8 cm, AC = 6 cm. Three circles are drawn with centre A, B and C touch each other at P, Q and R respectively.

We need to find the radii of the three circles.

In a square ABCD, its diagonal AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and bisector of angle ABD meets AC at N and AM at L. Show that -

i)

ii)

iii) ALOB is a cyclic quadrilateral.

ABCD is a square whose diagonals AC and BD intersect each other at right angles at O.

i)

In

But, (vertically opposite angles)

Now in

Adding (i) and (ii)

ii)

and

iii) In quadrilateral ALOB,

Therefore, ALOB is a cyclic quadrilateral.

The given figure shows a semicircle with centre O and diameter PQ. If PA = AB and ; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

Join PB.

i) In cyclic quadrilateral PBCQ,

Now in

In cyclic quadrilateral PQBA,

ii) Now in

iii) Arc AQ subtends at the centre and APQ at the remaining part of the circle.

We have,

From (1), (2) and (3), we have

Now in

But these are alternate angles.

Hence, AO is parallel to BQ.

The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20 and angle POQ = 100.

Calculate -

i) angle QTR

ii) angle QRP

iii) angle QRS

iv) angle STR

Join PQ, RQ and ST.

i)

Arc RQ subtends at the centre and QTR at the remaining part of the circle.

ii) Arc QP subtends at the centre and QRP at the remaining part of the circle.

iii) RS || QT

iv) Since RSTQ is a cyclic quadrilateral

(sum of opposite angles)

In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that:

i) Since PAT||BC

(alternate angles) .........(i)

In cyclic quadrilateral ABCD,

from (i) and (ii)

ii) Arc AB subtends at the centre and at the remaining part of the circle.

iii)

AB is a line segment and M is its midpoint. Three semicircles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semicircles. Show that: AB = 6 x r

Let O, P and Q be the centers of the circle and semicircles.

Join OP and OQ.

OR = OS = r

and AP = PM = MQ = QB =

Now, OP = OR + RP = r + (since PM=RP=radii of same circle)

Similarly, OQ = OS + SQ = r +

OM = LM -; OL = - r

Now in Rt.

Hence AB = 6 x r

TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.

Join PB.

In TAP and TBP,

TA = TB (tangents segments from an external points are equal in length)

Also, ATP = BTP. (since OT is equally inclined with TA and TB) TP = TP (common)

TAP TBP (by SAS criterion of congruency)

TAP = TBP (corresponding parts of congruent triangles are equal)

But TBP = BAP (angles in alternate segments)

Therefore, TAP = BAP.

Hence, AP bisects TAB.

Two circles intersect in points P and Q. A secant passing through P intersects the circle in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.

Join PQ.

AT is tangent and AP is a chord.

(angles in alternate segments) ........(i)

Similarly, .......(ii)

Adding (i) and (ii)

Now in

Therefore, AQBT is a cyclic quadrilateral.

Hence, A, Q, B and T lie on a circle.

Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle)

ABCDE is a regular pentagon.

In AED,

AE = ED (Sides of regular pentagon ABCDE)

EAD = EDA

In AED,

AED + EAD + EDA = 180º

108º + EAD + EAD = 180º

2EAD = 180º - 108º = 72º

EAD = 36º

EDA = 36º

BAD = BAE - EAD = 108º - 36º = 72º

In quadrilateral ABCD,

BAD + BCD = 108º + 72º = 180º

ABCD is a cyclic quadrilateral

Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm. Calculate the length of CD.

We know that XB.XA = XD.XC

Or, XB.(XB + BA) = XD.(XD + CD)

Or, 6(6 + 4) = 5(5 + CD)

Or, 60 = 5(5 + CD)

Or, 5 + CD = = 12

Or, CD = 12 - 5 = 7 cm.

In the given figure, find TP if AT = 16 cm and AB = 12 cm.

PT is the tangent and TBA is the secant of the circle.

Therefore, TP^{2} = TA x TB

TP^{2} = 16 x (16-12) = 16 x 4 = 64 = (8)^{2}

Therefore, TP = 8 cm

In the following figure, A circle is inscribed in the quadrilateral ABCD.

If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.

From the figure we see that BQ = BR = 27 cm (since length of the tangent segments from an

external point are equal)

As BC = 38 cm

CR = CB - BR = 38 - 27

= 11 cm

Again,

CR = CS = 11cm (length of tangent segments from an external point are equal)

Now, as DC = 25 cm

DS = DC - SC

= 25 -11

= 14 cm

Now, in quadrilateral DSOP,

PDS = 90 (given)

**OSD = 90, OPD = 90 (since tangent is perpendicular to the **

radius through the point of contact)

DSOP is a parallelogram

OP||SD and PD||OS

Now, as OP = OS (radii of the same circle)

OPDS is a square. DS = OP = 14cm

radius of the circle = 14 cm

In the figure, XY is the diameter of the circle, PQ is the tangent to the circle at Y. Given that and . Calculate and .

In AXB,

XAB + AXB + ABX=180 [Triangle property]

XAB + 50 + 70 = 180

XAB=180 - 120 = 60

XAY=90 [Angle of semi-circle]

BAY=XAY -XAB = 90 - 60 = 30

and BXY = BAY = 30 [Angle of same segment]

ACX = BXY + ABX [External angle = Sum of two interior angles]

= 30 + 70

= 100

also,

XYP=90 [Diameter ⊥ tangent]

APY = ACX -CYP

APY=100 - 90

APY=10

In the given figure, QAP is the tangent at point A and PBD is a straight line. If and ; find:

i)

ii)

iii)

iv)

PAQ is a tangent and AB is a chord of the circle.

i) (angles in alternate segment)

ii) In

iii) (angles in the same segment)

Now in

iv) PAQ is the tangent and AD is chord

In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.

If

i) AB is diameter of circle.

In

ii) QC is tangent to the circle

Angle between tangent and chord = angle in alternate segment

ABQ is a straight line

In the given figure, O is the centre of the circle. The tangets at B and D intersect each other at point P.

If AB is parallel to CD and , find:

i)

ii)

i)

ii) Since, BPDO is cyclic quadrilateral, opposite angles are supplementary.

In the figure given below PQ =QR, RQP = 68, PC and CQ are tangents to the circle with centre O. Calculate the values of:

i) QOP

ii) QCP

i) PQ = RQ

(opposite angles of equal sides of a triangle)

Now, QOP = 2PRQ (angle at the centre is double)

ii) PQC = PRQ (angles in alternate segments are equal)

QPC = PRQ (angles in alternate segments)

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

Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively.

To prove the given question, it is sufficient to prove AB = CD.

For this join OM, ON, OB and OD.

Let the radius of outer
and inner circles be *R* and *r* respectively.

AB touches the inner circle at M.

AB is a tangent to the inner circle

OMAB

BM = AB

AB = 2BM

Similarly ONCD, and CD = 2DN

Using Pythagoras theorem in OMB and OND

In the figure, given below, AC is a transverse common tangent to two circles with centers P and Q and of radii 6 cm and 3 cm respectively.

Given that AB = 8 cm, calculate PQ.

Since AC is tangent to the circle with center P at point A.

In

Also in Rt.

From (i) and (ii),

In the figure given below, O is the centre of the circum circle of triangle XYZ. Tangents at X and Y intersect at point T. Given XTY = 80 and XOZ = 140, calculate the value of ZXY.

In the figure, a circle with centre O, is the circum circle of triangle XYZ.

Tangents at X and Y intersect at point T, such that XTY = 80

In the given figure, AE and BC intersect each other at point D. If CDE=90, AB = 5 cm, BD = 4 cm and CD = 9 cm, find AE.

From Rt.

Now, since the two chords AE and BC intersect at D,

AD x DE = CD x DB

3 x DE = 9 x 4

Hence, AE = AD + DE = (3 + 12) = 15 cm

In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.

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

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