Class 10 SELINA Solutions Maths Chapter 18 - Tangents and Intersecting Chords

OP = 10 cm; radius OT = 8 cm

Length of tangent = 6 cm.

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,

AO² = AB² + OB²

Therefore, the length of radius of a circle is 11.25 cm.

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.

Radius of outer circle, OS = 5 cm.

Radius of inner circle, OT = 3 cm.

       (Angle between the radius and the tangent)

In Rt. triangle OTS, by Pythagoras Theorem,

Since OT is perpendicular to SP and OT bisects chord SP,

SP = 8 cm

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

Let radii of the circles having centers A, B and C be r₁, r₂ and r₃ respectively.

r₁ + r₃ = 8       ....(1)

r₃ + r₂ = 9       ....(2)

r₂ + r₁ = 6       ....(3)

Adding (1), (2) and (3),

r₁ + r₃ + r₃ + r₂ + r₂ + r₁ = 8 + 9 + 6

2(r₁ + r₂ + r₃) = 23

r₁ + r₂ + r₃ = 11.5 cm

r₁ + 9 = 11.5      (Since r₂ + r₃ = 9)

r₁ = 2.5 cm

r₂ + 8 = 11.5      (Since r₁ + r₃ = 8)

r₂ = 3.5 cm

r₃ + 6 = 11.5 (Since r₁ + r₂ = 6)

r₃ = 5.5 cm

Hence, r₁ = 2.5 cm, r₂ = 3.5 cm and r₃ = 5.5 cm

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 (i), (ii), (iii) and (iv),

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

AB + CD = AD + BC

Hence, AB + CD = AD + BC

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

But AB = CD and BC = AD      ....(v)      [Opposite sides of a parallelogram]

Therefore, AB + AB = BC + BC

2AB = 2BC

AB = BC      ....(vi)

From (v) and (vi)

AB = BC = CD = DA

Hence, ABCD is a rhombus.

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)    [Proved]

Adding (AP + BQ + CR) to both sides,

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

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, BQ = CQ

Radius of bigger circle = 6.3 cm

Radius 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

Distance between the centres = OO'

= 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

Distance between the centres = OO’

= OP - O’P

= 6.3 - 3.6

= 2.7 cm

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.

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,

In quadrilateral OPAQ,

Now,

In triangle OPQ,

OP = OQ (Radii of the same circle)

From (i) and (ii)

In

Therefore, LBMO is a square.

LB = BM = OM = OL = x

Since ABC is a right triangle

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.

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

OB = OC (Radii of the same circle)

Now in

(angles in alternate segment)

But OS = OR (Radii of the same circle)

Now,

OQ = OR   (radii of same circle)

But in

From (i) and (ii),

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

Now in

Join OC.

Now, PA and PC are the tangents to the circle with centre O.

In quadrilateral APCO,

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

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

Then, ∠CAQ = 2∠BAQ = 60°   

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

∴ ∠PAC = 180° - 60° = 120°

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

∠CAD = 60° 

Now,

∠DAB = ∠CAD + ∠CAB = 60° + 30° = 90° 

Thus, BD subtends a right angle on the circle.

So, BD is the diameter of the circle.

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

PQ is a tangent and OR is the radius.

But in

OT = OR (Radii of the same circle)

In

Join O'A and O'B.

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.

Join OC, OD and OA.

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.

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

Join AB.

           (angles in alternate segment)

Similarly,

Adding (i) and (ii),

From (iii) and (iv)

Hence, and are supplementary.

Join AB.

i) In Rt.

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

Therefore,

AD x DE = BD x DC

3 x DE = 4 x 9

DE = 12 cm

ii) Given, 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

AB is a tangent and hence, 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

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.

ABCD is a cyclic quadrilateral in which AD||BC

(Sum of opposite angles of a quadrilateral)

Now in

Now in

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

ABCD is a cyclic quadrilateral.

Similarly,

Hence,

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.

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)

Join OB.

BC = OD ... (given)

OD = OB ... (radii of the same circle)

Hence,

BC = OB

In 

In

Since DOC is a straight-line

Join AP and BP.

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

(angles in alternate segments)

But

arc PA = arc PB,

hence PA = PB

But these are alternate angles

i) PQ is tangent and CD is a chord

(angles in the alternate segment)

ii)

iii) In

Join OC.

BCD is the tangent and OC is the radius.

Substituting in (i)

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

Now,

∠BDC = ∠BDA  = 90^o ... angle subtended by a diameter

ii) In 

From (i) and (ii)

Now in

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)

i) Join OC and OB.

AB = BC = CD and

OA = OD = OC = OB all are radii of the circle

Hence,

All the 3 triangles are congruent

So,

∠AOB = ∠DOC = ∠COB [By C.P.C.T.C]

OB and OC are the bisectors of and respectively.

In

Hence,

∠AOB = ∠DOC = ∠COB = 60^o

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

ii) 

Similarly,

∠BED = 0.5 ∠BOD = 60^o

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:

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.

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)

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.

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.

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 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

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

i) AB is the diameter of the circle.

In

ii) QC is tangent to the circle

The angle between tangent and chord = angle in the alternate segment

ABQ is a straight-line

In ΔCBQ

i)

ii) Since, BPDO is a quadrilateral,

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)

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

In

Also in Rt.

From (i) and (ii),

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

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

Since a tangent at any point of a circle is perpendicular to the radius at the point of contact,

we have ∠OXT = ∠OYT = 90^o

So in quadrilateral XOYT

Sum of angles of a quadrilateral is 360 degrees,

∠XOY = 360 − (90 + 90 + 80)

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