Tangents & Intersecting Chords

Let tangent QR intersect circle at point D.

Given,

Perimeter of the △PQR = 18 cm.

⇒ PQ + QD + RD + PR = 18 .............(1)

We know that,

Two tangents drawn to a circle form an exterior point are equal in length.

∴ PA = PB = x (let), RD = RA and QD = QB.

Substituting above values in equation (1), we get :

⇒ PQ + QB + RA + PR = 18

⇒ PB + PA = 18

⇒ x + x = 18

⇒ 2x = 18

⇒ x = 9 cm.

∴ PA = 9 cm.

Hence, Option 3 is the correct option.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ OP ⊥ AP

In right angle triangle OAP,

By pythagoras theorem,

⇒ OA² = OP² + AP²

⇒ 10² = 6² + AP²

⇒ 100 = 36 + AP²

⇒ AP² = 100 - 36

⇒ AP² = 64

⇒ AP = $\sqrt{64}$ = 8 cm.

Since, AB is the chord to the bigger circle, with center O.

We know that,

Perpendicular from center to the chord, bisects it.

∴ PB = AP = 8 cm.

AB = AP + PB = 8 + 8 = 16 cm.

Hence, Option 1 is the correct option.

From figure,

AC = 9 + 6 = 15 cm

BC = 2 + 6 = 8 cm

In right angle triangle ACB,

⇒ AB² = AC² + BC²

⇒ AB² = 15² + 8²

⇒ AB² = 225 + 64

⇒ AB² = 289

⇒ AB = $\sqrt{289}$ = 17 cm.

Hence, Option 2 is the correct option.

From figure,

OD = OA (Radius of same circle)

In △OAD,

∠ODA = ∠OAD = x (let) (As angles opposite to equal sides are equal)

Since, exterior angle in a triangle is equal to the sum of two opposite interior angles.

∴ ∠DOC = ∠ODA + ∠OAD

⇒ 100° = 2x

⇒ x = $\frac{100°}{2}$

⇒ x = 50°.

⇒ ∠OAD = 50°.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠BCA = 90°.

In △ABC,

By angle sum property of triangle,

⇒ ∠ABC + ∠BCA + ∠CAB = 180°

⇒ ∠ABC + ∠BCA + ∠OAD = 180° [∵ From figure, ∠CAB = ∠OAD]

⇒ ∠ABC + 90° + 50° = 180°

⇒ ∠ABC + 140° = 180°

⇒ ∠ABC = 180° - 140° = 40°.

Hence, Option 3 is the correct option.

Join OA and OB.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAP = 90° and ∠OBP = 90°

In quadrilateral OAPB,

⇒ ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°

⇒ 90° + 70° + 90° + ∠BOA = 360°

⇒ ∠BOA + 250° = 360°

⇒ ∠BOA = 360° - 250° = 110°.

We know that,

The angle which an arc of a circle subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = $\frac{1}{2}$ ∠AOB

⇒ ∠ACB = $\frac{1}{2} \times 110°$ = 55°.

Hence, Option 4 is the correct option.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAP = 90°

From figure,

⇒ ∠PAB = ∠OAP - ∠OAB = 90° - 28° = 62°

PA and PB are tangents drawn to the circle from the external point P.

∴ PA = PB

We know that,

Angles opposite to equal sides of a triangle are equal.

∴ ∠PAB = ∠PBA = 62°

By angle sum property of triangle PAB,

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ 62° + 62° + ∠APB = 180°

⇒ ∠APB = 180° - 124° = 56°.

Hence, Option 2 is the correct option.

Join OB.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAP = 90° and ∠OBP = 90°

In quadrilateral OAPB,

⇒ ∠OAP + ∠APB + ∠PBO + ∠AOB = 360°

⇒ 90° + 50° + 90° + ∠AOB = 360°

⇒ ∠AOB + 230° = 360°

⇒ ∠AOB = 360° - 230° = 130°.

Reflex ∠AOB = 360° - ∠AOB = 360° - 130° = 230°.

Hence, Option 3 is the correct option.

Let the radius of the circle to be r cm.

So, AO = AC + OC = (7.5 + r) cm.

In right ∆AOB, we have

⇒ AO² = AB² + OB² [By Pythagoras Theorem]

⇒ (7.5 + r)² = 15² + r²

⇒ 56.25 + r² + 15r = 225 + r²

⇒ 15r = 225 - 56.25

⇒ 15r = 168.75

⇒ r = $\frac{168.75}{15}$ = 11.25 cm.

Hence, radius of circle = 11.25 cm.

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

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

As, AP and AS are tangents to the circle from an external point A, we have

AP = AS ......... (1)

Similarly, we also get

BP = BQ ......... (2)

CR = CQ ......... (3)

DR = DS ......... (4)

Adding (1), (2), (3) and (4), we get

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

From figure,

AP + BP = AB, CR + DR = CD, AS + DS = AD and BQ + CQ = BC

⇒ AB + CD = AD + BC

Hence, proved that AB + CD = AD + BC.

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From point B, BQ and BP are the tangents to the circle

BQ = BP ........ (1)

From point A, AP and AR are the tangents to the circle

AP = AR ........ (2)

From point C, CR and CQ are the tangents to the circle

CR = CQ ........ (3)

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

AP + BQ + CR = BP + CQ + AR ......... (4)

Hence, proved that AP + BQ + CR = BP + CQ + AR.

Now, adding AP + BQ + CR to both sides in (4), we get

2(AP + BQ + CR) = AP + BP + CQ + BQ + AR + CR ..........(5)

From figure,

AP + BP = AB, BQ + CQ = BC and AR + CR = AC.

Substituting above value in equation (5), we get :

⇒ 2(AP + BQ + CR) = AB + BC + CA

⇒ AP + BQ + CR = $\frac{1}{2}$(AB + BC + CA).

⇒ AP + BQ + CR = $\frac{1}{2}$ (Perimeter of △ABC). [∵ Perimeter = AB + BC + CA]

Hence, proved that AP + BQ + CR = $\frac{1}{2}$ (Perimeter of △ABC).

As, from point A, AP and AR are the tangents to the circle.

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

So, we have AP = AR ..........(1)

From point B, BP and BQ are the tangents to the circle.

∴ BP = BQ ..........(2)

From point C, CQ and CR are the tangents to the circle.

∴ CQ = CR ............(3)

Adding equations (1), (2) and (3), we get :

⇒ AP + BP + CQ = AR + BQ + CR

⇒ (AP + BP) + CQ = (AR + CR) + BQ

⇒ AB + CQ = AC + BQ

Given,

AB = AC

∴ BQ = CQ.

Hence, proved that BQ = CQ.

Let O be the center of the circle with radius = 6.3 cm and O' be the center of circle with radius = 3.6 cm.

(i) When the two circles touch each other at P externally. O' and O are the centers of the circles. Join O'P and OP.

So, O'P = 6.3 cm, OP = 3.6 cm

Hence, the distance between their centres (O'O) is given by

O'O = O'P + OP = 6.3 + 3.6 = 9.9 cm.

Hence, distance between their centers if they touch each other externally is 9.9 cm.

(ii) When the two circles touch each other at P internally, O and O' are the centers of the circles. Join OP and O'P.

So, O'P = 6.3 cm, OP = 3.6 cm.

Hence, the distance between their centres (O'O) is given by

O'O = O'P - OP = 6.3 - 3.6 = 2.7 cm.

Hence, distance between their centers if they touch each other internally is 2.7 cm.

(i) We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From figure,

TA and TP are the tangents to the circle with center O.

∴ TA = TP ...........(1)

TB and TP are the tangents to the circle with center O'.

∴ TB = TP ...........(2)

From (1) and (2) we get :

TA = TB.

Hence, proved that tangent at point P bisects AB.

(ii) In △ATP,

TA = TP [Proved above]

∴ ∠TAP = ∠TPA .........(1) [∵ angles opposite to equal sides are equal.]

In △BTP,

TB = TP [Proved above]

∴ ∠TBP = ∠TPB ..........(2) [∵ angles opposite to equal sides are equal.]

Adding (1) and (2), we get :

∠TAP + ∠TBP = ∠TPA + ∠TPB

∠TAP + ∠TBP = ∠APB ...........(3)

In △ABP,

⇒ ∠APB + ∠BAP + ∠ABP = 180° [Angle sum property of triangle]

⇒ ∠APB + ∠TAP + ∠TBP = 180° [From figure, ∠TAP = ∠BAP and ∠TBP = ∠ABP.]

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = 90°.

Hence, proved that ∠APB = 90°.

We know that,

The tangent at any point of a circle and the radius through this point are perpendicular to each other.

In quadrilateral OPAQ,

∠OPA = ∠OQA = 90°.

∠OPA + ∠OQA + ∠POQ + ∠PAQ = 360° [∵ Sum of angles in quadrilateral = 360°]

90° + 90° + ∠POQ + ∠PAQ = 360°

∠POQ + ∠PAQ = 360° - 180°

∠POQ + ∠PAQ = 180° ..........(1)

In △OPQ,

OP = OQ [Radius of same circle]

∴ ∠OPQ = ∠OQP [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OPQ + ∠OQP + ∠POQ = 180°

⇒ ∠OPQ + ∠OPQ + ∠POQ = 180°

⇒ 2∠OPQ + ∠POQ = 180° .........(2)

From (1) and (2) we get,

∠POQ + ∠PAQ = 2∠OPQ + ∠POQ

⇒ ∠PAQ = 2∠OPQ.

Hence, proved that ∠PAQ = 2∠OPQ.

Let AB touches the circle at L, AC at N and BC at M.

From figure,

LBMO is a square.

LB = BM = OM = OL = x.

AL = AB - LB = (12 - x) cm.

AL = AN = (12 - x) cm. [∵ Tangents from exterior point are equal in length.]

Since, ABC is a right angled triangle,

∴ AC² = AB² + BC² [By pythagoras theorem]

⇒ 13² = 12² + BC²

⇒ BC² = 13² - 12²

⇒ BC² = 169 - 144

⇒ BC² = 25

⇒ BC = $\sqrt{25}$

⇒ BC = 5 cm.

From figure,

MC = BC - BM = (5 - x) cm.

Also,

CN = CM = (5 - x) cm. [∵ Tangents from exterior point are equal in length.]

Also,

⇒ AC = AN + CN

⇒ 13 = (12 - x) + (5 - x)

⇒ 13 = 17 - 2x

⇒ 2x = 17 - 13

⇒ 2x = 4

⇒ x = $\frac{4}{2}$

⇒ x = 2 cm.

Hence, x = 2.

(i) From figure,

⇒ ∠QRP = ∠OSR = y° [Angles in alternate segment are equal]

⇒ OS = OR (Radius of same circle)

As, angles opposite to equal sides are equal,

∴ ∠ORS = ∠OSR = y°.

Hence, proved that ∠ORS = y°.

(ii) From figure,

∠ORP = 90° [As, tangent to a point and radius from that point are perpendicular to each other.]

⇒ ∠ORQ = ∠ORP - ∠QRP = 90° - y° ...........(1)

OQ = OR (Radius of same circle)

As, angles opposite to equal sides are equal,

∴ ∠OQR = ∠ORQ = 90° - y°

In △PQR,

⇒ ∠OQR = ∠QPR + ∠QRP (As exterior angle in a trinagle is equal to the sum of two opposite interior angles.)

⇒ 90° - y° = x° + y°

⇒ x° + 2y° = 90°.

Hence, x + 2y = 90°.

Join AT and BT.

(i) TC is the diameter of the circle.

Since, angle in a semi-circle is a right angle.

∴ ∠CBT = 90°.

Hence, ∠CBT = 90°.

(ii) In cyclic quadrilateral ATBC,

⇒ ∠CBT + ∠CAT = 180° (∵ Sum of opposite angles of a cyclic quadrilateral = 180°)

⇒ 90° + ∠CAT = 180°

⇒ ∠CAT = 180° - 90°

⇒ ∠CAT = 90°.

In △ABC,

⇒ ∠CBA + ∠CAB + ∠ACB = 180° [By angle sum property of triangle]

⇒ 70° + ∠CAB + 50° = 180°

⇒ ∠CAB + 120° = 180°

⇒ ∠CAB = 180° - 120°

⇒ ∠CAB = 60°.

From figure,

∠BAT = ∠CAT - ∠CAB = 90° - 60° = 30°.

Hence, ∠BAT = 30°.

(iii) From figure,

∠BTX = ∠BAT = 30° [Angle in same segment are equal]

∠PBT = ∠CBT - ∠CBA = 90° - 70° = 20°.

⇒ ∠PTB = 180° - ∠BTX = 180° - 30° = 150°.

In △PBT,

⇒ ∠PBT + ∠PTB + ∠APT = 180° [By angle sum property of triangle]

⇒ 20° + 150° + ∠APT = 180°

⇒ ∠APT + 170° = 180°

⇒ ∠APT = 180° - 170°

⇒ ∠APT = 10°.

Hence, ∠APT = 10°.

Join OC

∴ PA and PC are the tangents

∴ OA ⊥ PA and OC ⊥ PC

In quadrilateral APCO,

⇒ ∠APC + ∠AOC = 180°

⇒ 80° + ∠AOC = 180°

⇒ ∠AOC = 180° - 80°

⇒ ∠AOC = 100°

From figure,

∠BOC = 360° - (∠AOB + ∠AOC)

= 360° - (140° + 100°)

= 360° - 240° = 120°.

We know that,

The angle at the centre of a circle is twice the angle at the circumference, subtended by the same arc.

Now arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.

∴ ∠BAC = $\frac{1}{2}$∠BOC = $\frac{1}{2} \times 120°$ = 60°.

Hence, ∠BAC = 60°.

From figure,

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

⇒ ∠CAQ = 2∠BAQ = 60°.

From figure,

⇒ ∠CAQ + ∠PAC = 180° [Linear pair]

⇒ 60° + ∠PAC = 180°

⇒ ∠PAC = 180° - 60°

⇒ ∠PAC = 120°.

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

⇒ 120° = 2∠CAD

⇒ ∠CAD = $\frac{120°}{2}$

⇒ ∠CAD = 60°.

From figure,

∠DAB = ∠CAD + ∠CAB = 60° + 30° = 90°.

Thus BD, subtends 90° on the circle. Since, angle in semi-circle is a right angle.

Hence, BD is the diameter of the circle.

Given,

OA : OC = 4 : 7

We know that,

If two chords of a circle intersect internally or externally then the product of the lengths of their segments is equal.

From figure,

⇒ OA × OB = OC × OD

⇒ $\frac{OD}{OB} = \frac{OA}{OC}$

⇒ $\frac{OD}{OB} = \frac{4}{7}$

⇒ $\frac{OB}{OD} = \frac{7}{4}$.

⇒ OB : OD = 7 : 4.

Hence, Option 3 is the correct option.

We know that,

Sum of co-interior angles in a trapezium is 180°.

In trapezium ABDC,

⇒ ∠B + ∠BAC = 180°

⇒ 60° + ∠BAC = 180°

⇒ ∠BAC = 180° - 60° = 120°.

From figure,

⇒ ∠PAC + ∠BAC = 180°

⇒ ∠PAC + 120° = 180°

⇒ ∠PAC = 180° - 120° = 60°.

Given,

⇒ ∠PAC : ∠PCA = 4 : 5

$$\begin{aligned}\Rightarrow \frac{∠\text{PAC}}{∠\text{PCA}} = \frac{4}{5} \\[1em] \Rightarrow \frac{60°}{∠\text{PCA}} = \frac{4}{5} \\[1em] \Rightarrow ∠\text{PCA} = \frac{60° \times 5}{4} \\[1em] \Rightarrow ∠\text{PCA} = \frac{300°}{4} = 75°.\end{aligned}$$

In △PCA,

By angle sum property of triangle,

⇒ ∠APC + ∠PAC + ∠PCA = 180°

⇒ ∠APC + 60° + 75° = 180°

⇒ ∠APC + 135° = 180°

⇒ ∠APC = 180° - 135° = 45°.

Hence, Option 4 is the correct option.

From figure,

⇒ ∠ABE + ∠EBC = 180° [Linear pair]

⇒ ∠ABE + 45° = 180°

⇒ ∠ABE = 180° - 45° = 135°.

We know that,

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

⇒ ∠EDB = ∠ABE = 135°.

Hence, Option 4 is the correct option.

We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

⇒ PB × PC = PA²

⇒ 4 × PC = 8²

⇒ 4 × PC = 64

⇒ PC = $\frac{64}{4}$ = 16 cm.

From figure,

BC = PC - PB = 16 - 4 = 12 cm.

Hence, Option 2 is the correct option.

In △ABC,

∠BAC = 90° (Angle in semi-circle is a right angle)

⇒ ∠ABC + ∠BAC + ∠ACB = 180° (By angle sum property of triangle)

⇒ 60° + 90° + ∠ACB = 180°

⇒ 150° + ∠ACB = 180°

⇒ ∠ACB = 180° - 150° = 30°.

We know that,

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

⇒ ∠BAP = ∠ACB = 30°.

From figure,

⇒ ∠PBA + ∠ABC = 180° [Linear pairs]

⇒ ∠PBA + 60° = 180°

⇒ ∠PBA = 180° - 60° = 120°.

In △PBA,

⇒ ∠PBA + ∠BAP + ∠APB = 180° (By angle sum property of triangle)

⇒ 120° + 30° + ∠APB = 180°

⇒ 150° + ∠APB = 180°

⇒ ∠APB = 180° - 150° = 30°.

∴ ∠P = 30°.

Hence, Option 1 is the correct option.

(i) From figure,

PC = PD + CD = 5 + 7.8 = 12.8 cm.

We know that,

If two chords of a circle intersect internally of externally then the product of the lengths of their segments is equal.

Since, here AB and CD intersect externally at P.

∴ AP x PB = CP x PD

⇒ AP x 4 = 12.8 x 5

⇒ AP = $\frac{12.8 \times 5}{4} = \frac{64}{4}$ = 16 cm.

From figure,

AB = PA - PB = 16 - 4 = 12 cm.

Hence, AB = 12 cm.

(ii) We know that,

If a chord and a tangent intersect externally, then the product of lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ PA x PB = PT²

⇒ 16 x 4 = PT²

⇒ PT = $\sqrt{64}$ = 8 cm.

Hence, PT = 8 cm.

(i) ∠QAB = ∠ADB [∵ angles in alternate segment are equal.]

∴ ∠QAB = 30°.

Hence, the value of ∠QAB = 30°.

(ii) In △ADB,

∠DAB = 90° (∵ angle in semi-circle is 90°.)

Since, sum of angles in a triangle is 180°.

⇒ ∠ABD + ∠ADB + ∠DAB = 180°

⇒ ∠ABD + 30° + 90° = 180°

⇒ ∠ABD + 120° = 180°

⇒ ∠ABD = 60°.

From figure,

∠PAD = ∠ABD = 60° (∵ angles in alternate segment are equal)

Hence, the value of ∠PAD = 60°.

(iii) In △BCD,

∠BCD = 90° [∵ angle in a semi-circle is a right angle.]

∠CBD = 60°

∠CDB + ∠CBD + ∠BCD = 180° [By angle sum property of triangle]

⇒ ∠CDB + 60° + 90° = 180°

⇒ ∠CDB = 180° - 150° = 30°.

Hence, ∠CDB = 30°.

(i) Since, ST passes through O, so ST is the diameter of the circle.

We know that,

Angle in a semi-circle is a right angle.

∴ ∠SRT = 90°.

Since, PQ is a straight line.

∴ ∠PRS + ∠SRT + ∠TRQ = 180°

⇒ ∠PRS + 90° + 30° = 180°

⇒ ∠PRS + 120° = 180°

⇒ ∠PRS = 180° - 120°

⇒ ∠PRS = 60°.

Hence, ∠PRS = 60°.

(ii) We know that,

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

∠TSR = ∠TRQ = 30°.

Since, angle subtended by a segment at the center is double the angle suspended at the circumference.

∠ROT = 2∠TSR = 2 × 30° = 60°.

Hence, ∠ROT = 60°.

Join O'A and O'B.

As, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

CD is the tangent and AO is the chord.

∴ ∠OAC = ∠OBA ......... (1)

In ∆OAB,

OA = OB [Radius of the circle with center O.]

∠OAB = ∠OBA .......... (2) [Angles opposite to equal sides]

From (1) and (2), we have

∠OAC = ∠OAB

Hence, proved that OA is the bisector of ∠BAC.

Join OC, OD and BD.

Given,

∠BCG = 108°

From figure,

⇒ ∠BCG + ∠BCD = 180° [Linear pairs]

⇒ 108° + ∠BCD = 180°

⇒ ∠BCD = 180° - 108°

⇒ ∠BCD = 72°.

From figure,

⇒ ∠BDC = ∠DBC = x(let) [As, angles opposite to equal sides are equal]

In triangle BDC,

⇒ ∠DBC + ∠BDC + ∠BCD = 180° [Angle sum property of triangle]

⇒ x + x + 72° = 180°

⇒ 2x + 72° = 180°

⇒ 2x = 180° - 72°

⇒ 2x = 108°

⇒ x = $\frac{108°}{2}$

⇒ x = 54°.

From figure.

∠BCT = ∠BDC (Angles in alternate segment are equal)

∠BCT = 54°.

Hence, ∠BCT = 54°.

(ii) As, angle subtended by a segment on center is twice the angle subtended by it on any other part of circumference.

⇒ ∠DOC = 2∠DBC

⇒ ∠DOC = 2(54°) = 108°.

Hence, ∠DOC = 108°.

As, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

PA is a tangent and AB is a chord.

∴ ∠PAB = ∠C [Angles in alternate segment are equal]...........(1)

Given,

AD is bisector of ∠BAC.

∴ ∠BAD = ∠DAC .......(2)

We know that,

An exterior angle is equal to sum of two opposite interior angles.

⇒ ∠ADP = ∠C + ∠DAC

⇒ ∠ADP = ∠PAB + ∠BAD [From (1) and (2)]

⇒ ∠ADP = ∠PAD

Since, sides opposite to equal sides are equal.

∴ PA = PD

∴ PAD is an isosceles triangle.

In △ABC,

⇒ ∠PBA = ∠C + ∠BAC [Exterior angle is equal to sum of two opposite interior angles]

⇒ ∠BAC = ∠PBA - ∠C

⇒ ∠BAC = ∠PBA - ∠PAB [As, ∠C = ∠PAB]

⇒ 2∠CAD = ∠PBA - ∠PAB [As, AD bisects ∠BAC]

⇒ ∠CAD = $\frac{1}{2}$(∠PBA - ∠PAB).

Hence, proved that ∠CAD = $\frac{1}{2}$(∠PBA - ∠PAB).

Join AB.

As, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

PQ is a tangent and AB is a chord.

∴ ∠QPA = ∠PBA [Angles in alternate segment are equal] .........(1)

Also,

∴ ∠PQA = ∠QBA [Angles in alternate segment are equal] ..........(2)

Adding (1) and (2) we get,

⇒ ∠QPA + ∠PQA = ∠PBA + ∠QBA ..........(3)

⇒ ∠PBA + ∠QBA = ∠PBQ ...........(4)

In △PAQ,

⇒ ∠QPA + ∠PQA + ∠PAQ = 180° [Angle sum property of triangle]

⇒ ∠QPA + ∠PQA = 180° - ∠PAQ

⇒ ∠PBA + ∠QBA = 180° - ∠PAQ [From (3)] ..........(5)

From (4) and (5), we get :

⇒ ∠PBQ = 180° - ∠PAQ

⇒ ∠PBQ + ∠PAQ = 180°.

Hence, proved that PAQ and PBQ are supplementary.

(i) Join AB.

∠ADB = ∠CDE = 90° [Vertically opposite angles are equal.]

In right angle triangle ADB,

⇒ AB² = AD² + BD²

⇒ 5² = AD² + 4²

⇒ 25 = AD² + 16

⇒ AD² = 25 - 16

⇒ AD² = 9

⇒ AD = $\sqrt{9}$

⇒ AD = 3 cm.

We know that,

If two chords of a circle intersect internally or externally then the product of the lengths of their segment is equal.

From figure,

Chords AE and CB intersect internally at point D.

⇒ AD × DE = CD × BD

⇒ 3 × DE = 4 × 9

⇒ DE = $\frac{36}{3}$

⇒ DE = 12 cm.

Hence, DE = 12 cm.

(ii) Given,

AD = BD ........(1)

AD = BD = x (let)

We know that,

⇒ AD × DE = CD × BD

⇒ (x)DE = (x)CD

⇒ DE = CD ..........(2)

Adding (1) and (2), we get :

⇒ AD + DE = BD + CD

⇒ AE = BC.

Hence, proved that AE = BC.

From figure,

⇒ ∠ADE + ∠BDE = 180° [Linear pairs]

⇒ ∠ADE + 65° = 180°

⇒ ∠ADE = 180° - 65°

⇒ ∠ADE = 115° ..............(1)

∠DBO = 90° [As, DB is tangent and BC is diameter.]

In △BDC,

⇒ ∠BDC + ∠DBC + ∠DCB = 180° [Angle sum property of triangle]

⇒ 65° + 90° + ∠DCB = 180°

⇒ ∠DCB = 180° - 155° = 25°.

From figure,

OE = OC [Radius of same circle.]

⇒ ∠OCE = ∠OEC [As, angles opposite to equal sides are equal.]

⇒ ∠OEC = ∠DCB = 25°. [As, ∠OCE = ∠DCB]

In △ADE,

⇒ ∠ADE + ∠DEA + ∠DAE = 180° [Angle sum property of triangle]

⇒ 115° + 25° + ∠DAE = 180° [From figure, ∠DEA = ∠OEC [Vertically opposite angles are equal]]

⇒ ∠DAE = 180° - 140° = 40°.

From figure,

∠BAO = ∠DAE = 40°.

Hence, ∠BAO = 40°.

We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

From figure,

AC = AB + BC = 8 + 10 = 18 cm.

⇒ AB × AC = AP²

⇒ 8 × 18 = AP²

⇒ AP² = 144

⇒ AP = $\sqrt{144}$ = 12 cm.

Hence, Option 3 is the correct option.

From figure,

In △OAB,

OA = OB (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

⇒ ∠OBA = ∠OAB = x

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ OB ⊥ PQ

∴ ∠PBO = 90°

From figure,

∠ABP = ∠PBO - ∠OBA = 90° - x.

Hence, Option 4 is the correct option.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ OA ⊥ AB

∴ ∠OAB = 90°.

Let, ∠BAC = x

From figure,

In △OAC,

∠A = ∠OAB - ∠BAC = 90° - x.

Also,

OA = OC (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

∴ ∠C = ∠A = 90° - x.

By angle sum property of triangle,

⇒ ∠A + ∠O + ∠C = 180°

⇒ 90° - x + ∠O + 90° - x = 180°

⇒ ∠O + 180° - 2x = 180°

⇒ ∠O = 180° - 180° + 2x = 2x.

In △OAB,

By angle sum property of triangle,

⇒ ∠O + ∠A + ∠B = 180°

⇒ ∠O + ∠OAB + ∠B = 180°

⇒ 2x + 90° + 20° = 180°

⇒ 2x + 110° = 180°

⇒ 2x = 180° - 110°

⇒ 2x = 70°

⇒ x = $\frac{70°}{2}$ = 35°.

⇒ ∠BAC = 35°.

Hence, Option 3 is the correct option.

In △OAD,

OA = OD (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

∴ ∠A = ∠D = x (let)

⇒ ∠O + ∠A + ∠D = 180° (By angle sum property of triangle)

⇒ 84° + x + x = 180°

⇒ 2x = 180° - 84°

⇒ 2x = 96°

⇒ x = $\frac{96°}{2}$ = 48°.

From figure,

∠OAD = ∠A = 48°

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAQ = 90°

From figure,

∠DAQ = ∠OAQ - ∠OAD = 90° - 48° = 42°.

Hence, Option 4 is the correct option.

Let two perpendicular tangents from external point A touch the circle at points B and C.

Given,

Radius = $R\sqrt{2}$ units

From figure,

AC = OB = $R\sqrt{2}$,

AB = OC = $R\sqrt{2}$.

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ BC² = $(R\sqrt{2})^2 + (R\sqrt{2})^2$

⇒ BC² = 2R² + 2R²

⇒ BC² = 4R²

⇒ BC = $\sqrt{4R^2}$

⇒ BC = 2R units.

Hence, Option 4 is the correct option.

Let the circles intersect at points D, E and F.

From figure,

Perimeter of triangle ABC = AB + BC + CA

= (AD + BD) + (BE + CE) + (CF + FA)

= 5 + 2 + 2 + 6 + 6 + 5

= 26 cm.

On adding radii of three circles, we get :

5 + 2 + 6 = 13 cm, which is not equal to perimeter.

Sum of radii × 2 = 13 × 2 = 26 cm, which is equal to perimeter.

∴ A is false, R is true

Hence, Option 2 is correct option.

We know that,

The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOC = 2 x ∠ABC

⇒ 60° = 2 x ∠ABC

⇒ ∠ABC = $\frac{60°}{2}$ = 30°

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

∴ AP ⊥ OA

⇒ ∠OAP = 90°

⇒ ∠OAP = ∠BAP = 90°

In ΔABP, according to angle sum property,

∴ ∠ABP + ∠APB + ∠BAP = 180°

⇒ 30° + x + 90° = 180°

⇒ 30° + x = 180° - 90°

⇒ 30° + x = 90°.

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

In ΔPAB and ΔPCD,

⇒ ∠PBA = ∠PDC (Exterior angles of a cyclic quadrilateral is always equal to interior opposite angle)

⇒ ∠PAB = ∠PCD ((Exterior angles of a cyclic quadrilateral is always equal to interior opposite angle))

ΔPAB ~ ΔPCD (By A.A. axiom of similarity)

We know that,

Corresponding sides of similar triangles are proportional.

$\Rightarrow \frac{PD}{PB} = \frac{PC}{PA}$

∴ Reason (R) is true.

⇒ PD x PA = PC x PB

∴ Assertion (A) is false.

∴ A is false, R is true.

Hence, option 2 is the correct option.

Join OP.

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From figure,

O is the point from which, OA and OP are two tangents to the circle with centre Q'.

So, OA = OP .......(1)

Similarly, from point O, OB and OP are two tangents to the circle with centre Q.

So, OB = OP ......(2)

From (1) and (2), we have

⇒ OA = OB

⇒ OB = 10 cm

∴ Both the statements are true.

Hence, option 1 is the correct option.

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

∴ OB ⊥ BP and OC ⊥ CP

⇒ ∠OBP = 90° and ∠OCP = 90°

OCPB is a quadrilateral.

∴ ∠OBP + ∠BPC + ∠OCP + ∠BOC = 360°

⇒ 90° + 50° + 90° + ∠BOC = 360°

⇒ 230° + ∠BOC = 360°

⇒ ∠BOC = 360° - 230°

⇒ ∠BOC = 130°

We know that,

The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠BOC = 2 x ∠BAC

⇒ 130° = 2 x ∠BAC

⇒ ∠BAC = $\frac{130°}{2}$ = 65°.

So, Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

As ABCD is a cyclic quadrilateral, we have

⇒ ∠BCD + ∠BAD = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

⇒ ∠BCD + 70° = 180°

⇒ ∠BCD = 180° - 70° = 110°

In ∆BCD,

⇒ ∠CBD + ∠BCD + ∠BDC = 180° [By angle sum property of triangle]

⇒ 30° + 110° + ∠BDC = 180°

⇒ ∠BDC = 180° - 140° = 40°.

Hence, ∠BDC = 40°.

Given, cyclic quadrilateral ABCD

So, ∠A + ∠C = 180° [Opposite angles in a cyclic quadrilateral is supplementary]

⇒ 3∠C + ∠C = 180° [As ∠A = 3∠C]

⇒ 4∠C = 180°

⇒ ∠C = $\frac{180°}{4}$

⇒ ∠C = 45°.

Now,

⇒ ∠A = 3∠C = 3 x 45° = 135°.

Similarly,

⇒ ∠B + ∠D = 180°

⇒ ∠B + 5∠B = 180° [As, ∠D = 5∠B]

⇒ 6∠B = 180°

⇒ ∠B = $\frac{180°}{6}$

⇒ ∠B = 30°.

Now,

⇒ ∠D = 5∠B = 5 x 30° = 150°.

Hence, ∠A = 135°, ∠B = 30°, ∠C = 45° and ∠D = 150°.

Let circle be drawn on one of the equal sides AB of the isosceles triangle ABC as shown in the figure below:

We have ∠ADB = 90° [Angle in a semi-circle is a right angle]

But,

⇒ ∠ADB + ∠ADC = 180° [Linear pair]

⇒ 90° + ∠ADC = 180°

⇒ ∠ADC = 180° - 90°

⇒ ∠ADC = 90°.

In ∆ABD and ∆ACD, we have

⇒ ∠ADB = ∠ADC [Each 90°]

⇒ AB = AC [Given]

⇒ AD = AD [Common]

Hence, ∆ABD ≅ ∆ACD by RHS congruence criterion.

By, C.P.C.T we get :

BD = DC

Hence, the circle bisects base BC at D.

BE is the bisector of ∠B.

⇒ ∠ABE = $\frac{∠B}{2}$

From figure,

⇒ ∠ADE = ∠ABE [Angles in same segment are equal]

⇒ ∠ADE = $\frac{∠B}{2}$ ...........(1)

Also,

FC is the bisector of ∠C.

⇒ ∠ACF = $\frac{∠C}{2}$

⇒ ∠ACF = ∠ADF [Angles in same segment are equal]

⇒ ∠ADF = $\frac{∠C}{2}$ .............(2)

From figure,

⇒ ∠D = ∠ADE + ∠ADF

⇒ ∠D = $\frac{∠B + ∠C}{2}$ ...........(3)

In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° - ∠A

Substituting above value in (3), we get :

⇒ ∠D = $\frac{180° - ∠A}{2}$

⇒ ∠D = 90° - $\frac{1}{2}$∠A.

From figure,

∠EDF = ∠D.

Hence, proved that ∠EDF = $90° - \frac{1}{2}∠A$.

Join OB.

Given,

BC = OD = radius of circle

⇒ BC = OB.

As, angles opposite to equal sides are equal.

⇒ ∠BOC = ∠BCO = 20°.

An exterior angle is equal to the sum of two opposite interior angles.

∴ ∠ABO = ∠BCO + ∠BOC = 20° + 20° = 40° ............(1)

Now in ∆OAB,

OA = OB [Radii of the same circle]

As, angles opposite to equal sides are equal.

∠OAB = ∠ABO = 40° [from (1)]

By angle sum property of triangle,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 40° + 40° = 180°

⇒ ∠AOB + 80° = 180°

⇒ ∠AOB = 180° - 80°

⇒ ∠AOB = 100°

As DOC is a straight line,

⇒ ∠AOD + ∠AOB + ∠BOC = 180°

⇒ ∠AOD + 100° + 20° = 180°

⇒ ∠AOD = 180° - 120° = 60°.

Hence, ∠AOD = 60°.

We know that,

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

As, TPS is a tangent and PA is the chord of the circle.

∠BPT = ∠PAB [Angles in alternate segments are equal] ..........(1)

But,

∠PBA = ∠PAB [Since, PA = PB as P is mid-point of arc APB.] ........(2)

From (1) and (2), we get :

∠BPT = ∠PBA

The above angles are alternate angles,

∴ TPS || AB

Hence, proved that the tangent drawn at P will be parallel to the chord AB.

(i) We know that,

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

PQ is a tangent and CD is a chord.

⇒ ∠DBC = ∠DCQ [Angles in the alternate segment are equal]

⇒ ∠DBC = 40°.

Hence, ∠DBC = 40°.

(ii) In △DCB

⇒ ∠DBC + ∠DCB + ∠CDB = 180° [By angle sum property of triangle]

⇒ 40° + 90° + ∠CDB = 180° [∠DCB = 90°, as angle in a semi-circle is a right angle]

⇒ ∠CDB = 180° - 130° = 50°.

From figure,

⇒ ∠BCP = ∠CDB = 50°. [Angles in the alternate segment are equal]

Hence, ∠BCP = 50°.

(iii) In ∆ABD,

∠BAD = 90° [Angle in a semi-circle is a right angle]

∠ABD = 60° [Given]

⇒ ∠ADB + ∠BAD + ∠ABD = 180° [By angle sum property of triangle]

⇒ ∠ADB + 90° + 60° = 180°

⇒ ∠ADB = 180° - 150° = 30°

Hence, ∠ADB = 30°.

Join OC.

We know that,

The radius from the center of the circle to the point of tangent is perpendicular to the tangent line.

BCD is the tangent and OC is the radius.

As, OC ⊥ BD

∠OCD = 90°

∴ ∠OCA + ∠ACD = 90° ............. (1)

In ∆OCA,

⇒ OA = OC [Radius of the same circle]

∴ ∠OCA = ∠OAC [As, angles opposite to equal sides are equal]

Substituting in (1), we get

⇒ ∠OAC + ∠ACD = 90°

⇒ ∠BAC + ∠ACD = 90° [From figure, ∠BAC = ∠OAC]

Hence, proved that ∠ACD + ∠BAC = 90°.

(i) In ∆ABC, we have

∠B = 90° and BC is the diameter of the circle.

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

We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ AB² = AD x AC.

Hence, proved that AB² = AD x AC.

(ii) From figure,

∠BDC = 90° [Angle in a semi-circle is a right angle.]

From figure,

⇒ ∠ADB + ∠BDC = 180° [Linear pairs]

⇒ ∠ADB + 90° = 180°

⇒ ∠ADB = 180° - 90°

⇒ ∠ADB = 90°

In ∆ADB,

⇒ ∠ADB + ∠A + ∠ABD = 180° [By angle sum property of triangle]

⇒ 90° + ∠A + ∠ABD = 180°

⇒ ∠A + ∠ABD = 90° ...............(1)

In ∆ABC, ∠ABC = 90°.

⇒ ∠ABC + ∠A + ∠ACB = 180° [By angle sum property of triangle]

⇒ 90° + ∠A + ∠ACB = 180°

⇒ ∠A + ∠ACB = 90° ...............(2)

From (1) and (2),

⇒ ∠A + ∠ABD = ∠A + ∠ACB

⇒ ∠ABD = ∠ACB.

From figure,

⇒ ∠ACB = ∠BCD

∴ ∠ABD = ∠BCD

Now in ∆ABD and ∆CBD, we have

∠BDA = ∠BDC [Both equal to 90°]

∠ABD = ∠BCD

Hence, ∆ABD ~ ∆CBD by AA postulate.

We know that,

Ratio of corresponding sides of similar triangles are same.

$\frac{BD}{DC} = \frac{AD}{BD}$

∴ BD² = AD x DC.

Hence, proved that BD² = AD x DC.

(i) In ∆ADC and ∆ABE,

⇒ ∠ACD = ∠AEB [Angles in the same segment are equal]

⇒ AC = AE [Given]

⇒ ∠A = ∠A [Common]

Hence, ∆ADC ≅ ∆ABE by ASA axiom.

So, by C.P.C.T we have

⇒ AD = AB ..............(1)

Given,

⇒ AE = AC .............(2)

Subtracting equation (1) from (2), we get :

⇒ AE - AD = AC - AB

⇒ DE = BC

In ∆BPC and ∆DPE,

⇒ ∠C = ∠E [Angles in the same segment are equal]

⇒ BC = DE [Proved above]

⇒ ∠CBP = ∠PDE [Angles in the same segment are equal]

Hence, ∆BPC ≅ ∆DPE by ASA axiom.

So, by C.P.C.T we have

⇒ CP = EP

Hence, proved that CP = EP.

(ii) Proved above,

∆BPC ≅ ∆DPE

∴ BP = DP [By C.P.C.T]

Hence, proved that BP = DP.

(i) In ∆OAC and ∆OBC, we have

⇒ OC = OC [Common]

⇒ OA = OB [Radii of the same circle]

⇒ CA = CB [Tangents to the circle from an exterior point are equal.]

Hence, ∆OAC ≅ ∆OBC by SSS congruence criterion.

∴ ∠ACO = ∠BCO = 30° [By C.P.C.T.]

Hence, ∠BCO = 30°.

(ii) From figure,

∠ACB = ∠ACO + ∠BCO = 30° + 30° = 60°.

Sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠AOB + ∠ACB = 180°

⇒ ∠AOB = 180° - 60° = 120°.

Hence, ∠AOB = 120°.

(iii) Arc AB subtends ∠AOB at the center and ∠APB is the remaining part of the circle.

We know that.

When two angles are subtended by the same arc, the angle at the centre of a circle is twice the angle at the circumference.

⇒ ∠APB = $\frac{1}{2}$∠AOB = $\frac{1}{2}$ x 120° = 60°.

Hence, ∠APB = 60°.

Join PB.

In cyclic quadrilateral PBCQ,

⇒ ∠BPQ + ∠BCQ = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

⇒ ∠BPQ + 140° = 180°

⇒ ∠BPQ = 180° - 140°

⇒ ∠BPQ = 40° ............(1)

In △PBQ,

∠PBQ = 90° [Angle in a semi-circle is a right angle.]

⇒ ∠PBQ + ∠BPQ + ∠PQB = 180° [Angle sum property of triangle]

⇒ 90° + 40° + ∠PQB = 180°

⇒ 130° + ∠PQB = 180°

⇒ ∠PQB = 180° - 130°

⇒ ∠PQB = 50°.

In cyclic quadrilateral PQBA,

⇒ ∠PQB + ∠PAB = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

⇒ 50° + ∠PAB = 180°

⇒ ∠PAB = 180° - 50°

⇒ ∠PAB = 130°.

In △PAB,

⇒ ∠PAB + ∠PBA + ∠BPA = 180° [Angle sum property of triangle]

⇒ 130° + ∠PBA + ∠BPA = 180°

⇒ ∠PBA + ∠BPA = 180° - 130°

⇒ ∠PBA + ∠BPA = 50° ................(2)

Given, PA = PB

Angles opposite to equal sides are equal.

∴ ∠PBA = ∠BPA = x (let)

Substituting above value in (2), we get :

⇒ x + x = 50°

⇒ 2x = 50°

⇒ x = $\frac{50°}{2}$

⇒ x = 25°.

From figure,

∠AQB = ∠APB = 25° [Angles in same segment are equal.]

∠APQ = ∠APB + ∠BPQ = 25° + 40° = 65.

We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc AQ subtends ∠AOQ at the center and ∠APQ at the remaining part of the circle.

∠AOQ = 2∠APQ = 2 × 65° = 130°.

In △AOQ,

OA = OQ [Radii of same circle]

As, angles opposite to equal sides are equal.

∠OAQ = ∠OQA = y (let)

⇒ ∠OAQ + ∠OQA + ∠AOQ = 180° [Angle sum property of triangle]

⇒ y + y + 130° = 180°

⇒ 2y = 180° - 130°

⇒ 2y = 50°

⇒ y = $\frac{50°}{2}$

⇒ y = 25°.

∴ ∠OAQ = 25°.

Since, ∠OAQ = ∠AQB = 25°.

∠OAQ and ∠AQB are alternate angles.

Thus, AO and BQ are parallel.

Hence, ∠AQB = 25° and ∠PAB = 130°.

(i) From figure,

⇒ ∠POQ + ∠QOR = 180° [Linear pairs]

⇒ 100° + ∠QOR = 180°

⇒ ∠QOR = 180° - 100°

⇒ ∠QOR = 80°.

We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc RQ subtends ∠QOR at the center and ∠QTR at the remaining part of the circle.

⇒ ∠QOR = 2∠QTR

⇒ ∠QTR = $\frac{1}{2}$∠QOR = $\frac{1}{2} \times 80° = 40°.$

Hence, ∠QTR = 40°.

(ii) We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc QP subtends ∠QOP at the center and ∠QRP at the remaining part of the circle.

⇒ ∠QOP = 2∠QRP

⇒ ∠QRP = $\frac{1}{2}$∠QOP = $\frac{1}{2} \times 100° = 50°.$

Hence, ∠QRP = 50°.

(iii) Given,

RS || QT

⇒ ∠SRT = ∠QTR = 40° (Alternate angles are equal)

From figure,

∠QRS = ∠QRP + ∠PRT + ∠SRT = 50° + 20° + 40° = 110°.

Hence, ∠QRS = 110°.

(iv) Since, RSTQ is a cyclic quadrilateral and sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠QRS + ∠QTS = 180°

⇒ ∠QRS + ∠QTR + ∠STR = 180°

⇒ 110° + 40° + ∠STR = 180°

⇒ ∠STR = 180° - 150°

⇒ ∠STR = 30°.

Hence, ∠STR = 30°.

In △AXB,

⇒ ∠AXB + ∠XAB + ∠ABX = 180° [Angle sum property of triangle]

⇒ 50° + XAB + 70° = 180°

⇒ ∠XAB = 180° - 120° = 60°.

From figure,

∠XAY = 90° [Angle in a semi-circle is a right angle.]

∠BAY = ∠XAY - ∠XAB = 90° - 60 = 30°.

∠BXY = ∠BAY = 30° [Angles in same segment are equal]

We know that,

An exterior angle is equal to the sum of two opposite interior angles.

⇒ ∠ACX = ∠BXC + ∠CBX

⇒ ∠ACX = ∠BXY + ∠ABX [From figure, ∠BXC = ∠BXY and ∠CBX = ∠ABX]

⇒ ∠ACX = 30° + 70° = 100°.

We know that,

Diameter is perpendicular to tangent.

⇒ ∠XYP = 90°

An exterior angle in a triangle is equal to sum of two opposite interior angles.

⇒ ∠ACX = ∠APY + ∠CYP

⇒ ∠APY = ∠ACX - ∠CYP = 100° - 90° = 10°.

Hence, ∠APY = 10° and ∠BAY = 30°.

(i) We know that,

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

∴ ∠BAP = ∠ACB = 36°.

Hence, ∠BAP = 36°.

(ii) We know that,

An exterior angle in a triangle is equal to the sum of two opposite interior angles.

In △APB,

∠ABD = ∠APB + ∠BAP = 42° + 36° = 78°.

Hence, ∠ABD = 78°.

(iii) From figure,

∠ADB = ∠ACB = 36° (Angles in same segment are equal)

In △PAD,

∠QAD = ∠APB + ∠ADB = 42° + 36° = 78°. [Exterior angle is equal to sum of two opposite interior angles.]

Hence, ∠QAD = 78°.

(iv) We know that,

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

∴ ∠ACD = ∠QAD = 78°.

From figure,

∠BCD = ∠ACB + ∠ACD = 36° + 78° = 114°.

Hence, ∠BCD = 114°.

(i) From figure,

∠ACB = 90° (Angle in a semi-circle is a right angle.)

In △ACB,

⇒ ∠ACB + ∠CAB + ∠CBA = 180° [By angle sum property of triangle]

⇒ 90° + 34° + ∠CBA = 180°

⇒ ∠CBA = 180° - 124° = 56°.

Hence, ∠CBA = 56°.

(ii) From figure,

∠QCB = ∠CAB = 34° [Angle in alternate segment are equal.]

⇒ ∠CBQ + ∠CBA = 180° [Linear pairs]

⇒ ∠CBQ + 56° = 180°

⇒ ∠CBQ = 124°.

In △CBQ,

⇒ ∠CBQ + ∠QCB + ∠CQB = 180° [By angle sum property of triangle]

⇒ 124° + 34° + ∠CQB = 180°

⇒ ∠CQB = 180° - 158° = 22°.

Hence, ∠CQB = 22°.

(i) From figure,

∠BCD = ∠ABC = 55° [Alternate angles are equal.]

We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠BOD = 2∠BCD = 2 x 55° = 110°.

Hence, ∠BOD = 110°.

(ii) We know that,

A tangent line is always at a right angle to the radius of the circle at the point of tangency.

∴ ∠OBP = 90° and ∠ODP = 90°.

In quadrilateral ODPB,

⇒ ∠BOD + ∠OBP + ∠ODP + ∠BPD = 360° [Angle sum property of quadrilateral]

⇒ 110° + 90° + 90° + ∠BPD = 360°

⇒ ∠BPD = 360° - 290° = 70°.

Hence, ∠BPD = 70°.

(i) Given,

PQ = QR

∴ ∠PRQ = ∠QPR [Angles opposite to equal sides are equal in a triangle.]

In △PQR,

⇒ ∠PRQ + ∠QPR + ∠RQP = 180°

⇒ ∠PRQ + ∠PRQ + 68° = 180°

⇒ 2∠PRQ = 180° - 68°

⇒ 2∠PRQ = 112°

⇒ ∠PRQ = $\frac{112°}{2}$

⇒ ∠PRQ = 56°.

We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠QOP = 2∠PRQ = 2 x 56 = 112°.

Hence, ∠QOP = 112°.

(ii) We know that,

A tangent line is always at a right angle to the radius of the circle at the point of tangency.

∴ ∠OPC = 90° and ∠OQC = 90°.

In quadrilateral OQCP,

⇒ ∠QOP + ∠OPC + ∠OQC + ∠QCP = 360° [Angle sum property of quadrilateral]

⇒ 112° + 90° + 90° + ∠QCP = 360°

⇒ ∠QCP = 360° - 292° = 68°.

Hence, ∠QCP = 68°.

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

∴ ∠PAB = 90°.

Since, AC is a tangent to the circle with center Q at point C.

∴ ∠QCB = 90°.

In △PAB and △QCB,

⇒ ∠PAB = ∠QCB (Both equal to 90°)

⇒ ∠PBA = ∠QBC (Vertically opposite angles are equal)

⇒ △PAB ~ △QCB.

In right angle △PAB,

$$\begin{aligned}\phantom{\Rightarrow} \text{PB}^2 = \text{PA}^2 + \text{AB}^2 \\[1em] \Rightarrow \text{PB} = \sqrt{\text{PA}^2 + \text{AB}^2} \\[1em] \Rightarrow \text{PB} = \sqrt{6^2 + 8^2} \\[1em] \Rightarrow \text{PB} = \sqrt{36 + 64} \\[1em] \Rightarrow \text{PB} = \sqrt{100} \\[1em] \Rightarrow \text{PB} = 10 \text{ cm}\end{aligned}$$

We know that,

In similar triangles ratio of corresponding sides are equal.

$$\begin{aligned}\Rightarrow \frac{PA}{QC} = \frac{PB}{QB} \\[1em] \Rightarrow \frac{6}{3} = \frac{10}{QB} \\[1em] \Rightarrow QB = \frac{3 \times 10}{6} \\[1em] \Rightarrow QB = \frac{30}{6} \\[1em] \Rightarrow QB = 5 \text{ cm}.\end{aligned}$$

From figure,

QP = QB + PB = 5 + 10 = 15 cm.

Hence, QP = 15 cm.

YT and XT are tangents to the circle.

∴ ∠OYT = 90° and ∠OXT = 90°.

In quadrilateral OYTX,

⇒ ∠XOY + ∠OYT + ∠OXT + ∠XTY = 360°

⇒ ∠XOY + 90° + 90° + 80° = 360°

⇒ ∠XOY = 360° - 260° = 100°.

From figure,

⇒ ∠XOZ + ∠YOZ + ∠XOY = 360°

⇒ 140° + ∠YOZ + 100° = 360°

⇒ ∠YOZ = 360° - 240° = 120°.

We know that,

When two angles are subtended by the same arc, the angle at the centre of a circle is twice the angle at the circumference.

∴ ∠YOZ = 2∠ZXY

⇒ ∠ZXY = $\frac{1}{2}$∠YOZ = $\frac{1}{2} \times 120°$ = 60°.

Hence, ∠ZXY = 60°.

From figure,

⇒ ∠ADC + ∠ABC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ ∠ADC + 100° = 180°

⇒ ∠ADC = 180° - 100°

⇒ ∠ADC = 80°.

In △ADC,

⇒ ∠ADC + ∠CAD + ∠ACD = 180° [By angle sum property of triangle]

⇒ 80° + ∠CAD + 40° = 180°

⇒ ∠CAD = 180° - 120° = 60°.

From figure,

⇒ ∠DCT = ∠CAD = 60° [Angles in alternate segment are equal].

Hence, ∠DCT = 60° and ∠ADC = 80°.

As, SP is tangent to the circle.

∴ ∠TSR = 90°.

In △TSR,

⇒ ∠TSR + ∠STR + ∠SRT = 180° [By angle sum property of triangle]

⇒ 90° + x + 65° = 180°

⇒ x = 180° - 155° = 25°.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠SOQ = 2∠STQ

⇒ y = 2x = 2(25°) = 50°.

In △OSP,

⇒ ∠OSP + ∠SOP + ∠SPO = 180° [By angle sum property of triangle]

⇒ 90° + y + z = 180°

⇒ 90° + 50° + z = 180°

⇒ z = 180° - 140° = 40°.

Hence, x = 25°, y = 50° and z = 40°.

(a)

The two circles touch each other externally at one point. Hence, 3 common tangents can be drawn — two direct common tangents and one common tangent at the point of contact.

(b)

The two circles lie apart and neither touch nor intersect. Hence, 4 common tangents can be drawn — two direct common tangents and two transverse common tangents.

(c)

The two circles touch each other internally at one point. Hence, 1 common tangent can be drawn, at the point of contact.

(d)

The two circles intersect each other at two points. Hence, 2 common tangents can be drawn — two direct common tangents.