Tangent Properties of Circles

Consider a circle with centre O and radius 8 cm.

Let P be an external point from where a tangent is drawn to meet the circle at T. Join OT.

∴ OP = 10 cm and OT = 8 cm

We know that,

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

In right angled ∆OTP, we have

⇒ OP² = OT² + PT²

⇒ 10² = 8² + PT²

⇒ PT² = 100 - 64 = 36

⇒ PT = 6 cm

Hence, the length of tangent = 6 cm.

Consider a circle with centre O,

Let P be an external point from where a tangent is drawn to meet the circle at A.

∴ OP = 17 cm and AP = 15 cm

We know that,

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

In right angled ∆OAP, we have

⇒ OP² = OA² + AP²

⇒ 17² = OA² + 15²

⇒ 289 = OA² + 225

⇒ OA² = 289 - 225

⇒ OA² = 64

⇒ OA = $\sqrt{64}$

⇒ OA = 8 cm

Hence, the radius of circle is 8 cm.

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

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

In the right angled triangle OPB, by pythagoras theorem,

⇒ OB² = OP² + PB²

⇒ 26² = 10² + PB²

⇒ 676 = 100 + PB²

⇒ PB² = 676 - 100

⇒ PB² = 576

⇒ PB = $\sqrt{576}$

⇒ PB = 24 cm

As perpendicular line from centre bisects the chord of the circle so,

AP = PB = 24 cm.

AB = AP + PB = 24 + 24 = 48 cm.

Hence, the length of chord (AB) = 48 cm.

From figure,

In △ABC,

By pythagoras theorem,

⇒ AB² = AC² + BC²

⇒ 17² = (r + 9)² + (r + 2)²

⇒ 289 = r² + 81 + 18r + r² + 4 + 4r

⇒ 289 = 2r² + 85 + 22r

⇒ 2r² + 22r + 85 - 289 = 0

⇒ 2r² + 22r - 204 = 0

⇒ 2(r² + 11r - 102) = 0

⇒ r² + 11r - 102 = 0

⇒ r² + 17r - 6r - 102 = 0

⇒ r(r + 17) - 6(r + 17) = 0

⇒ (r - 6)(r + 17) = 0

⇒ r - 6 = 0 or r + 17 = 0

⇒ r = 6 or r = -17.

Since, radius cannot be negative.

⇒ r = 6 cm.

Hence, equation is r² + 11r - 102 = 0 and r = 6 cm.

The tangents from a point outside the circle are equal.

The point P is also external to the right circle. The segments PB and PC are tangents drawn from P to this circle.

PB = PC .........(1)

The point P is external to the left circle. The segments PA and PC are tangents drawn from P to this circle.

PA = PC .........(2)

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

∴ PA = PB.

Hence, proved that PA = PB.

As tangents drawn from an external point to a circle are equal in length.

From T, TA and TP are tangents to the circle with centre O.

TA = TP .....(1)

From T, TB and TP are tangents to the circle with centre O'.

TB = TP ........(2)

From (1) and (2),

TA = TB.

Hence, proved that tangents drawn to two circles from any point on common tangent are equal in length.

Let there be two circles with centre A and B with radius 18 cm and 8 cm respectively.

Let TT' be the length of common tangent.

Construct a right-angled triangle Δ ADB by drawing a line from center B parallel to TT' to intersect radius AT at D.

From figure,

DT = BT' = 8cm.

AD = AT - DT = 18 - 8 = 10 cm.

AB = 18 + 8 = 26 cm

In right angled triangle ADB,

⇒ AB² = AD² + DB²

⇒ 26² = 10² + DB²

⇒ 676 = 100 + DB²

⇒ DB² = 676 - 100

⇒ DB² = 576

⇒ DB = 24 cm

Since, TDBT' is a rectangle,

So, TT' = DB = 24 cm.

Hence, the length of direct common tangent is 24 cm.

Let there be two circles with centre A and B with radius 8 cm and 3 cm respectively.

Let TT' be the length of common tangent.

Construct a right-angled triangle ΔADB by drawing a line from center B parallel to TT' to intersect radius AT at D.

From figure,

DT = BT' = 3cm.

AD = AT - DT = 8 - 3 = 5 cm.

In right angled triangle ADB,

⇒ AB² = AD² + DB²

⇒ 13² = 5² + DB²

⇒ DB² = 13² - 5²

⇒ DB² = 169 - 25

⇒ DB² = 144

⇒ DB = 12 cm

Since, TDBT' is a rectangle,

So, TT' = DB = 12 cm.

Hence, the length of direct common tangent is 12 cm.

Let there be two circles with centre A and B with radius 8 cm and 3 cm respectively.

Let TT' be the length of common tangent.

Construct a right-angled triangle Δ ADB by drawing a line from center B parallel to TT' to intersect radius AT at D.

From figure,

⇒ TT' = BD = 10 cm

⇒ DT = BT' = 3 cm.

⇒ AD = AT - DT = 8 - 3 = 5 cm.

In right angled triangle ADB,

⇒ AB² = AD² + DB²

⇒ AB² = 10² + 5²

⇒ AB² = 100 + 25

⇒ AB² = 125

⇒ AB = $\sqrt{125}$

⇒ AB = 11.18 cm.

Hence, the distance between the centres 11.18 cm.

Let radius of circles with center P, Q and R be r₁, r₂ and r₃.

⇒ PQ = r₁ + r₂ = 7 …………(1)

⇒ PR = r₁ + r₃ = 8 …………(2)

⇒ QR = r₂ + r₃ = 11 ………….(3)

Adding all the above equations, we get

⇒ r₁ + r₂ + r₁ + r₃ + r₂ + r₃ = 7 + 8 + 11

⇒ 2(r₁ + r₂ + r₃) = 26

⇒ r₁ + r₂ + r₃ = $\frac{26}{2}$ ​ ⇒ r₁ + r₂ + r₃ = 13 cm ………..(4)

Substituting value of r₂ + r₃ = 11 in equation (4) we get :

⇒ r₁ + 11 = 13

⇒ r₁ = 13 - 11

⇒ r₁ = 2 cm.

Substituting value of r₁ + r₂ = 7 in equation (4) we get :

⇒ 7 + r₃ = 13

⇒ r₃ = 13 - 7

⇒ r₃ = 6 cm.

Substituting value of r₁ + r₃ = 8 in equation (4) we get :

⇒ 8 + r₂ = 13

⇒ r₂ = 13 - 8

⇒ r₂ = 5 cm.

Hence, the radii of the circles with center P, Q and R are 2 cm, 5 cm and 6 cm.

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

From figure,

ANLO is a square

AL = LO = ON = AN = x

NC = AC - AN = (12 - x) cm

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

Since, ABC is a right angled triangle,

∴ BC² = AC² + AB² [By pythagoras theorem]

⇒ 13² = 12² + AB²

⇒ AB² = 13² - 12²

⇒ AB² = 169 - 144

⇒ AB² = 25

⇒ AB = $\sqrt{25}$

⇒ AB = 5 cm.

From figure,

LB = AB - AL = (5 - x) cm.

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

Then,

⇒ BC = BM + CM

⇒ 13 = (5 - x) + (12 - x)

⇒ 13 = 17 - 2x

⇒ 2x = 17 - 13

⇒ 2x = 4

⇒ x = $\frac{4}{2}$

⇒ x = 2 cm.

Hence, x = 2 cm.

Let the inscribed circle touch the sides PQ, QR and PR at A, B and C respectively.

PQR is a right-angled triangle with PQ = 3 cm and QR = 4 cm.

By pythagoras theorem,

⇒ PR² = PQ² + QR²

⇒ PR² = 3² + 4²

⇒ PR² = 9 + 16

⇒ PR² = 25

⇒ PR = 5 cm.

By tangent property we have,

∠OAQ = ∠OBQ = 90°

∠AQB = 90°

Since, all the angles of OAQB equals to 90°.

From figure,

OAQB is a square.

OA = OB = AQ = BQ = x (let)

PA = PQ - AQ = (3 - x) cm

PA = PC = (3 - x) cm.[∵ Tangents from exterior point are equal in length.]

RB = RQ - QB = (4 - x) cm

RC = RB = (4 - x) cm.[∵ Tangents from exterior point are equal in length.]

PR = PC + RC = 3 - x + 4 - x = 7 - 2x

5 = 7 - 2x

2x = 7 - 5

2x = 2

x = 1 cm.

Hence, radius of the circle inscribed in the triangle equals to 1 cm.

Given,

O is the centre of two concentric circles of radii OA = 6 cm and OB = 4 cm.

PA and PB are the two tangents to the outer and inner circles respectively and PA = 10 cm.

We know that,

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

∠OAP = ∠OBP = 90°

Since, OAP is a right angled triangle,

∴ OP² = OA² + PA² [By pythagoras theorem]

⇒ OP² = 6² + 10²

⇒ OP² = 36 + 100

⇒ OP² = 136

⇒ OP = $\sqrt{136}$

Since, OBP is a right angled triangle,

⇒ OP² = OB² + PB²

⇒ PB² = OP² - OB²

⇒ PB² = $(\sqrt{136})^2 - 4^2$

⇒ PB² = 136 - 16

⇒ PB² = 120

⇒ PB = $\sqrt{120}$

⇒ PB = 10.95 cm

Hence, length of PB is 10.95 cm.

We know that,

Tangents from an exterior point to a circle are equal in length.

From A,

⇒ AP = AR = 5 cm

From figure,

⇒ AC = AR + RC

⇒ 14 = 5 + RC

⇒ RC = 14 - 5 = 9 cm

From B,

BP = BQ = 7 cm

From C,

CQ = CR = 9 cm

From figure,

⇒ BC = BQ + QC = 7 + 9 = 16 cm

⇒ x = 16 cm.

Hence, x = 16 cm.

We know that,

Tangents from an exterior point to a circle are equal in length.

From A,

AP = AS = 9 cm

From B,

BP = BQ = 7 cm

From C,

CQ = CR = 5 cm

From D,

DR = DS = 6 cm

From figure,

⇒ AB = AP + PB = 9 + 7 = 16 cm

⇒ BC = BQ + QC = 7 + 5 = 12 cm

⇒ CD = CR + RD = 5 + 6 = 11 cm

⇒ DA = AS + SD = 9 + 6 = 15 cm

Perimeter of ABCD = AB + BC + CD + DA

= 16 + 12 + 11 + 15

= 54 cm.

Hence, perimeter of ABCD = 54 cm.

We know that,

Tangents from an exterior point to a circle are equal in length.

From A,

⇒ AP = AS = 6 cm

From C,

⇒ CR = CQ = 4 cm

⇒ BP = AB - AP = 11 - 6 = 5 cm.

From B,

⇒ BQ = BP = 5 cm

From figure,

BC = BQ + CQ = 5 + 4 = 9 cm

x = 9 cm

Hence, x = 9 cm.

We know that,

Tangents from an exterior point to a circle are equal in length.

From A,

⇒ AQ = AR ....(1)

From B,

⇒ BQ = BP ....(2)

From C,

⇒ CP = CR .....(3)

Perimeter of triangle ABC = AB + BC + CA

= AB + (BP + PC) + (AR - CR)

= (AB + BP) + PC + (AQ - CP)    ...[From equation (1) and (3)]

= (AB + BQ) + PC + (AQ - CP)    ...[From equation (2)]

= AQ + PC + AQ - PC

= 2AQ

= 2(15)

= 30 cm.

Hence, perimeter of triangle ABC = 30 cm.

Join OA and OB.

We know that,

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

∠PAO = ∠PBO = 90°

In quadrilateral AOPB, By angle sum property of quadrilateral,

⇒ ∠OAP + ∠APB + ∠PBO + ∠AOB = 360°

⇒ 90° + 40° + 90° + ∠AOB = 360°

⇒ 220° + ∠AOB = 360°

⇒ ∠AOB = 360° - 220°

⇒ ∠AOB = 140°.

Arc AB subtends ∠AOB at center and ∠AQB on the remaining part of the circle.

⇒ ∠AQB = $\frac{1}{2}$ ∠AOB

⇒ ∠AQB = $\frac{140°}{2}$

⇒ ∠AQB = 70°.

Sum of opposite angles in cyclic quadrilateral is 180°.

⇒ ∠AQB + ∠AMB = 180°

⇒ 70° + ∠AMB = 180°

⇒ ∠AMB = 180° - 70°

⇒ ∠AMB = 110°.

Hence, ∠AMB = 110° and ∠AQB = 70°.

(i) We know that,

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

∠OBP = ∠OAP = 90°

In quadrilateral AOPB, By angle sum property of quadrilateral,

⇒ ∠OBP + ∠OAP + ∠AOB + ∠APB = 360°

⇒ 90° + 90° + ∠AOB + 50° = 360°

⇒ 230° + ∠AOB = 360°

⇒ ∠AOB = 360° - 230°

⇒ ∠AOB = 130°.

Hence, ∠AOB = 130°.

(ii) From figure,

OA = OB (Radii)

In triangle OAB,

∠OBA = ∠OAB [Angles opposite to equal sides]

In triangle OAB,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 2∠OAB + 130° = 180°

⇒ 2∠OAB = 180° - 130°

⇒ 2∠OAB = 50°

⇒ ∠OAB = 25°.

Hence, ∠OAB = 25°.

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

⇒ ∠ACB = $\frac{1}{2}$ ∠AOB

⇒ ∠ACB = $\frac{130°}{2}$

⇒ ∠ACB = 65°.

Hence, ∠ACB = 65°.

Arc PR subtends ∠POR at center and ∠PQR on the remaining part of the circle.

⇒ ∠PQR = $\frac{1}{2}$ ∠POR

⇒ ∠PQR = $\frac{72°}{2}$

⇒ ∠PQR = 36°.

We know that,

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

∠QPT = 90°

In triangle QPT,

⇒ ∠QPT + ∠PQR + ∠PTR = 180°

⇒ 90° + 36° + ∠PTR = 180°

⇒ 126° + ∠PTR = 180°

⇒ ∠PTR = 180° - 126°

⇒ ∠PTR = 54°.

Hence, ∠PTR = 54°.

From figure,

TA and TB are the tangents.

∴ OA ⊥ TA and OB ⊥ TB

In quadrilateral AOBT,

By angle sum property,

⇒ ∠ATB + ∠TAO + ∠TBO + ∠AOB = 360°

⇒ ∠ATB + ∠AOB + 90° + 90° = 360°

⇒ ∠ATB + ∠AOB = 180°

⇒ 80° + ∠AOB = 180°

⇒ ∠AOB = 180° - 80°

⇒ ∠AOB = 100°.

From figure,

∠BOC = 360° - (∠AOC + ∠AOB)

= 360° - (130° + 100°)

= 360° - 230° = 130°.

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 ∠COB at the centre and ∠CAB at the remaining part of the circle.

∴ ∠CAB = $\frac{1}{2}$ ∠COB

= $\frac{1}{2}$ × 130° = 65°.

Hence, ∠CAB = 65°.

(a) From figure,

∠AOC = ∠AOM + ∠COM = 65° + 65° = 130°.

We know that,

Angle at the center is twice the angle formed by the same arc at any other point of the circle.

⇒ ∠AOC = 2∠ABC

⇒ ∠ABC = $\frac{1}{2} \times ∠AOC = \frac{1}{2} \times 130°$ = 65°.

Hence, ∠ABC = 65°.

(b) In △ABC,

⇒ AB = AC (Given)

⇒ ∠ACB = ∠ABC = 65° (Opposite angles of equal sides are equal)

By angle sum property of triangle,

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 65° + 65° + ∠BAC = 180°

⇒ ∠BAC = 180° - 65° - 65° = 50°.

Hence, ∠BAC = 50°.

(c) We know that,

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

∴ ∠BCQ = ∠BAC = 50°.

Hence, ∠BCQ = 50°.

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

Thus, OA ⊥ EF and OB ⊥ CD

Since the tangents are perpendicular to the radius,

⇒ ∠CBO = 90°, ∠EAO = 90°

⇒ ∠FAO = 90°, ∠OBD = 90°

∴ ∠DBO = ∠OAE, ∠CBO = ∠OAF

These are pair of alternate interior angles.

If the alternate interior angles are equal, then lines CD and EF should be parallel.

CD and EF are the tangents drawn to the circle at the ends of the diameter AB.

Hence, proved that tangents drawn at the ends of a diameter of a circle are parallel.

Let AB be a chord of a circle with centre O, and AP, BP be the tangents at A and B respectively.

From figure,

PA = PB [∵ Tangents from an external point to a circle are equal]

In triangle PAB,

∠PAB = ∠PBA [Angles opposite to equal sides in a triangle are equal]

∴ ∠PAC = ∠PBC.

Hence, proved any chord make equal angles with the chord.

Let AB and CD are parallel tangents of circle with centre O.

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

∠APO = 90°

⇒ ∠APO + ∠EOP = 180° (sum of adjacent interior angles)

⇒ ∠EOP = 180° - 90°

⇒ ∠EOP = 90°

Similarly,

⇒ ∠EOQ + ∠CQO = 180° (sum of adjacent interior angles)

⇒ ∠EOQ = 180° - 90°

⇒ ∠EOQ = 90°

∠EOP + ∠EOQ = 90° + 90° = 180°

∴ POQ is a straight line.

Hence, line segment joining the points of contact of two parallel tangents passes through the centre.

From figure,

AP ⊥ PQ (∵ tangent at a point and radius through the point are perpendicular to each other.)

In right angled triangle PAC.

⇒ CA² = PC² + AP²

⇒ CA² = 12² + 5²

⇒ CA² = 144 + 25

⇒ CA² = 169

⇒ CA = $\sqrt{169}$

⇒ CA = 13 cm.

Considering triangles PAC and BCQ,

⇒ ∠APC = ∠BQC = 90°

⇒ ∠PCA = ∠QCB (Vertically opposite angles are equal)

△PAC ~ △QBC by AA axiom.

Since triangles are similar hence, the ratio of their corresponding sides are equal.

$\frac{CA}{CB} = \frac{PA}{QB} = \frac{5}{3}$

$\frac{13}{CB} = \frac{5}{3}$

⇒ CB = 7.8 cm.

From figure,

AB = AC + CB = 13 + 7.8 = 20.8 cm

Hence, AB = 20.8 cm.

We know that,

Tangents from exterior point are equal in length.

We have,

AR = AP, BQ = BP and CQ = CR

Now, AB = AC

⇒ AP + PB = AR + RC

⇒ AR + PB = AR + RC [∵ AR = AP]

⇒ PB = RC

⇒ BQ = CQ.

It means BC is bisected at point Q.

Hence, proved that the base is bisected by the point of contact.

From figure,

OS = OP (As both are radius of circle.)

Since,

Adjacent sides are equal and each angle is a right angle.

∴ AP = OS.

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

∴ AP = AS = 10 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

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

∴ CQ = CR = 27 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From figure,

BQ = BC - CQ = 38 - 27 = 11 cm.

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

BP = BQ = 11 cm.

⇒ AB = x = AP + BP = 10 + 11 = 21 cm.

Hence, x = 21 cm.

From figure,

⇒ BR = BQ = 27 cm [∵ Length of tangents form an external point to a circle are equal.]

⇒ CR = BC - BR = 38 - 27 = 11 cm.

⇒ CR = CS = 11 cm [∵ Length of tangents form an external point to a circle are equal.]

⇒ DS = DC - CS = 25 - 11 = 14 cm.

In quadrilateral DSOP,

⇒ ∠SDP + ∠DPO + ∠OSD + ∠POS = 360°

⇒ 90° + 90° + 90° + ∠POS = 360°

⇒ ∠POS = 360° - 270° = 90°.

Since, all angles are 90° and OS = OP [∵ Both equal to radius of same circle]

Hence, proved that DPOS is a square.

OP = DS = 14 cm.

Hence, radius of circle = 14 cm.

From figure,

In △SRT,

SP ⊥ ST (∵ tangent is perpendicular to radius from that point.)

so, ∠TSR = 90°

Since, sum of angles in a triangle = 180°

⇒ ∠TSR + ∠SRT + ∠STR = 180°

⇒ 90° + 65° + x = 180°

⇒ x + 155° = 180°

⇒ x = 25°.

SQ subtends ∠SOQ at the centre and ∠STQ on point T.

∴ ∠SOQ = 2∠STQ (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle.)

y = 2x = 2 × 25° = 50°.

In △OSP,

Since, sum of angles in a triangle = 180°

⇒ ∠OSP + ∠SOP + ∠SPO = 180°

⇒ 90° + y + z = 180°

⇒ 90° + 50° + z = 180°

⇒ z + 140° = 180°

⇒ z = 180° - 140° = 40°.

Hence, the value of x = 25°, y = 50° and z = 40°.

(i) ∠OAP = 90° [As TP is tangent at point A]

From figure,

⇒ ∠OAB = ∠OAP - ∠BAP = 90° - 60° = 30°.

⇒ OA = OB (Radius of same circle)

⇒ ∠OBA = ∠OAB = 30° (Angles opposite to equal sides are equal)

∠OCQ = 90° [As TQ is tangent at point C]

From figure,

∠OCB = ∠OCQ - ∠BCQ = 90° - 55° = 35°.

⇒ OB = OC (Radius of same circle)

⇒ ∠OBC = ∠OCB = 35° (Angles opposite to equal sides are equal)

Hence, ∠OBA = 30° and ∠OBC = 35°.

(ii) 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.

⇒ ∠AOC = 2∠ABC

⇒ ∠AOC = 2(∠OBA + ∠OBC)

⇒ ∠AOC = 2(30° + 35°)

⇒ ∠AOC = 2 × 65° = 130°.

Hence, ∠AOC = 130°.

(iii) In quadrilateral OATC,

⇒ ∠AOC + ∠OAT + ∠OCT + ∠ATC = 360°

⇒ 130° + 90° + 90° + ∠ATC = 360°

⇒ 310° + ∠ATC = 360°

⇒ ∠ATC = 360° - 310° = 50°.

Hence, ∠ATC = 50°.

(i) We know that,

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

In quadrilateral ORPQ,

∠OQP = ∠ORP = 90° [∵ The tangent at any point of a circle and the radius through this point are perpendicular to each other]

∠QPR = 70° [Given]

⇒ ∠OQP + ∠ORP + ∠QPR + ∠QOR = 360° [By angle sum property of quadrilateral]

⇒ 90° + 90° + 70° + ∠QOR = 360°

⇒ 250° + ∠QOR = 360°

⇒ ∠QOR = 110°.

Hence, ∠QOR = 110°.

(ii) Let M be a point on circumference of circle.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended at the circumference.

⇒ ∠QMR = $\frac{1}{2}$ ∠QOR

= $\frac{110°}{2}$

= 55°.

Sum of opposite angles in cyclic quadrilateral is 180°.

⇒ ∠QSR + ∠QMR = 180°.

⇒ ∠QSR = 180° - 55°

⇒ ∠QSR = 125°.

Hence, ∠QSR = 125°.

(i) We know that,

Angle in a semi-circle is a right angle.

∴ ∠BDA = 90°.

Hence, ∠BDA = 90°.

(ii) In △ BAD,

By angle sum property of triangle,

⇒ ∠BDA + ∠BAD + ∠ABD = 180°

⇒ 90° + ∠BAD + 26° = 180°

⇒ ∠BAD + 116° = 180°

⇒ ∠BAD = 180° - 116° = 64°.

Hence, ∠BAD = 64°.

(iii) From figure,

CE is tangent to the circle.

Angle between tangent and radius of the circle is 90°.

From figure,

⇒ ∠CAB = 90°

⇒ ∠CAD + ∠BAD = 90°

⇒ ∠CAD + 64° = 90°

⇒ ∠CAD = 90° - 64° = 26°.

Hence, ∠CAD = 26°.

(iv) In △ OBD,

⇒ OD = OB (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

⇒ ∠ODB = ∠OBD = 26°.

Hence, ∠ODB = 26°.

In △ PST,

⇒ PS = PT (Tangents from an external point to a circle are equal in length)

⇒ ∠PST = ∠PTS = a (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠PST + ∠PTS + ∠SPT = 180°

⇒ a + a + 80° = 180°

⇒ 2a = 180° - 80°

⇒ 2a = 100°

⇒ a = $\frac{100°}{2}$ = 50°.

From figure,

⇒ ∠QST = ∠STP = 50° (Alternate angles are equal)

Hence, ∠QST = 50°.

(i) From figure,

⇒ ∠OQP = 90° (Tangent is perpendicular to radius at the point of contact)

⇒ ∠OQM = ∠OQP - ∠PQM

⇒ ∠OQM = 90° - 42° = 48°.

Hence, ∠OQM = 48°.

(ii) From figure,

⇒ ∠QNM = ∠PQM = 42° (By alternate segment theorem)

⇒ ∠SNM = ∠PSM = 25° (By alternate segment theorem)

⇒ ∠QNS = ∠QNM + ∠SNM

⇒ ∠QNS = 42° + 25° = 67°.

Hence, ∠QNS = 67°.

(iii) We know that,

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

⇒ ∠QOS = 2∠QNS

⇒ ∠QOS = 2 × 67° = 134°.

Hence, ∠QOS = 134°.

(iv) From figure,

QMSN is a cyclic quadrilateral.

We know that,

Sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠QMS + ∠QNS = 180°

⇒ ∠QMS + 67° = 180°

⇒ ∠QMS = 180° - 67° = 113°.

Hence, ∠QMS = 113°.

We know that,

If two chords of a circle intersect internally, then the products of the length of segments are equal.

(i) PA × PB = PC × PD

5 × 5.6 = 3.5 × x

x = $\frac{28}{3.5}$

x = 8 cm.

Hence, x = 8 cm.

(ii) PA × PB = PC × PD

x × 9 = 8.1 × 5

x = $\frac{40.5}{9}$

x = 4.5 cm

Hence, x = 4.5 cm.

We know that,

If two chords of a circle intersect externally, then the products of the length of segments are equal.

(iii) PA × PB = PC × PD

⇒ PB = PA + AB = 7 + 9 = 16

⇒ PD = PC + CD = 8 + x

⇒ 7 × 16 = 8 × (8 + x)

⇒ 112 = 64 + 8x

⇒ 8x = 112 - 64

⇒ x = $\frac{48}{8}$

⇒ x = 6 cm.

Hence, x = 6 cm.

(iv) 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.

∴ PT² = AP × BP

x² = 4.5 × 18

x² = 81

x = 9 cm.

Hence, x = 9 cm.

(v) 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.

∴ PT² = AP × BP

⇒ BP = AP + AB = x + 10

⇒ 12² = x × (x + 10)

⇒ 144 = x² + 10x

⇒ x² + 10x - 144 = 0

⇒ x² + 18x - 8x - 144 = 0

⇒ x(x + 18) - 8(x + 18) = 0

⇒ (x - 8)(x + 18) = 0

⇒ (x - 8) = 0 or (x + 18) = 0      [Using Zero-product rule]

⇒ x = 8 or x = -18

∴ x = 8 because length cannot be negative.

Hence, x = 8 cm.

We know that,

If two chords of a circle intersect externally, then the products of the length of segments are equal.

From figure,

EA × EB = EC × ED ....(1)

EB = EA + AB = 3 + 5 = 8 cm

Substituting values in equation (1) we get,

⇒ 3 × 8 = 2 × ED

⇒ 24 = 2 × ED

⇒ ED = $\frac{24}{2}$

⇒ ED = 12 cm.

⇒ CD = ED - EC = 12 - 2 = 10 cm

Hence, CD = 10 cm.

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.

∴ PT² = AP × BP

From figure,

BP = AP - AB = 16 - 12 = 4 cm.

Substituting values we get,

PT² = 16 × 4

PT² = 64

PT = $\sqrt{64}$

PT = 8 cm.

Hence, PT = 8 cm.

We know that,

If two chords of a circle intersect internally, then the products of the length of segments are equal.

AP × PB = CP × PD

PB = AB - AP = 12 - 2.4 = 9.6 cm

Substituting values we get,

⇒ 2.4 × 9.6 = CP × 7.2

⇒ CP = $\frac{23.04}{7.2}$

⇒ CP = 3.2 cm

⇒ CD = CP + PD = 3.2 + 7.2 = 10.4 cm

Hence, CD = 10.4 cm.

We know that,

If two chords of a circle intersect externally, then the products of the length of segments are equal.

PA × PB = CP × PD ........(1)

PB = PA - AB

PB = 12 - 4 = 8 cm

Let length of PD be x.

PC = x + CD = x + 10

Substituting values in equation (1) we get,

⇒ 12 × 8 = (x + 10) × x

⇒ 96 = x² + 10x

⇒ x² + 10x - 96 = 0

⇒ x² + 16x - 6x - 96 = 0

⇒ x(x + 16) - 6(x + 16) = 0

⇒ (x - 6)(x + 16) = 0

⇒ x = 6 [Length cannot be negative]

⇒ PD = 6 cm.

Hence, PD = 6 cm.

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.

For circle 1:

∴ PQ² = PA × PB

For circle 2:

∴ PR² = PA × PB

Thus,

PQ² = PR²

Taking square root on both sides,

PQ = PR

Hence, proved that PQ = PR.

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.

For circle 1:

∴ PA² = PC × PD

For circle 2:

∴ PB² = PC × PD

Thus,

PA² = PB²

Taking square root on both sides,

PA = PB

Point P divides AB into two equal parts.

Hence, proved P is the mid-point of AB.

(i) ∠TAB = ∠ACB = 50° [Angles in the alternate segment are equal]

Hence, ∠TAB = 50°.

(ii) ∠ADB + ∠ACB = 180° [Opposite angles of cyclic quadrilateral]

∠ADB = 180° - 50°

∠ADB = 130°.

Hence, ∠ADB = 130°.

From figure,

∠ACB = ∠TAB = 70° [Angles in the alternate segment are equal]

In ΔABC,

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

⇒ ∠ABC = 180° - (∠BAC + ∠ACB)

⇒ ∠ABC = 180° - (45° + 70°)

⇒ ∠ABC = 65°.

Hence, ∠ABC = 65°.

(i) ∠TAC = ∠ABC = 35° [Angles in the alternate segment are equal]

Hence, ∠TAC = 35°.

(ii) Since BC is a diameter, ∠BAC is an angle in a semicircle.Therefore, ∠BAC = 90°

⇒ ∠ACB + ∠BAC + ∠ABC = 180°

⇒ ∠ACB + 90° + 35° = 180°

⇒ ∠ACB = 180° - 90° - 35°

⇒ ∠ACB = 55°

⇒ ∠PAB = ∠ACB [Angles in the alternate segment are equal]

⇒ ∠PAB = 55°

Hence, ∠PAB = 55°.

(i) ∠TAD = ∠ABD = 28° [Angles in the alternate segment are equal]

Hence, ∠TAD = 28°.

(ii) ∠BAD = 90° [Angle in the semicircle]

Hence, ∠BAD = 90°.

(iii) ∠PAB = ∠ADB [Angles in the alternate segment]

By angle sum property of triangle :

⇒ ∠ADB + ∠ABD + ∠BAD = 180°

⇒ ∠ADB + 28° + 90° = 180°

⇒ ∠ADB = 180° - (28° + 90°)

⇒ ∠ADB = 62°

⇒ ∠PAB = 62°.

Hence, ∠PAB = 62°.

(iv) In ΔBCD, since BD is the diameter ∠BCD = 90°.

By angle sum property of triangle,

⇒ ∠CBD + ∠BCD + ∠BDC = 180°

⇒ ∠CBD + 90° + 52° = 180°

⇒ ∠CBD = 180° - 90° - 52°

⇒ ∠CBD = 38°.

Hence, ∠CBD = 38°.

Given,

PQ = PR

∠PRQ = ∠PQR [Angles opposite to equal sides in a triangle are equal]

∠TPR = ∠PQR [Angles in alternate segment]

∴ ∠PRQ = ∠TPR

These are pair of alternate interior angles.

If the alternate interior angles are equal, then lines P and QR should be parallel.

Hence, proved tangent at P is parallel to QR.

In △OAB,

OA = OB (∵ both are radius of the common circle.)

So, △OAB is a isosceles triangle with,

∠OBA = ∠OAB = 35°.

Since sum of angles in a triangle = 180°.

In △OAB,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 35° + 35° + ∠AOB = 180°

⇒ 70° + ∠AOB = 180°

⇒ ∠AOB = 180° - 70°

⇒ ∠AOB = 110°.

Arc AB subtends ∠AOB at centre and ∠ACB at remaining part of circle.

∴ ∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of the circle.)

⇒ 110° = 2y

⇒ y = $\frac{110}{2}$

⇒ y = 55°.

From figure,

∠ABT = ∠ACB = 55° (∵ Angles in alternate segments are equal.)

∴ x = 55°.

Hence, the value of x = 55 and y = 55.

(i) In △ PTB and △ PAT,

⇒ ∠PTA = ∠PBT (Alternate segment theorem)

⇒ ∠TPA = ∠BPT (Common angle)

∴ △ PTB ~ △ PAT (By A.A. axiom)

Hence, proved that △ PTB ~ △ PAT.

(ii) We know that,

If a chord and a tangent intersect externally, then the product of the lengths of 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 × PB = PT²

⇒ PA × (PA + AB) = PT²

⇒ 16 × (16 + AB) = 20²

⇒ 16 × (16 + AB) = 400

⇒ 16 + AB = 25

⇒ AB = 25 - 16 = 9 cm.

Hence, AB = 9 cm.

(i) In ΔABC and ΔADB,

⇒ ∠ABC = ∠ADB = 90°

⇒ ∠BAC = ∠DAB [Common angles]

△ABC ∼ △ADB [By AA similarity]

Since, corresponding sides of similar triangle are proportional to each other.

$\frac{AB}{AD} = \frac{AC}{AB}$

AB² = AC × AD

Hence, AB² = AC × AD.

(ii) In ΔABC and ΔBDC

⇒ ∠ABC = ∠CDB = 90°

⇒ ∠C [Common angles]

△ABC ∼ △BDC [By AA similarity]

$\frac{BC}{CD} = \frac{AC}{BC}$

BC² = AC × CD

Hence, BC² = AC × CD.

(i) From figure,

⇒ ∠DCE + ∠DCB = 180° [Linear pair]

⇒ ∠DCB = 180° - ∠DCE

⇒ ∠DCB = 180° - 110°

⇒ ∠DCB = 70°.

Given,,

CB = CD

∠BDC = ∠DBC = 55° [Angles opposite to equal sides in a triangle are equal]

Arc BC subtends ∠BOC at center and ∠BDC on the remaining part of the circle.

⇒ ∠BOC = 2(∠BDC)

⇒ ∠BOC = 2(55°) = 110°

⇒ ∠BCT = ∠BDC = 55° [Angles in alternate segments]

From figure,

⇒ ∠DCT = ∠DCB + ∠BCT

⇒ ∠DCT = 70° + 55° = 125°.

Hence, ∠DCT = 125°.

(ii) From part (i), we get :

∠BOC = 110°

Hence, ∠BOC = 110°.

We know that,

Angle between the radius and tangent at the point of contact is right angle.

∴ ∠A = 90°.

Also in △OBE, OB = OE = radius of the circle.

∴ ∠B = ∠OEB …..(i)

In △ABD,

⇒ ∠A + ∠B + ∠ADB = 180°

⇒ 90° + ∠B + 55° = 180°

⇒ ∠B + 145° = 180°

⇒ ∠B = 180° - 145° = 35°.

∴ ∠OEB = 35°.

From figure,

∠DEC = ∠OEB = 35° (∵ vertically opposite angles are equal.)

∠EDC + ∠ADE = 180° (∵ both form a linear pair)

∠EDC + 55° = 180°

∠EDC = 180° - 55°

∠EDC = 125°.

In △EDC,

⇒ ∠DEC + ∠EDC + ∠DCE = 180°

⇒ 35° + 125° + x° = 180°

⇒ x° + 160° = 180°

⇒ x° = 180° - 160° = 20°.

In △AOC,

⇒ ∠AOC + ∠OAC + ∠ACO = 180°

⇒ y° + 90° + x° = 180°

⇒ y° + 90° + 20° = 180°

⇒ y° + 110° = 180°

⇒ y° = 180° - 110° = 70°.

Hence, the value of x = 20° and y = 70°.

(i) In △ APB and △ CPD,

⇒ ∠APB = ∠CPD (Vertically opposite angles are equal)

⇒ ∠BAP = ∠DCP (Angles in same segment are equal)

∴ △ APB ~ △ CPD (By A.A. axiom)

Hence, proved that △ APB ~ △ CPD.

(ii) We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{CD}{AB} = \frac{PD}{PB} \\[1em] \Rightarrow \frac{CD}{9} = \frac{2}{3} \\[1em] \Rightarrow CD = 9 \times \frac{2}{3} \\[1em] \Rightarrow CD = 6\text{ cm}.\end{aligned}$$

Hence, CD = 6 cm.

(iii) We know that,

Ratio of area of similar triangles is equal to the ratio of square of the corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APB}}{\text{Area of △CPD}} = \frac{PB^2}{PD^2} \\[1em] = \frac{3^2}{2^2} \\[1em] = \frac{9}{4} \\[1em] = 9 : 4.\end{aligned}$$

Hence, area △ APB : area △ CPD = 9 : 4.

(i) Given,

In ΔBOX, OX ⟂ BX (radius ⟂ tangent at its point of contact)

⇒ ∠OXB = 90°.

By angle‑sum property of triangle,

⇒ ∠BOX + ∠OBX + ∠OXB = 180°

⇒ ∠BOX + 32° + 90° = 180°

⇒ ∠BOX + 122° = 180°

⇒ ∠BOX = 180° - 122°

⇒ ∠BOX = 58°.

Hence, ∠BOX = 58°.

(ii) From figure,

∠COX = ∠BOX = 58°

We know that,

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

$$\begin{aligned}\Rightarrow ∠CYX = \frac{1}{2}∠COX \\[1em] \Rightarrow ∠CYX = \frac{58°}{2} \\[1em] \Rightarrow ∠CYX = 29°\end{aligned}$$

Hence, ∠CYX = 29°.

(iii) We know that,

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

∠ZYX = ∠AXZ = 66°

Hence, ∠ZYX = 66°.

(iv) From figure,

In isosceles ΔZXY,

ZY = XY

We know that,

The angles opposite to equal side of a triangle are equal.

∠ZXY = ∠XZY

By angle sum property in ΔXYZ,

$$\begin{aligned}\Rightarrow ∠XZY + ∠ZXY + ∠ZYX = 180° \\[1em] \Rightarrow ∠ZYX + 2∠XZY = 180° \\[1em] \Rightarrow 2∠XZY = 180° - ∠ZYX \\[1em] \Rightarrow ∠XZY = \frac{180° - ∠ZYX}{2} \\[1em] = \frac{180° - 66°}{2} \\[1em] = \frac{114°}{2} \\[1em] = 57° .\end{aligned}$$

We know that,

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

∠YXB = ∠XZY = 57°.

Also,

∠ZXY = ∠XZY = 57°.

From figure,

⇒ ∠OXY = ∠OXB - ∠YXB

⇒ ∠OXY = 90° - 57° = 33°.

Hence, ∠OXY = 33°.

(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 of the circle.

∴ ∠PTR = ∠PST

⇒ ∠RPT = ∠TPS [Common angles]

Therefore, by AA similarity, ΔPTR ~ ΔPST.

Hence, proved that ΔPTR ~ ΔPST.

(ii) Since, corresponding sides of similar triangles are proportional we have :

⇒ $\frac{PT}{PS} = \frac{PR}{PT}$

⇒ PT² = PR × PS.

Hence, proved that PT² = PR × PS.

(iii) Given,

PR = 4 cm and PS = 16 cm.

$$\begin{aligned}\Rightarrow PT^2 = PR \times PS \\[1em] \Rightarrow PT = \sqrt{PR \times PS} \\[1em] \Rightarrow PT = \sqrt{4 \times 16} \\[1em] \Rightarrow PT = \sqrt{64} \\[1em] \Rightarrow PT = 8 \text{ cm}.\end{aligned}$$

Hence, PT = 8 cm.

In two concentric circles, a chord of larger circle which is tangent to smaller circle is bisected at the point of contact.

Hence, option 2 is the correct option.

We know that,

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

In right ∆OTP, we have

OT² = PT² + OP²

PT² = OT² - OP²

PT² = 100 - 64

PT² = 36

PT = $\sqrt{36}$ = 6 cm.

Hence, option 1 is the correct option.

A standard theorem in circle states:

From a point lying outside a circle, exactly two tangents can be drawn to the circle.

Hence, option 3 is the correct option.

When two tangents are drawn from an external point to a circle, the line joining the external point and the centre of the circle bisects the angle between the two tangents.

Hence, option 2 is the correct option.

We know that,

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

In right ∆MOT, we have

MO² = MT² + OT²

OT² = MO² - MT²

OT² = 25² - 24²

OT² = 625 - 576

OT² = 49

OT = $\sqrt{49}$

OT = 7 cm.

Hence, option 1 is the correct option.

Let QP = OP

We know that,

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

∠OPQ = 90°

∠QOP = ∠OQP = x [Anglers opposite to equal sides are equal in a triangle]

By angles sum triangle property,

x + x + 90° = 180°

2x + 90° = 180°

2x = 180° - 90°

2x = 90°

x = $\frac{90°}{2}$

x = 45°

∠QOP = 45°

Hence, option 3 is the correct option.

From figure,

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

∠OZX = 90°

∠OYX = 90°

∠ZXY + ∠ZOY + ∠OZX + ∠OYX = 360° [Angle sum property of quadrilateral]

40° + 90° + 90° + ∠ZOY = 360°

220° + ∠ZOY = 360°

∠ZOY = 360° - 220°

∠ZOY = 140°.

We know that,

Angle at the centre is double the angle at a point on the remaining part of the circle.

∠ZCY = $\frac{1}{2}$ ∠ZOY

∠ZCY = $\frac{140°}{2}$

∠ZCY = 70°.

Hence, option 2 is the correct option.

Tangents drawn from the same external point to a circle are equal.

From B,

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

From C,

CR = CQ ..........(2)

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

∴ CR + PB = CQ + BQ = BC

Hence, option 1 is the correct option.

Tangents drawn from the same external point to a circle are equal.

From M,

MA = CM = 6 cm

From N,

NA = NB = 5 cm

From L,

LB = LC = 4 cm

MN = MA + AN = 6 + 5 = 11

NL = NB + BL = 5 + 4 = 9

LM = LC + CM = 4 + 6 = 10

Perimeter of ΔLMN = MN + LN + LM

= 11 + 9 + 10

= 30 cm.

Hence, option 1 is the correct option.

We know that if two tangents are drawn from an external point to a circle then, the lengths of the tangents are equal.

From figure,

PX and PY are the tangents to the first circle.

∴ PX = PY = 3.8 cm

PX and PZ are tangents to the second circle.

∴ PZ = PX = 3.8 cm

From figure,

YZ = PZ + PY = 3.8 + 3.8 = 7.6 cm.

Hence, option 3 is the correct option.

From figure,

∠BAP = ∠AXB = 50° [Angle in alternate segment are equal]

Hence, option 2 is the correct option.

We know that,

Angle between tangent and the chord at the point of contact is equal to angle of the alternate segment.

∴ ∠PQS = ∠PST = 30°

In △ PQS,

By angle sum property of triangle,

⇒ ∠PQS + ∠QPS + ∠PSQ = 180°

⇒ 30° + 60° + ∠PSQ = 180°

⇒ ∠PSQ + 90° = 180°

⇒ ∠PSQ = 180° - 90° = 90°.

Hence, option 4 is the correct option.

In a circle, radius through the point of contact is perpendicular to the tangent.

∠OAM = 90°

∠OAB + ∠BAM = 90°

∠OAB + 50° = 90°

∠OAB = 90° - 50°

∠OAB = 40°

OA = OB (Radii of same circle)

∠OBA = ∠OAB = 40° [Angles opposite to equal sides in a triangle are equal]

In △AOB,

By angle sum property of triangle,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 40° + 40° + ∠AOB = 180°

⇒ ∠AOB + 80° = 180°

⇒ ∠AOB = 180° - 80° = 100°.

Hence, option 1 is the correct option.

∠XYO = 25°

In a circle, radius through the point of contact is perpendicular to the tangent.

∠OXY = 90°

In △ XOY,

By angle sum property of triangle,

⇒ ∠XYO + ∠OXY + ∠XOY = 180°

⇒ 25° + 90° + ∠XOY = 180°

⇒ ∠XOY + 115° = 180°

⇒ ∠XOY = 180° - 115° = 65°.

From figure,

∠XOY + x = 180° [Linear pairs]

x = 180° - 65°

x = 115°.

Hence, option 2 is the correct option.

From figure,

⇒ ∠QOT + ∠TOP = 180° (Linear Pair)

⇒ 110° + ∠TOP = 180°

⇒ ∠TOP = 180° - 110° = 70°.

⇒ ∠OPT = 90° (Tangent is perpendicular to radius at point of intersection)

In △ TOP,

⇒ ∠OPT + ∠TOP + ∠PTO = 180°

⇒ 90° + 70° + x°= 180°

⇒ x° + 160° = 180°

⇒ x° = 180° - 160° = 20°.

Hence, option 1 is the correct option.

We know that,

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

∴ ∠SQP = ∠SPT = 45° and ∠SPQ = ∠SQT = 30°.

In △ SQP,

⇒ ∠SQP + ∠SPQ + ∠QSP = 180°

⇒ 45° + 30° + ∠QSP = 180°

⇒ 75° + ∠QSP = 180°

⇒ ∠QSP (x) = 180° - 75° = 105°.

Hence, option 4 is the correct option.

From figure,

MQ is chord and XQY is a tangent.

∠P = ∠MQY (∵ angles in alternate segment are equal.)

As PM || XQY

∠MQY = ∠M (∵ alternate angles are equal)

∴ ∠P = ∠M = 70°

We know that sum of angles in a triangle = 180°.

In △PQM,

⇒ ∠P + ∠M + ∠PQM = 180°

⇒ 70° + 70° + ∠PQM = 180°

⇒ 140° + ∠PQM = 180°

⇒ ∠PQM = 180° - 140°

⇒ ∠PQM = 40°.

Hence, option 3 is the correct option.

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.

∴ ∠AOC (y) = 2∠ABC = 2 × 50° = 100°.

In △ AOC,

⇒ OA = OC (Radii of same circle)

⇒ ∠OAC = ∠OCA = z (Angle opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ z + z + 100° = 180°

⇒ 2z = 180° - 100°

⇒ 2z = 80°

⇒ z = $\frac{80°}{2}$ = 40°.

We know that,

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

⇒ ∠CAQ (x) = ∠OAQ - ∠OAC = 90° - 40° = 50°.

Hence, option 1 is the correct option.

Join OA and AP.

Given,

Radius of circle (OB) = 3 cm

PB and AP are at right angles.

We know that,

Radius and tangent at point of contact are perpendicular to each other.

OA ⊥ AP

From figure,

⇒ OA = OB = 3 cm [Radius of circle]

⇒ OB is perpendicular to PB

⇒ ∠OBP = 90°

This shows APBO is square.

OA = PB = AP = OB = 3 cm.

In right angled triangle OBP,

⇒ OP² = OB² + PB²

⇒ OP² = 3² + 3²

⇒ OP² = 9 + 9

⇒ OP² = 18

⇒ OP = $\sqrt{18}$

⇒ OP = $3\sqrt{2}$ cm.

Hence, option 3 is the correct option.

Given,

∠OAC = 25°

We know that,

Radius from the center and tangent are perpendicular to each other at point of contact.

⇒ ∠OAQ = ∠OAP = 90°

From figure,

⇒ ∠OAQ = ∠OAC + ∠CAQ

⇒ 90° = 25° + ∠CAQ

⇒ ∠CAQ = 90° - 25°

⇒ ∠CAQ = 65°.

We know that,

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

⇒ ∠ABC = ∠CAQ

⇒ ∠ABC = 65°.

Hence, option 2 is the correct option.

The lengths of tangents to a circle from an exterior point are equal.

∴ Reason (R) is true.

∴ AB = AC = 7 cm

⇒ Perimeter of ΔADE = AD + AE + DE

We can write DE as (DP + PE)

⇒ Perimeter of ΔADE = AD + AE + (DP + PE)

Substituting DP with DB and PE with EC:

⇒ Perimeter of ΔADE = AD + AE + (DB + EC)

⇒ Perimeter of ΔADE = (AD + DB) + (AE + EC)

⇒ Perimeter of ΔADE = AB + AC

⇒ Perimeter of ΔADE = 7 + 7

⇒ Perimeter of ΔADE = 14 cm.

Assertion (A) is true. The equal-tangent property stated in Reason (R) is exactly what is used to prove the perimeter, so R is the correct explanation of A.

Hence, option 1 is the correct option.

In a circle, radius through the point of contact is perpendicular to the tangent.

In right angled triangle OPT,

OP² = OT² + PT²

10² = OT² + 8²

OT² = 100 - 64

OT² = 36

OT = 6 cm.

Assertion (A) is false.

In a circle, the radius through the point of contact is perpendicular to the tangent is true.

Reason (R) is true.

Assertion (A) is false, but Reason (R) is true.

Hence, option 4 is the correct option.

In concentric circles, a chord of the larger circle touching the smaller circle is bisected at the point of contact.

Assertion (A) is true.

A line from the centre bisects a chord only when it is perpendicular to the chord, not to any point on the chord.

Reason (R) is false.

Assertion (A) is true, but Reason (R) is false.

Hence, option 3 is the correct option.

We know that,

The length of tangents drawn from an external point to the circle are equal.

⇒ AE = AH = x (let)

⇒ BE = BF = y (let)

⇒ CF = CG = z (let)

⇒ DG = DH = k (let)

Given,

⇒ AB = 15

⇒ AE + EB = 15

⇒ x + y = 15 .......(1)

⇒ BC = 18

⇒ BF + CF = 18

⇒ y + z = 18 ........(2)

⇒ AD = 24

⇒ AH + DH = 24

⇒ x + k = 24 ..........(3)

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

⇒ y + z + x + k = 18 + 24

⇒ x + y + z + k = 42

⇒ 15 + z + k = 42 [From equation (1)]

⇒ z + k = 42 - 15

⇒ z + k = 27.

⇒ CD = CG + DG = k + z = 27 cm.

Hence, length of CD = 27 cm.

(a) By formula,

Each interior angle of a n sided polygon = $\frac{(n - 2) \times 180°}{n}$

$$\begin{aligned}= \frac{(6 - 2) \times 180°}{6} \\[1em] = \frac{4 \times 180°}{6} \\[1em] = \frac{2}{3} \times 180° \\[1em] = 120°.\end{aligned}$$

∴ ∠FAB = 120°

From figure,

∠FAG and ∠FAB form a linear pair.

∴ ∠FAB + ∠FAG = 180°

⇒ 120° + ∠FAG = 180°

⇒ ∠FAG = 180° - 120° = 60°.

Hence, ∠FAG = 60°.

(b) Each interior angle of regular hexagon = 120°.

Hence, ∠BCD = 120°.

(c) In triangle DEF,

⇒ ∠DEF = 120° (Each interior angle of a regular hexagon equals to 120°)

⇒ DE = EF (Sides of a regular hexagon)

⇒ ∠EDF = ∠EFD = a (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠DEF + ∠EDF + ∠EFD = 180°

⇒ 120° + a + a = 180°

⇒ 2a = 180° - 120°

⇒ 2a = 60°

⇒ a = $\frac{60}{2}$ = 30°

⇒ ∠EFD = 30°.

By alternate segment theorem,

The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle subtended by that chord in the alternate segment of the circle.

∴ ∠PDE = ∠EFD = 30°.

Hence, ∠PDE = 30°.

(a) We know that,

The radius of a circle and tangent are perpendicular at the point of contact.

∴ ∠PQO = 90°

In right-angled triangle PQO,

⇒ PO² = PQ² + OQ²

⇒ 25² = PQ² + 7²

⇒ 625 = PQ² + 49

⇒ PQ² = 625 - 49

⇒ PQ² = 576

⇒ PQ = $\sqrt{576}$ = 24 cm.

In △ POQ,

⇒ tan θ = $\frac{OQ}{PQ}$

⇒ tan θ = $\frac{7}{24}$ ...............(1)

From figure,

⇒ ∠PAS = 90°

⇒ PA = OP - OA = 25 - 7 = 18 cm.

⇒ tan θ = $\frac{AS}{PA}$

⇒ tan θ = $\frac{AS}{18}$ ...........(2)

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

$$\begin{aligned}\Rightarrow \frac{7}{24} = \frac{AS}{18} \\[1em] \Rightarrow AS = \frac{7}{24} \times 18 \\[1em] \Rightarrow AS = \frac{7 \times 3}{4} = \frac{21}{4} \text{ cm}.\end{aligned}$$

From figure,

⇒ ST = 2 × AS = $2 \times \frac{21}{4} = \frac{21}{2}$ = 10.5 cm

Hence, ST = 10.5 cm.

(b) From equation (1),

⇒ tan θ = $\frac{7}{24}$

⇒ tan θ = 0.292

⇒ tan θ = tan 16° 16'

⇒ θ = 16° 16'.

Hence, θ = 16° 16'.

(c) In △ POQ,

⇒ ∠POQ + ∠PQO + ∠QPO = 180°

⇒ ∠POQ + 90° + 16° 16' = 180°

⇒ ∠POQ + 106° 16' = 180°

⇒ ∠POQ = 180° - 106° 16' = 73° 44' = 74°.

We know that,

Tangent from an external point to the circle are equal in length.

∴ PQ = PR.

Also,

OR = OQ (Both equal to radius of circle)

In △ POQ and △ POR,

⇒ PQ = PR (Proved above)

⇒ OQ = OR (Radius of same circle)

⇒ PO = PO (Common side)

∴ △ POQ ≅ △ POR (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠POR = ∠POQ = 74°

From figure,

⇒ ∠QOR = ∠POR + ∠POQ = 74° + 74° = 148°.

We know that,

The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.

⇒ ∠QOR = 2∠QUR

⇒ ∠QUR = $\frac{∠QOR}{2} = \frac{148°}{2}$ = 74°.

Hence, ∠QUR = 74°.

Join AT and BT.

(a) We know that,

Angle in a semicircle is a right angle.

∴ ∠CBT = 90°.

Hence, ∠CBT = 90°.

(b) In cyclic quadrilateral ATBC,

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

⇒ 90° + ∠CAT = 180°

⇒ ∠CAT = 180° - 90° = 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°.

(c) From figure,

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

∠PBT = ∠CBT - ∠CBA = 90° - 70° = 20°.

Hence, ∠PBT = 20°.

(d) Since, ∠PTB and ∠BTX are linear pairs.

⇒ ∠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°.