RS Aggarwal Solutions: Circles- 2 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

The maximum number of tangents that can be drawn from an external point to a circle is two and they are equal in length.

As SQ is diameter and OQ is radius in the given circle,
∴ 2OQ = SQ [As 2 × radius) = diameter]
2OQ = 6 cm
OQ = 3 cm
Now, QR is tangent
∴ OQ ⏊ QR
In right - angled △OQR,
By Pythagoras Theorem,
[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)² ]
(QR)² + (OQ)² = (OR)²
(4)² + (3)² = (OR)²
16 + 9 = (OR)²
(OR)² = 25
OR = 5 cm

We have given, PT is a tangent drawn at point T on the circle.
∴ OT ⏊ TP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, In △OTP we have,
By Pythagoras Theorem,
[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]
(OP)² = (OT)² + (PT)²
(OP)² = (7)² + (24)²
(OP)² = 49 + 576
(OP)² = 625
⇒ OP = 25 cm

As all diameters of a circle passes through center O it is not possible to have two parallel diameters in a circle.

Let us consider a circle with center O and AB be any chord that subtends 90° angle at its center.
Now, In △OAB
OA = OB = 10 cm
And as ∠AOB = 90°,
By Pythagoras Theorem,
[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]
(OA)² + (OB)² = (AB)²
(10)²+ (10)² = (AB)²
100 + 100 = (AB)²
⇒ AB = √200 = 10√2
So, Correct option is C .

We have given, PT is a tangent drawn at point T on the circle.
∴ OT ⏊ TP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, In △OTP we have,
By Pythagoras Theorem,
[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]
(OP)² = (OT)² + (PT)²
(10)² = (6)² + (PT)²
(PT)² = 100 - 36
(PT)² = 64
⇒ PT = 8 cm

We have given, PT is a tangent drawn at point T on the circle and OP = 26 cm and PT = 24 cm
Join OT
∴ OT ⏊ TP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, In △OTP we have,
By Pythagoras Theorem,
[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)²]
(OP)² = (OT)² + (PT)²
(26)² = (OT)² + (24)²
(OT)² = 676 - 576
(OT)² = 100
OT = 10 cm
Hence, radius of circle is 10 cm.

Let us consider a circle with center O and PQ is a tangent on the circle, Joined OP and OQ
But OPQ is an isosceles triangle,
∴ OP = PQ
∠OQP = ∠POQ
[Angles opposite to equal sides are equal]
In △OQP
∠OQP + ∠OPQ + ∠POQ = 180°
[Angle sum property of triangle]
∠OQP + 90° + ∠OPQ = 180°
2 ∠OPQ = 90°
∠OPQ = 45°

As AB and AC are tangents to given circle,
We have,
OB ⏊ AB and OC ⏊ AC
[∵ Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OBA = ∠OCA = 90°
In quadrilateral AOBC, By angle sum property of quadrilateral, we have,
∠OBA + ∠OCA + ∠BOC + ∠BAC = 360°
90° + 90° + ∠BOC + 40° = 360°
∠BOC = 140°

Let us consider a circle with center O and AB be a chord such that ∠AOB = 60°
AP and BP are two intersecting tangents at point P at point A and B respectively on the circle.
To find : Angle between tangents, i.e. ∠APB
As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP, By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
90° + 90° + ∠APB + 60° = 360°
∠APB = 120°

Given: Two concentric circles (say C₁ and C₂) with common center as O and radius r₁ = 6 cm(inner circle) and r₂ = 10 cm (outer circle) respectively.
To Find : Length of the chord AB.
As AB is tangent to circle C₁ and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"
So, we have,
OP ⏊ AB
∴ OPB is a right - angled triangle at P,
By Pythagoras Theorem in △OPB
[i.e. (hypotenuse)² = (perpendicular)² + (base)²]
We have,
(OP)² + (PA)² = (OA)²
r₁² + (PA)² = r₂²
(6)² + (PA)²= (10)²
36 + (PA)² = 100
(PA)² = 64
PA = 8 cm
Now, PA = PB ,
[as perpendicular from center to chord bisects the chord and OP ⏊ AB]
So,
AB = PA + PB = PA + PA = 2PA = 2(8) = 16 cm

As AB is tangent to the circle at point B
OB ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
In right angled triangle AOB,
By Pythagoras Theorem,
[i.e. (Hypotenuse)² = (Base)² + (Perpendicular)² ]
(OA)² = (OB)² + (AB)²
(17)² = (8)² + (AB)²
[As OA = 17 cm is given and OB is radius]
289 = 64 + (AB)²
(AB)² = 225
AB = 15 cm
Now, AB = AC [Tangents drawn from an external point are equal]
∴ AC = 15 cm

In △ABC
∠ABC = 90°
[Angle in a semicircle is a right angle]
∠ACB = 50° [Given]
By angle sum Property of triangle,
∠ACB + ∠ABC + ∠CAB = 180°
90° + 50° + ∠CAB = 180°
∠CAB = 40°
Now,
∠CAT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠CAB + ∠BAT = 90°
40° + ∠BAT = 90°
∠BAT = 50°

In △OPQ
∠POQ = 70° [Given]
OP = OQ [radii of same circle]
∠OQP = ∠OPQ [Angles opposite to equal sides are equal]
By angle sum Property of triangle,
∠POQ + ∠OQP + ∠OPQ = 180°
70° + ∠OPQ + ∠OPQ = 180°
2 ∠OPQ = 110°
∠OPQ = 55°
Now,
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPQ + ∠TPQ = 90°
55° + ∠TPQ = 90°
∠TPQ = 35°

Given: AT is a tangent to the circle with center O such that OT = 4 cm and ∠OTA = 30°.
To find: The value of AT.

In △ OAT ,
OA ⏊ AT [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴OAT is a right - angled triangle at A and



AT = 2√3 cm

As AP and BP are tangents to given circle,
We have,
OA ⏊ AP and OB ⏊ BP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AOBP,
By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠AOB + ∠APB = 360°
90° + 90° + 110° + ∠APB = 360°
∠APB = 70°

As we know,
Tangents drawn from an external point are equal, We have
AF = AE = 4 cm
[Tangents from common point A]
BF = BD = 3 cm
[Tangents from common point B]
CE = CD = x (say)
[Tangents from common point C]
Now,
AC = AE + CE
11 = 4 + x
x = 7 cm [1]
and, BC = BD + BC
BC = 3 + x = 3 + 7 = 10 cm

We know that the sum of angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180º
Therefore,
∠AOD + ∠BOC = 180º
135º + ∠BOC = 180º
∠BOC = 45º

In the given figure PT is a tangent to circle ∴ we have
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPQ + ∠QPT = 90°
∠OPQ + 50° = 90°
∠OPQ = 40°
Now, In △POQ
OP = OQ
∠PQO = ∠QPO = 40°
[Angles opposite to equal sides are equal]
Now,
∠ PQO + ∠QPO + ∠ POQ = 180°
[By angle sum property of triangle]
40° + 40° + ∠POQ = 180°
∠POQ = 100°

In the given figure, PA and PB are two tangents from common point P
∴ PA = PB
[Tangents drawn from an external point are equal]
∠PBA = ∠PAB…[1]
[Angles opposite to equal angles are equal]
By angle sum property of triangle in △APB
∠APB + ∠PBA + ∠PAB = 180°
60° + ∠PAB + ∠PAB = 180° [From 1]
2∠PAB = 120°
∠PAB = 60°…[2]
Now,
∠OAP = 90° [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAB + ∠PAB = 90°
∠OAB + 60° = 90° [From 2]
∠OAB = 30°

Let us consider a circle with center O and AP and BP are two tangents such that angle of inclination i.e. ∠APB = 60°
Joined OA, OB and OP.
To Find : Length of tangents
Now,
PA = PB [Tangents drawn from an external point are equal] [1]
In △AOP and △BOP
PA = PB [By 1]
OP = OP [Common]
OA = OB [radii of same circle]
△AOP ≅△BOP
[By Side - Side - Side Criterion]
∠OPA = ∠OPB
[Corresponding parts of congruent triangles are congruent]
Now,
∠APB = 60° [Given]
∠OPA + ∠OPB = 60°
∠OPA + ∠OPA = 60°
2 ∠OPA = 60°
∠OPA = 30°
In △AOP
OA ⏊ PA
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact
∴ AOP is a right - angled triangle.
So, we have



⇒ PA = 3√3 cm
From [1]
PA = PB = 4 cm
i.e. length of each tangent is 3√3 cm

In Given Figure,
PQ = PR…[1]
[Tangents drawn from an external point are equal]
In △AOP and △BOP
PQ = PR [By 1]
AP = AP [Common]
AQ = AR [radii of same circle]
△AQP ≅△ARP [By Side - Side - Side Criterion]
∠QPA = ∠RPA
[Corresponding parts of congruent triangles are congruent]
Now,
∠QPA + ∠RPA = ∠QPR
∠QPA + ∠QPA = ∠QPR
2 ∠QPA = ∠QPR
∠QPR = 2(27) = 54°
As PQ and PQ are tangents to given circle,
We have,
AQ ⏊ PQ and AR ⏊ PR
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠AQP = ∠ARP = 90°
In quadrilateral AQRP, By angle sum property of quadrilateral, we have
∠AQP + ∠ARP + ∠QAR + ∠QPR = 360°
90° + 90° + ∠QAR + 54° = 360°
∠QAR = 126°

Join AC, BC and CP
To Find: Length of tangents
Now,
PA = PB…[1]
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
In △ACP and △BCP
PA = PB [By 1]
CP = CP [Common]
CA = CB [radii of same circle]
△ACP ≅△BCP [By Side - Side - Side Criterion]
∠CPA = ∠CPB
[Corresponding parts of congruent triangles are congruent]
Now,
∠APB = 90°
[Given that PA ⏊ PB]
∠CPA + ∠CPB = 90°
∠CPA + ∠CPA = 90°
2 ∠CPA = 90°
∠CPA = 45°
In △ACP
CA ⏊ PA [Tangents drawn at a point on circle is perpendicular to the radius through point of contact
∴ ACP is a right - angled triangle.
So, we have


1 = 4/PA
⇒ PA = 4 cm
From [1]
PA = PB = 4 cm
i.e. length of each tangent is 4 cm

In Given Figure,
PA = PB…[1]
[Tangents drawn from an external point are equal]
In △AOP and △BOP
PA = PB [By 1]
OP = OP [Common]
OA = OB
[radii of same circle]
△AOP ≅△BOP
[By Side - Side - Side Criterion]
∠OPA = ∠OPB
[Corresponding parts of congruent triangles are congruent]
Now,
∠APB = 80° [Given]
∠OPA + ∠OPB = 80°
∠OPA + ∠OPA = 80°
2 ∠OPA = 80°
∠OPA = 40°
In △AOP,
∠OAP = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
And
∠OAP + ∠OPA + ∠AOP = 180°
90° + 40° + ∠AOP = 180°
∠AOP = 50°

In the given Figure, Join OP
Now, OP ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴∠OPA = 90°
∠OPQ + ∠APQ = 90°
∠OPQ + 58° = 90°
[Given, ∠APQ = 58°]
∠OPQ = 32°
In △OPQ
OP = OQ
[Radii of same circle]
∠OQP = ∠OPQ
[Angles opposite to equal sides are equal]
∠PQB = 32°
[As ∠OQP = ∠PQB ]

In given Figure, Join OP
In △OPC,
OP = OC [Radii of same circle]
∠OCP = ∠OPC
[Angles opposite to equal sides are equal]
∠ACP = ∠OPC
[As ∠OCP = ∠ACP] …[1]
Now,
∠OPB = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPC + ∠CPB = 90°
∠ACP + ∠CPB = 90° [By 1]
So,
∠CPB + ∠ACP = 90°

In the given Figure, Join OA
Now,
OA ⏊ PQ
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAP = ∠OAQ = 90° [1]
∠OAB + ∠PAB = 90°
∠OAB + 67° = 90°
∠OAB = 23°
Now,
∠BAC = 90°
[Angle in a semicircle is a right angle]
∠OAB + ∠OAC = 90°
23° + ∠OAC = 90°
∠OAC = 67°
∠OAQ = 90° [From 1]
∠OAC + ∠CAQ = 90°
67° + ∠CAQ = 90°
∠CAQ = 23° [2]
Now,
OA = OC
[radii of same circle]
∠OCA = ∠OAC
[Angles opposite to equal sides are equal]
∠OCA = 67°
∠OCA + ∠ACQ = 180° [Linear Pair]
67° + ∠ACQ = 180°
∠ACQ = 113° [3]
Now, In △ACQ By Angle Sum Property of triangle
∠ACQ + ∠CAQ + ∠AQC = 180°
113° + 23° + ∠AQC = 180° [By 2 and 3]
∠AQC = 44°

In Given Figure,
PQ = PR…[1]
[Tangents drawn from an external point are equal]
In △QOP and △ROP
PQ = PR [By 1]
OP = OP [Common]
OQ = OR [radii of same circle]
△QOP ≅△ROP
[By Side - Side - Side Criterion]
area(Δ QOP) = area(Δ ROP)
[Congruent triangles have equal areas]
area(PQOR) = area(Δ QOP) + area(Δ ROP)
area(PQOR) = area(Δ QOP) + area(Δ QOP) = 2[area(Δ QOP)]
Now,
OQ ⏊PQ
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, QOP is a right - angled triangle at Q with OQ as base and PQ as height.
In △QOP,
By Pythagoras Theorem in △OPB
[i.e. (hypotenuse)² = (perpendicular)² + (base)²]
(OQ)² + (PQ)² = (OP)²
(5)² + (PQ)² = (13)²
25 + (PQ)² = 169
(PQ)² = 144
PQ = 12 cm
Area(ΔQOP) = 1/2 × Base × Height
= 1/2 × OQ × PQ
= 1/2 × 5 × 12
= 30 cm²
So,
Area(PQOR) = 2(30) = 60 cm²

In given figure, as PR is a tangent
OQ ⏊ PR
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
⇒ LQ ⏊ PR
⇒ LQ ⏊ AB
[As, AB || PR]
AL = LB
[Perpendicular from center to the chord bisects the chord]
Now,
∠LQR = 90°
∠LQB + ∠BQR = 90°
∠LQB + 70° = 90°
∠LQB = 20°…[1]
In △AQL and △BQL
∠ALQ = ∠BLQ [Both 90° as LQ ⏊ AB]
AL = LB [Proved above]
QL = QL [Common]
△AQL ≅ △BQL
[Side - Angle - Side Criterion]
∠LQA = ∠LQB
[Corresponding parts of congruent triangles are congruent]
∠AQB = ∠LQA + ∠LQB = ∠LQB + ∠LQB
= 2∠LQB = 2(20) = 40° [By 1]

Let us consider a circle with center O and TP be a tangent at point A on the circle, Joined OT and OP
Given Length of tangent, TP = 10 cm, and OT = 5 cm [radius]
To Find : Distance of center O from P i.e. OP
Now,
OP ⏊ TP
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So OPT is a right - angled triangle,
By Pythagoras Theorem in ΔOPB
[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]
(OT)² + (TP)² = (OP)²
(OP)² = (5)² + (10)²
(OP)² = 25 + 100 = 125
OP = √125 cm

In △BOP
OB = OP [radii of same circle]
∠OPB = ∠PBO
[Angles opposite to equal sides are equal]
As, ∠PBO = 30°
∠OPB = 30°
Now,
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠BPT = ∠OPB + ∠OPT = 30° + 90° = 120°
Now, In ΔBPT
∠BPT + ∠PBO + ∠PTB = 180°
120° + 30° + ∠PTB = 180°
∠PTB = 30°
∠PTA = ∠PTB = 30°

Given : In the given figure, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively and EK = 9 cm
To Find : Perimeter of △EDF
As we know that, Tangents drawn from an external point to a circle are equal.
So, we have
KD = DH …[1]
[Tangents from point D]
HF = FM …[2]
[Tangents from point F]
Now Perimeter of Triangle PCD
= ED + DF + EF
= ED + DH + HF + EF
= ED + KD + FM + EF [From 1 and 2]
= EK + EM
Now,
EK = EM = 9 cm as tangents drawn from an external point to a circle are equal
So, we have
Perimeter = EK + EM = 9 + 9 = 18 cm

Let us consider a circle with center O and PA and PB are two tangents from point P, given that angle of inclination i.e. ∠APB = 45°
As PA and PB are tangents to given circle,
We have,
OA ⏊ PA and OB ⏊ PB [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAP = ∠OBP = 90°
In quadrilateral AQRP,
By angle sum property of quadrilateral, we have
∠OAP + ∠OBP + ∠ABP + ∠AOB = 360°
90° + 90° + 45° + ∠AOB = 360°
∠AOB = 135°

As PL and PM are tangents to given circle,
We have,
OR ⏊ PM and OQ ⏊ PL
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠ORM = ∠OQL = 90°
∠ORM = ∠ORS + ∠SRM
90° = ∠ORS + 60°
∠ORS = 30°
And ∠OQL = ∠OQS + ∠SQL
90° = ∠OQS + 50°
∠OQS = 40°
Now, In △SOR
OS = OQ [radii of same circle]
∠ORS = ∠OSR
[Angles opposite to equal sides are equal]
∠OSR = 30°
[as ∠ORS = 30°]
Now, In △SOR
OS = SQ [radii of same circle]
∠OQS = ∠OSQ
[Angles opposite to equal sides are equal]
∠OSQ = 40° [as ∠OQS = 40°]
As,
∠QSR = ∠OSR + ∠OSQ
∠QSR = 30° + 40° = 70°

Given : △PQR that is drawn to circumscribe a circle with radius r = 6 cm and QT = 12 cm QR = 9cm
Also, area(△PQR) = 189 cm²
Let tangents PR and PQ touch the circle at X and Y respectively.
To Find : PQ and QR
Now,
As we know tangents drawn from an external point to a circle are equal.
Then,
QT = QY = 12 cm
[Tangents from same external point B]
TR = RX = 9 cm
[Tangents from same external point C]
PX = PY = x (let)
[Tangents from same external point A]
Using the above data we get
PQ = PY + QT = x + 12 cm
PR = PC + RX = x + 9 cm
QR = QT + TR = 12 + 9 = 21 cm
Now we have heron's formula for area of triangles if its three sides a, b and c are given

Where,

So for △PQR
a = PQ = x + 12
b = PR = x + 9
c = QR = 21 cm

And

Squaring both side
189(189) = 108(x + 21)
7(189) = 4(x + 21)
4x² + 84x - 1323 = 0
As we know roots of a quadratic equation in the form ax2 + bx + c = 0 are,

So, roots of this equation are,

but x = - 31.5 is not possible as length can't be negative.
So
PQ = x + 12 = 10.5 + 12 = 22.5 cm

Let the bigger circle be C1 and Smaller be C2,
Now,
PQ and PT are two tangents to circle C1,
∴ PT = QP
[Tangents drawn from an external point are equal]
QP = 3.8 cm
[ As PT = 3.8 cm is given]
Also,
PR and PT are two tangents to circle C2,
∴ PT = PR
[Tangents drawn from an external point are equal]
PR = 3.8 cm
[ As PT = 3.8 cm is given]
QR = QP + PR = 3.8 + 3.8 = 7.6 cm

As we know Tangents drawn from an external point are equal]
In the given Figure, we have
AP = AQ = 5 cm
[Tangents from point A] [AP = 5 cm is given]
BQ = BR = x(say)
[Tangents from point B]
CR = CS = 3 cm [∵ CS = 3 cm is given]
[Tangents from point C]
Given,
BC = 7 cm
CR + BR = 7
3 + x = 7 cm
x = 4 cm
Now,
AB = AQ + BQ = 5 + x = 5 + 4 = 9 cm

As we know Tangents drawn from an external point are equal]
In the given Figure, we have
AP = AS = 6 cm [AP = 6 cm is given]
[∵ Tangents from point A]
BP = BQ = 5 cm [BP = 5 cm is given]
[∵ Tangents from point B]
CR = CQ = 3 cm [CQ = 3 cm is given]
[∵ Tangents from point C]
DR = DS = 4 cm ][DR = 4 cm is given]
[∵ Tangents from point D]
Now,
Perimeter of ABCD = AB + BC + CA + DA
= AP + BP + BQ + CQ + CR + DR + DS + AS
= 6 + 5 + 5 + 3 + 3 + 4 + 4 + 6 = 36 cm

In △AOB
OA = OB [radii of same circle]
∠OBA = ∠OAB [Angles opposite to equal sides are equal]
Also, By Triangle sum Property
∠AOB + ∠OBA + ∠OAB = 180°
100 + ∠OAB + ∠OAB = 180°
2 ∠OAB = 90°
∠OAB = 40°
As AT is tangent to given circle,
We have,
OA ⏊ AT
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAT = 90°
∠OAB + ∠BAT = 90°
40° + ∠BAT = 90°
∠BAT = 50°

Let AB, BC and AC touch the circle at points P, Q and R respectively.
As ABC is a right triangle,
By Pythagoras Theorem
[i.e. (hypotenuse)² = (perpendicular)² + (base)²]
(AC)² = (BC)² + (AB)²
(AC)² = (12)² + (5)²
(AC)² = 144 + 25 = 169
AC = 13 cm
Let O be the center of circle, Join OP, OQ and PR
Let the radius of circle be r,
We have
r = OP = OQ = OR
[radii of same circle] [1]
Now,
ar(△ABC) = ar(△AOB) + ar(△BOC) + ar(△AOC)
As we know,
Area of triangle is 1/2 × Base × Height (Altitude)
Now,
OP ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴ OP is the altitude in △AOB
OQ ⏊ BC
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴ OQ is the altitude in △BOC
OR ⏊ AC
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴ OR is the altitude in △AOC
So, we have
1/2 × BC × AB = (1/2 × AB × OP) + (1/2 × BC × OQ) + (1/2 × AC × OR)
12(5) = 5(r) + 12(r) + 13(r) [Using 1]
60 = 30r
r = 2 cm

In quadrilateral ORDS
∠ORD = 90°
[∵ Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OSD = 90°
[∵ Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠SDR = 90° [AD ⏊ CD]
By angle sum property of quadrilateral PQOB
∠ORD + ∠OSD + ∠SDR + ∠SOR = 360°
90° + 90° + 90° + ∠SOR = 360°
∠SOR = 90°
As all angles of this quadrilaterals are 90° The quadrilateral is a rectangle
Also, OS = OR = r
i.e. adjacent sides are equal, and we know that a rectangle with adjacent sides equal is a square
∴ POQB is a square
And OS = OR = DR = DS = r = 10 cm [1]
Now,
As we know that tangents drawn from an external point to a circle are equal
In given figure, We have
CQ = CR …[2]
[∵ tangents from point C]
PB = BQ = 27 cm
[∵Tangents from point B and PB = 27 cm is given]
BC = 38 cm [Given]
BQ + CQ = 38
27 + CQ = 38
CQ = 11 cm
From [2]
CQ = CR = 11 cm
Now,
CD = CR + DR
CD = 11 + 10 = 21 cm [from 1, DR = 10 cm]

As ABC is a right triangle,
By Pythagoras Theorem
[i.e. (hypotenuse)² = (perpendicular)² + (base)²]
(AC)² = (BC)² + (AB)²
(AC)² = (6)² + (8)²
(AC)² = 36 + 64 = 100
AC = 10 cm
Now,
ar(△ABC) = ar(△AOB) + ar(△BOC) + ar(△AOC)
As we know,
Area of triangle is 1/2 × Base × Height(Altitude)
Now,
OP ⏊ AB [Given]
∴ OP is the altitude in △AOB
OQ ⏊ BC [Given]
∴ OQ is the altitude in △BOC
OR ⏊ AC [Given]
∴ OR is the altitude in △AOC
So, we have
1/2 × BC × AB = (1/2 × AB × OP) + (1/2 × BC × OQ) + (1/2 × AC × OR)
6(8) = 8(x) + 6(x) + 10(x)
[∵ OP = OQ = OR = x, Given]
48 = 24x
x = 2 cm

Let sides AB, BC, CD, and AD touches circle at P, Q, R and S respectively.
As we know that tangents drawn from an external point to a circle are equal,
So, we have,
AP = AS = w (say)
[∵ Tangents from point A]
BP = BQ = x (say)
[∵Tangents from point B]
CP = CR = y (say)
[∵Tangents from point C]
DR = DS = z (say)
[∵Tangents from point D]
Now,
Given,
AB = 6 cm
AP + BP = 6
w + x = 6 …[1]
BC = 7 cm
BP + CP = 7
x + y = 7 …[2]
CD = 4 cm
CR + DR = 4
y + z = 4 …[3]
Also,
AD = AS + DS = w + z …[4]
Add [1] and [3] and substracting [2] from the sum we get,
w + x + y + z - (x + y) = 6 + 4 - 7
w + z = 3 cm
From [4]
AD = 3 cm

In the given Figure
AR = AP = x(let) [Radii of same circle]
BP = BQ = y(let) [Radii of same circle]
CR = CQ = z(let) [Radii of same circle]
Now,
AB = 5 cm [Given]
AP + BP = 5
x + y = 5
y = 5 - x …[1]
BC = 7 cm [Given]
BQ + CQ = 7
y + z = 7
5 - x + z = 7 [using 1]
z = 2 + x …[2]
and
AC = 6 cm [Given]
x + z = 6
x + 2 + x = 6 [Using 2]
2x = 4
x = 2 cm

Let tangent BC touch the circle at point R
As we know tangents drawn from an external point to a circle are equal.
We have
AP = AQ
[tangents from point A]
BP = BR …[1]
[tangents from point B]
CQ = CR …[2]
[tangents from point C]
Now,
AP = AQ
⇒ AB + BP = AC + CQ
⇒ 5 + BR = 6 + CR [From 1 and 2]
⇒CR = BR - 1 …[3]
Now,
BC = 4 cm
BR + CR = 4
BR + BR - 1 = 4 [Using 3]
2BR = 5 cm
BR = 2.5 cm
BP = BR = 2.5 cm [Using 2]
AP = AB + BP = 5 + 2.5 = 7.5 cm

In given Figure,
OA ⏊ AP [Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △OAP,
By Pythagoras Theorem
[i.e. (hypotenuse)² = (perpendicular)² + (base)²]
(OP)² = (OA)² + (PA)²
Given, PA = 12 cm and OA = radius of outer circle = 5 cm
(OP)² = (5)² + (12)²
(OP)² = 25 + 144 = 136
OP = 13 cm …[1]
Also,
OB ⏊ BP [Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △OBP,
By Pythagoras Theorem
[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]
(OP)² = (OB)² + (PB)²
Now, OB = radius of inner circle = 3 cm
And, from [2] (OP) = 13 cm
(13)² = (3)² + (PB)²
(PB)² = 169 - 9 = 160
PB = 4√10 cm

A circle cannot have more than two tangents parallel, because tangents to be parallel they should be at diametrically ends and a diameter has two ends only.

A straight line can meet a circle at two points in case if it is a chord or diameter or a line intersecting the circle at two points.

If a tangent is drawn from a point inside a circle, it will intersect the circle at two points, so no tangent can be drawn from a point inside the circle.

Let us consider a circle with center O and radius 12 cm
A tangent PQ is drawn at point P such that PQ = 16 cm
To Find : Length of OQ
Now, OP ⏊ PQ [Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right - angled △POQ,
By Pythagoras Theorem
[i.e. (hypotenuse)² = (perpendicular)² + (base)² ]
(OQ)² = (OP)² + (PQ)²
(OQ² = (12)² + (16)²
625 = 144 + 256
(OQ)² = 400
OQ = 20 cm
So,
Assertion is correct, and Reason is also correct.

Let PT and PQ are two tangents from external point P to a circle with center O
In △OPT and △OQT
OP = OQ
[radii of same circle]
OT = OT
[common]
PT = PQ
[Tangents drawn from an external point are equal]
△OPT ≅ △OQT
[By Side - Side - Side Criterion]
∠POT = ∠QOT
[Corresponding parts of congruent triangles are congruent]
i.e. Assertion is true
Now,

Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.
As ABCD is a parallelogram
AB = CD and BC = AD
[opposite sides of a parallelogram are equal] [1]
Now, As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
AB + AB = BC + BC [From 1]
AB = BC …[2]
From [1] and [2]
AB = BC = CD = AD
And we know,
A parallelogram with all sides equal is a rhombus
So, reason is also true, but not a correct reason for assertion.
Hence, B is correct option .

For Assertion :
In the given Figure,
As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
So, assertion is not true
For Reason,

Consider two concentric circles with common center O and AB is a chord to outer circle and is tangent to inner circle P.
Now,
OP ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
We know, that perpendicular from center to chord bisects the chord.
So, P bisects AB.
Reason is true
Hence, Assertion is false, But Reason is true.