Q.1. Find the length of tangent drawn to a circle with radius 8 cm from a point 17 cm away from the center of the circle.
Let us consider a circle with center O and radius 8 cm.
The diagram is given as:Consider a point A 17 cm away from the center such that OA = 17 cm
A tangent is drawn at point A on the circle from point B such that OB = radius = 8 cm
To Find: Length of tangent AB = ?
As seen OB ⏊ AB
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∴ In right  angled ΔAOB, By Pythagoras Theorem
[i.e. (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2} ]
(OA)^{2 }= (OB)^{2} + (AB)^{2}
(17)^{2} = (8)^{2} + (AB)^{2}
289 = 64 + (AB)^{2}
(AB)^{2} = 225
AB = 15 cm
∴ The length of the tangent is 15 cm.
Q.2. A point P is 25 cm away from the center of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
Let us consider a circle with center O.
Consider a point P 25 cm away from the center such that OP = 25 cm
A tangent PQ is drawn at point Q on the circle from point P such that PQ = 24 cm
To Find : Length of radius OQ = ?
Now, OQ ⏊ 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} ]
(OP)^{2} = (OQ)^{2} + (PQ)^{2}
(25)^{2} = (OQ)^{2} + (24)^{2}
625 = (OQ)^{2 }+ 576
(OQ)^{2} = 49
OQ = 7 cm
Q.3. Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Given: Two concentric circles (say C_{1} and C_{2}) with common center as O and radius r_{1} = 6.5 cm and r_{2} = 2.5 cm respectively.
To Find: Length of the chord of the larger circle which touches the circle C_{2}. i.e. Length of AB.
As AB is tangent to circle C_{2} 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} + (PB)^{2} = (OB)^{2}
r_{2}^{2} + (PB)^{2} = r_{1}^{2}
(2.5)^{2} + (PB)^{2}= (6.5)^{2}
6.25 + (PB)^{2} = 42.25
(PB)^{2} = 36
PB = 6 cm
Now, AP = PB ,
[ as perpendicular from center to chord bisects the chord and OP ⏊ AB ]
So,
AB = AP + PB = PB + PB
= 2PB = 2(6)
= 12 cm
Q.4. In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF.
Let AD = x cm, BE = y cm and CF = z cm
As we know that,
Tangents from an external point to a circle are equal,
In given Figure we have
AD = AF = x [Tangents from point A]
BD = BE = y [Tangents from point B]
CF = CE = z [Tangents from point C]
Now, Given: AB = 12 cm
AD + BD = 12
x + y = 12
y = 12 – x…. [1]
and BC = 8 cm
BE + EC = 8
y + z = 8
12  x + z = 8 [From 1]
z = x – 4…. [2]
and
AC = 10 cm
AF + CF = 10
x + z = 10 [From 2]
x + x  4 = 10
2x = 14
x = 7 cm
Putting value of x in [1] and [2]
y = 12  7 = 5 cm
z = 7  4 = 3 cm
So, we have AD = 7 cm, BE = 5 cm and CF = 3 cm
Q.5. In the given figure, PA and PB are the tangent segments to a circle with center 0. Show that the points A, O, B and P are concyclic.
Given: PA and PB are tangents to a circle with center O
To show : A, O, B and P are concyclic i.e. they lie on a circle i.e. AOBP is a cyclic quadrilateral.
Proof:
OB ⏊ PB and OA ⏊ AP
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OBP = ∠OAP = 90°
∠OBP + ∠OAP = 90 + 90 = 180°
AOBP is a cyclic quadrilateral i.e. A, O, B and P are concyclic.
[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic]
Hence Proved.
Q.6. In the given figure, the chord AB of the larger of the two concentric circles, with center O, touches the smaller circle at C. Prove that AC = CB.
Given: Two concentric circles with common center as O
To Prove: AC = CB
Construction: Join OC, OA and OB
Proof :
OC ⏊ AB
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
In △OAC and △OCB, we have
OA = OB
[∵ radii of same circle]
OC = OC
[∵ common]
∠OCA = ∠OCB
[∵ Both 90° as OC ⏊ AB]
△OAC ≅ △OCB
[By Right Angle  Hypotenuse  Side]
AC = CB
[Corresponding parts of congruent triangles are congruent]
Hence Proved.
Q.7. From an external point P, tangents PA and PB are drawn to a circle with center O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of △PCD.
Given : From an external point P, two tangents, PA and PB are drawn to a circle with center O. At a point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. And PA = 14 cm
To Find : Perimeter of △PCD
As we know that, Tangents drawn from an external point to a circle are equal.
So we have
AC = CE …[1] [Tangents from point C]
ED = DB …[2] [Tangents from point D]
Now Perimeter of Triangle PCD
= PC + CD + DP
= PC + CE + ED + DP
= PC + AC + DB + DP [From 1 and 2]
= PA + PB
Now,
PA = PB = 14 cm as tangents drawn from an external point to a circle are equal
So we have
Perimeter = PA + PB = 14 + 14 = 28 cm
Q.8. A circle is inscribed in a LABC touching AB, BC and AC at P, Q and R respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.
As we know that tangents drawn from an external point to a circle are equal ,
In the Given image we have,
AP = AR = 7 cm ….[1]
[tangents from point A]
CR = QC = 5 cm ….[2]
[tangents from point C]
BQ = PB …[3]
[tangents from point B]
Now,
AB = 10 cm [Given]
AP + PB = 10 cm
7 + PB = 10 [From 1]
PB = 3 cm
BQ = 3 cm …..[4]
[From 3]
BC = BQ + QC = 5 + 3 = 8 cm [ From 2 and 4]
Q.9. In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
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,
In the given image 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
Q.10. In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.
As we know that tangents drawn from an external point to a circle are equal,
BR = BP [ Tangents from point B] [1]
QC = CP [ Tangents from point C] [2]
AR = AQ [ Tangents from point A] [3]
As ABC is an isosceles triangle,
AB = BC [Given] [4]
Now substract [3] from [4]
AB  AR = BC  AQ
BR = QC
BP = CP [ From 1 and 2]
∴ P bisects BC
Hence Proved.
Q.11. In the given figure, O is the center of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. If PA = 10 cm, find the length of PB up to one place of decimal.
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 = 10 cm and OA = radius of outer circle = 6 cm
(OP)^{2} = (6)^{2} + (100)^{2}
(OP)^{2} = 36 + 100 = 136 [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 = 4 cm
And from [2]
(OP)^{2} = 136
136 = (4)^{2 }+ (PB)^{2}
(PB)^{2} = 136  16 = 120
PB = 10.9 cm
Q.12. In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ABC = 54 cm^{2} then find the lengths of sides AB and AC.
Given : △ABC that is drawn to circumscribe a circle with radius r = 3 cm and BD = 6 cm DC = 9cm
Also, area(△ABC) = 54 cm^{2}
To Find : AB and AC
Now,
As we know tangents drawn from an external point to a circle are equal.
Then,
FB = BD = 6 cm [Tangents from same external point B]
DC = EC = 9 cm [Tangents from same external point C]
AF = EA = x (let) [Tangents from same external point A]
Using the above data, we get
AB = AF + FB = x + 6 cm
AC = AE + EC = x + 9 cm
BC = BD + DC = 6 + 9 = 15 cm
Now we have heron's formula for area of triangles if its three sides a, b and c are given
Where,
⇒
So for △ABC
a = AB = x + 6
b = AC = x + 9
c = BC = 15 cm
⇒
And
Squaring both sides, we get,
54(54) = 54x(x + 15)
x^{2} + 15x  54 = 0
x^{2} + 18x  3x  54 = 0
x(x + 18)  3(x + 18) = 0
(x  3)(x + 18) = 0
x = 3 or  18
but x =  18 is not possible as length can't be negative.
So
AB = x + 6 = 3 + 6 = 9 cm
AC = x + 9 = 3 + 9 = 12 cm
Q.13. PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
Given : A circle with center O and radius 3 cm and PQ is a chord of length 4.8 cm. The tangents at P and Q intersect at point T
To Find : Length of TP
Construction : Join OQ
Now in △OPT and △OQT
OP = OQ [radii of same circle]
PT = PQ
[tangents drawn from an external point to a circle are equal]
OT = OT [Common]
△OPT ≅ △OQT [By Side  Side  Side Criterion]
∠POT = ∠OQT
[Corresponding parts of congruent triangles are congruent]
or ∠POR = ∠OQR
Now in △OPR and △OQR
OP = OQ [radii of same circle]
OR = OR [Common]
∠POR = ∠OQR [Proved Above]
△OPR ≅ △OQT [By Side  Angle  Side Criterion]
∠ORP = ∠ORQ
[Corresponding parts of congruent triangles are congruent]
Now,
∠ORP + ∠ORQ = 180° [Linear Pair]
∠ORP + ∠ORP = 180°
∠ORP = 90°
⇒ OR ⏊ PQ
⇒ RT ⏊ PQ
As OR ⏊ PQ and Perpendicular from center to a chord bisects the chord we have
∴ In right  angled △OPR,
By Pythagoras Theorem
[i.e. (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}]
(OP)^{2} = (OR)^{2} + (PR)^{2}
(3)^{2} = (OR)^{2} + (2.4)^{2}
9 = (OR)^{2 }+ 5.76
(OR)^{2}= 3.24
OR = 1.8 cm
Now,
In right angled △TPR,
By Pythagoras Theorem
(PT)^{2} = (PR)^{2} + (TR)^{2} …[1]
Also, OP ⏊ OT
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
In right angled △OPT, By Pythagoras Theorem
(PT)^{2} + (OP)^{2} = (OT)^{2}
(PR)^{2} + (TR)^{2} + (OP)^{2}= (TR + OR)^{2} …[From 1]
(2.4)^{2} + (TR)^{2} + (3)^{2 }= (TR + 1.8)^{2}
4.76 + (TR)^{2} + 9 = (TR)^{2} + 2(1.8)TR + (1.8)^{2}
13.76 = 3.6TR + 3.24
3.6TR = 10.52
TR = 2.9 cm [Appx]
Using this in [1]
PT^{2} = (2.4)^{2} + (2.9)^{2}
PT^{2} = 4.76 + 8.41
PT^{2} = 13.17
PT = 3.63 cm [Appx]
Q.14. Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its center.
Given: A circle with center O and AB and CD are two parallel tangents at points P and Q on the circle.
To Prove: PQ passes through O
Construction: Draw a line EF parallel to AB and CD and passing through O
Proof :
∠OPB = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
Now, AB  EF
∠OPB + ∠POF = 180°
90° + ∠POF = 180°
∠POF = 90° …[1]
Also,
∠OQD = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
Now, CD  EF
∠OQD + ∠QOF = 180°
90° + ∠QOF = 180°
∠QOF = 90° [2]
Now From [1] and [2]
∠POF + ∠QOF = 90 + 90 = 180°
So, By converse of linear pair POQ is a straight Line
i.e. O lies on PQ
Hence Proved.
Q.15. In the given figure, a circle with center O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, LB = 90° and DS = 5 cm then find the radius of the circle.
In quadrilateral POQB
∠OPB = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OQB = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠PQB = 90° [Given]
By angle sum property of quadrilateral PQOB
∠OPB + ∠OQB + ∠PBQ + ∠POQ = 360°
90° + 90° + 90° + ∠POQ = 360°
∠POQ = 90°
As all angles of this quadrilaterals are 90° The quadrilateral is a rectangle
Also, OP = OQ = 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 OP = PB = BQ = OQ = r [1]
Now,
As we know that tangents drawn from an external point to a circle are equal
In given figure, We have
DS = DR = 5 cm
[Tangents from point D and DS = 5 cm is given]
AD = 23 cm [Given]
AR + DR = 23
AR + 5 = 23
AR = 18 cm
Now,
AR = AQ = 18 cm
[Tangents from point A]
AB = 29 cm [Given]
AQ + QB = 29
18 + QB = 29
QB = 11 cm
From [1]
QB = r = 11 cm
Hence Radius of circle is 11 cm.
Q.16. In the given figure, O is the center of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA : AT = 2 : 1.
In Given Figure, we have a circle with center O let the radius of circle be r.
Construction : Join OP
Now, In △APB
∠ABP = 30°
∠APB = 90°
[Angle in a semicircle is a right angle]
By angle sum Property of triangle,
∠ABP + ∠APB + ∠PAB = 180
30° + 90° + ∠PAB = 180
∠PAB = 60°
OP = OA = r [radii]
∠PAB = ∠OPA = 60°
[Angles opposite to equal sides are equal]
By angle sum Property of triangle
∠OPA + ∠OAP + ∠AOP = 180°
60° + ∠PAB + ∠AOP = 180
60 + 60 + ∠AOP = 180
∠AOP = 60°
As all angles of △OPA equals to 60°, △OPA is an equilateral triangle
So, we have, OP = OA = PA = r [1]
∠OPT = 90°
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OPA + ∠APT = 90
60 + ∠APT = 90
∠APT = 30°
Also,
∠PAB + ∠PAT = 180° [Linear pair]
60° + ∠PAT = 180°
∠PAT = 120°
In △APT
∠APT + ∠PAT + ∠PTA = 180°
30° + 120° + ∠PTA = 180°
∠PTA = 30°
So,
We have
∠APT = ∠PTA = 30°
AT = PA
[Sides opposite to equal angles are equal]
AT = r [From 1] [2]
Now,
AB = OA + OB = r + r = 2r [3]
From [2] and [3]
AB : AT = 2r : r = 2 : 1
Hence Proved !
Q.1. In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD.
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 ,
In the given image 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 = 9 cm
BP + CP = 9
x + y = 9 [2]
CD = 8 cm
CR + DR = 8
y + z = 8 [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 + 8  9
w + z = 5 cm
From [4]
AD = 5 cm
Q.2. In the given figure, PA and PB are two tangents to the circle with center O. If ∠APB = 50° then what is the measure of ∠OAB.
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
[Angles opposite to equal angles are equal] [1]
By angle sum property of triangle in △APB
∠APB + ∠PBA + ∠PAB = 180°
50° + ∠PAB + ∠PAB = 180° [From 1]
2∠PAB = 130°
∠PAB = 65° [2]
Now,
∠OAP = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAB + ∠PAB = 90°
∠OAB + 65° = 90° [From 2]
∠OAB = 25°
Q.3. In the given figure, O is the center of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.
Given: In the figure, PT and PQ are two tangents and ∠TPQ = 70°
To Find: ∠TRQ
Construction: Join OT and OQ
In quadrilateral OTPQ
∠OTP = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OQP = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠TPQ = 70° [Common]
By Angle sum of Quadrilaterals,
In quadrilateral OTPQ we have
∠OTP + ∠OQP + ∠TPQ + ∠TOQ = 360°
90° + 90° + 70° + ∠TOQ = 360°
250° + ∠TOQ = 360
∠TQO = 110°
Now,
As we Know the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
∴ we have
∠TOQ = 2∠TRQ
110° = 2 ∠TRQ
∠TRQ = 55°
Q.4. In the given figure, common tangents AB and CD to the two circles with centers O1 and O2 intersect at E. Prove that AB = CD.
Given: AB and CD are two tangents to two circles which intersects at E .
To Prove: AB = CD
Proof:
As
AE = CE …[1]
[Tangents drawn from an external point to a circle are equal]
And
EB = ED …[2]
[Tangents drawn from an external point to a circle are equal]
Adding [1] and [2]
AE + EB = CE + ED
AB = CD
Hence Proved.
Q.5. If PT is a tangent to a circle with center O and PQ is a chord of the circle such that ∠QPT = 70°, then find the measure of ∠POQ.
Given: PT is a tangent to a circle with center O and PQ is a chord of the circle such that ∠QPT = 70°
To Find: ∠POQ = ?
Now,
∠OPT = 90°
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OPQ + ∠QPT = 90°
∠OPQ + 70° = 90°
∠OPQ = 20°
Also,
OP = OQ [Radii of same circle]
∠OQP = ∠OPQ = 20°
[Angles opposite to equal sides are equal]
In △OPQ By Angle sum property of triangles,
∠OPQ + ∠OQP + ∠POQ = 180°
20° + 20° + ∠POQ = 180°
∠POQ = 140°
Q.6. In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 4 cm and 3 cm respectively. If the area of ΔABC = 21 cm^{2 }then find the lengths of sides AB and AC.
Given: △ABC that is drawn to circumscribe a circle with radius r = 2 cm and BD = 4 cm DC = 3cm
Also, area(△ABC) = 21 cm^{2}
To Find: AB and AC
Now,
As we know tangents drawn from an external point to a circle are equal.
Then,
FB = BD = 4 cm [Tangents from same external point B]
DC = EC = 3 cm [Tangents from same external point C]
AF = EA = x (let) [Tangents from same external point A]
Using the above data, we get
AB = AF + FB = x + 4 cm
AC = AE + EC = x + 3 cm
BC = BD + DC = 4 + 3 = 7 cm
Now we have heron's formula for area of triangles if its three sides a, b and c are given
Where,
So, for △ABC
a = AB = x + 4
b = AC = x + 3
c = BC = 7 cm
⇒
And
Squaring both sides,
21(21) = 12x(x + 7)
12x^{2} + 84x  441 = 0
4x^{2} + 28x  147 = 0
As we know roots of a quadratic equation in the form ax^{2} + bx + c = 0 are,
So roots of this equation are,
but x =  10.5 is not possible as length can't be negative.
So
AB = x + 4 = 3.5 + 4 = 7.5 cm
AC = x + 3 = 3.5 + 3 = 6.5 cm
Q.7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle (in cm) which touches the smaller circle.
Given : Two concentric circles (say C_{1} and C_{2}) with common center as O and radius r_{1} = 5 cm and r_{2} = 3 cm respectively.
To Find : Length of the chord of the larger circle which touches the circle C_{2}. i.e. Length of AB.
As AB is tangent to circle C_{2} 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} + (PB)^{2} = (OB)^{2}
r_{2}^{2} + (PB)^{2} = r_{1}^{2}
(3)^{2} + (PB)^{2}= (5)^{2}
9 + (PB)^{2} = 25
(PB)^{2} = 16
PB = 4 cm
Now, AP = PB,
[ as perpendicular from center to chord bisects the chord and OP ⏊ AB ]
So,
AB = AP + PB = PB + PB
= 2PB = 2(4) = 8 cm
Q.8. Prove that the perpendicular at the point of contact of the tangent to a circle passes through the center.
Let us consider a circle with center O and XY be a tangent
To prove : Perpendicular at the point of contact of the tangent to a circle passes through the center i.e. the radius OP ⏊ XY
Proof :
Take a point Q on XY other than P and join OQ .
The point Q must lie outside the circle. (because if Q lies inside the circle, XY will become a secant and not a tangent to the circle).
∴ OQ is longer than the radius OP of the circle. That is,
OQ > OP.
Since this happens for every point on the line XY except the point P, OP is the
shortest of all the distances of the point O to the points of XY.
So OP is perpendicular to XY.
[As Out of all the line segments, drawn from a point to points of a line not passing through the point, the smallest is the perpendicular to the line.]
Q.9. In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with center 0. If ∠PRQ = 120°, then prove that OR = PR + RQ.
Given : In the figure ,
Two tangents RQ and RP are drawn from an external point R to the circle with center O and ∠PRQ = 120°
To Prove: OR = PR + RQ
Construction: Join OP and OQ
Proof :
In △△OPR and △OQR
OP = OQ [radii of same circle]
OR = OR [Common]
PR = PQ …[1]
[Tangents drawn from an external point are equal]
△OPR ≅ △OQR
[By Side  Side  Side Criterion]
∠ORP = ∠ORQ
[Corresponding parts of congruent triangles are congruent]
Also,
∠PRQ = 120°
∠ORP + ∠ORQ = 120°
∠ORP + ∠ORP = 120°
2∠ORP = 120°
∠ORP = 60°
Also, OP ⏊ PR
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, In right angled triangle OPR,
∴ OR = 2PR
OR = PR + PR
OR = PR + RQ [From 1]
Hence Proved.
Q.10. In the given figure, a circle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. If AB = 14 cm, BC = 8 cm and CA = 12 cm. Find the lengths AD, BE and CF.
Let AD = x cm, BE = y cm and CF = z cm
As we know that,
Tangents from an external point to a circle are equal,
In given Figure we have
AD = AF = x
[Tangents from point A]
BD = BE = y
[Tangents from point B]6CF = CE = z [Tangents from point C]
Now, Given: AB = 14 cm
AD + BD = 14
x + y = 14
y = 14  x … [1]
and BC = 8 cm
BE + EC = 8
y + z = 8
14  x + z = 8 … [From 1]
z = x  6 [2]
and
CA = 12 cm
AF + CF = 12
x + z = 12 [From 2]
x + x  6 = 12
2x = 18
x = 9 cm
Putting value of x in [1] and [2]
y = 14  9 = 5 cm
z = 9  6 = 3 cm
So, we have AD = 9 cm, BE = 5 cm and CF = 3 cm
Q.11. In the given figure, 0 is the center of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral.
Given : PA and PB are tangents to a circle with center O
To show : AOBP is a cyclic quadrilateral.
Proof :
OB ⏊ PB and OA ⏊ AP
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OBP = ∠OAP = 90°
∠OBP + ∠OAP = 90 + 90 = 180°
AOBP is a cyclic quadrilateral
[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic ]
Hence Proved.
Q.12. In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then find the radius of the smaller circle.
Let us consider circles C_{1} and C_{2} with common center as O. Let AB be a tangent to circle C_{1} at point P and chord in circle C_{2}. Join OB
In △OPB
OP ⏊ AB
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∴ OPB is a right  angled triangle at P,
By Pythagoras Theorem,
[i.e. (Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}]
(OB)^{2} = (OP)^{2} + (PB)^{2}
Now, 2PB = AB
[As we have proved above that OP ⏊ AB and Perpendicular drawn from center to a chord bisects the chord]
2PB = 8 cm
PB = 4 cm
(OB)^{2} = (5)^{2} + (4)^{2}
[As OP = 5 cm, radius of inner circle]
(OB)^{2} = 41
⇒ OB = √41 cm
Q.13. In the given figure, PQ is a chord of a circle with center 0 and PT is a tangent. If ∠QPT = 60°, find ∠P
Given : , PQ is a chord of a circle with center 0 and PT is a tangent and ∠QPT = 60°.
To Find : ∠PRQ
∠OPT = 90°
∠OPQ + ∠QPT = 90°
∠OPQ + 60° = 90°
∠OPQ = 30° … [1]
Also.
OP = OQ [radii of same circle]
∠OQP = ∠OPQ [Angles opposite to equal sides are equal]
From [1], ∠OQP = ∠OQP = 30°
In △OPQ , By angle sum property
∠OQP + ∠OPQ + ∠POQ = 180°
30° + 30° + ∠POQ = 180°
∠POQ = 120°
As we know, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
So, we have
2∠PRQ = reflex ∠POQ
2∠PRQ = 360°  ∠POQ
2∠PRQ = 360°  120° = 240°
∠PRQ = 120°
Q.14. In the given figure, PA and PB are two tangents to the circle with center O. If ∠APB = 60° then find the measure of ∠OAB.
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
[∵ Angles opposite to equal angles are equal] …[1]
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°
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