Chapter 9 Circles NCERT Exemplar Solutions Exercise 9.4 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 9 Circles Exercise 9.4 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises   Exercise 9.1
 Exercise 9.2
 Exercise 9.3

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 9.4 Solutions
Long Answer Questions
1. If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.
Solution
As given in the question,
A Hexagon ABCDEF circumscribe a circle.
To prove:
AB + CD + EF = BC + DE + FA
We know that,
Tangents drawn from an external point to a circle are equal.
So,
AM = RA ...i [tangents from point A]
BM = BN ...ii [tangents from point B]
CO = NC ...iii [tangents from point C]
OD = DP ...iv [tangents from point D]
EQ = PE ...v [tangents from point E]
QF = FR ...vi [tangents from point F]
Now, adding,
[eq i]+[eq ii]+[eq iii]+[eq iv]+[eq v]+[eq vi]
AM + BM + CO + OD + EQ + QF = RA + BN + NC + DP + PE + FR
On solving, we get,
(AM + BM) + (CO + OD) + (EQ + QF) = (BN + NC) + (DP + PE) + (FR + RA)
⇒ AB + CD + EF = BC + DE + FA
Hence Proved!
2. Let s denote the semiperimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b.
Solution
As given in the question,
A triangle ABC with BC = a , CA = b and AB = c . Also, a circle is inscribed which touches the sides BC, CA and AB at D, E and F respectively and s is semi perimeter of the triangle
To Prove:
BD = s – b
We have,
Semi Perimeter = s
Perimeter = 2s
2s = AB + BC + AC ...(i)
We know that,
Tangents drawn from an external point to a circle are equal
So,
AF = AE ...[ii] [Tangents from point A]
BF = BD ...[iii] [Tangents From point B]
CD = CE ...[iv] [Tangents From point C]
Adding [ii] [iii] and [iv],
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
Adding BD both side,
AB + CD + BD = AC + BD + BD
⇒ AB + BC – AC = 2BD
⇒ AB + BC + AC – AC – AC = 2BD
⇒ 2s – 2AC = 2BD [From i]
⇒ 2BD = 2s – 2b [as AC = b]
⇒ BD = s – b
Hence Proved.
3. From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD.
Solution
As given in the question,
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 = 10 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 ...[i] [Tangents from point C]
ED = DB ...[ii] [Tangents from point D]
Now,
Perimeter of Triangle PCD = PC + CD + DP
= PC + CE + ED + DP
= PC + AC + DB + DP ...[From i and ii]
= PA + PB
Now,
PA = PB = 10 cm as tangents drawn from an external point to a circle are equal
So ,
Perimeter = PA + PB
= 10 + 10
= 20 cm
4. If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Fig.. Prove that ∠BAT = ∠ACB.
Solution
As given in the question,
A circle with center O and AC as a diameter and AB and BC as two chords also AT is a tangent at point A
To Prove : ∠BAT = ∠ACB
Proof :
∠ABC = 90° [Angle in a semicircle is a right angle]
In △ABC By angle sum property of triangle
∠ABC + ∠ BAC + ∠ACB = 180 °
⇒ ∠ACB + 90° = 180° – ∠BAC
⇒ ∠ACB = 90 – ∠BAC ...[i]
Now,
OA ⏊ AT
[Tangent at a point on the circle is perpendicular to the radius through point of contact]
∠OAT = ∠CAT = 90°
⇒ ∠BAC + ∠BAT = 90°
⇒ ∠BAT = 90° – ∠BAC [ii]
From [i] and [ii],
∠BAT = ∠ACB [Proved]
5. Two circles with centres O and O‘ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O‘P are tangents to the two circles. Find the length of the common chord PQ.
Solution
We have,
Two circles with centers O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’P are tangents to the two circles and PQ is a common chord.
To Find: Length of common chord PQ
∠OPO’ = 90°
[Tangent at a point on the circle is perpendicular to the radius through point of contact]
Therefore,
OPO is a rightangled triangle at P
By Pythagoras in △ OPO’, we have
(OO’)^{2} = (O’P)^{2} + (OP)^{2}
⇒ (OO’)^{2} = (4)^{2} + (3)^{2}
⇒ (OO’)^{2} = 25
⇒ OO’ = 5 cm
Let ON = x cm and
NO’ = 5 – x cm
In right angled triangle ONP
(ON)^{2} + (PN)^{2} = (OP)^{2}
⇒ x^{2} + (PN)^{2} = (3)^{2}
⇒ (PN)^{2} = 9 – x^{2} ...[i]
In right angled triangle O’NP
(O’N)^{2} + (PN)^{2} = (O’P)^{2}
⇒ (5 – x)^{2} + (PN)^{2} = (4)^{2}
⇒ 25 – 10x + x^{2} + (PN)^{2} = 16
⇒ (PN)^{2} = x^{2} + 10x – 9 ...[ii]
From [i] and [ii]
9 – x^{2} = x^{2} + 10x – 9
⇒ 10x = 18 x = 1.8
From (1) we have
(PN)^{2} = 9 – (1.8)^{2}
= 9 – 3.24
= 5.76
PN = 2.4 cm
PQ = 2PN
= 2(2.4)
= 4.8 cm
6. In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.Solution
As given in the question,
In a right angle Î”ABC is which ∠B = 90°, a circle is drawn with AB as diameter intersecting
the hypotenuse AC at P. Also PQ is a tangent at P
To Prove:
PQ bisects BC or, BQ = QC
We have,
∠APB = 90° [Angle in a semicircle is a rightangle]
∠BPC = 90° [Linear Pair]
∠3 + ∠4 = 90 ...[i]
Now,
∠ABC = 90°
In △ABC,
∠ABC + ∠BAC + ∠ACB = 180°
⇒ 90 + ∠1 + ∠5 = 180
⇒ ∠1 + ∠5 = 90 ...[ii]
Now,
∠ 1 = ∠ 3
[angle between tangent and the chord equals angle made by the chord in alternate segment]
Using this in [ii] we have,
∠3 + ∠5 = 90 ...[iii]
From [i] and [iii] we have
∠3 + ∠4 = ∠3 + ∠5
⇒ ∠4 = ∠5
⇒ QC = PQ [Sides opposite to equal angles are equal]
Also
PQ = BQ [Tangents drawn from an external point to a circle are equal]
So,
BQ = QC
Therefore, PQ bisects BC
7. In Fig., tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find the ∠RQS.
[Hint: Draw a line through Q and perpendicular to QP.]
Solution
As given in the question,
Tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn
parallel to the tangent PQ.
To Find :
∠RQS
PQ = PR [Tangents drawn from an external point to a circle are equal]
∠PRQ = ∠PQR [Angles opposite to equal sides are equal] ..[i]
In △PQR
∠PRQ + ∠PQR + ∠QPR = 180°
⇒ ∠PQR + ∠PQR + ∠QPR = 180° [Using 1]
⇒ 2∠PQR + ∠RPQ = 180°
⇒ 2∠PQR + 30 = 180
⇒ 2∠PQR = 150
⇒ ∠PQR = 75°
∠QRS = ∠PQR = 75° [Alternate interior angles]
⇒ ∠QSR = ∠PQR = 75°
[angle between tangent and the chord equals angle made by the chord in alternate segment]
Now
In △RQS
∠RQS + ∠QRS + ∠QSR = 180
⇒ ∠RQS + 75 + 75 = 180
⇒ ∠RQS = 30°
8. AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC 30° . The tangent at C intersects extended AB at a point D.
Prove that BC = BD.
Solution
Given,
AB is a diameter and AC is a chord of circle with centre O, ∠BAC = 30°
To prove : BC = BD
Construction: Join BC
∠BCD = ∠CAB [Angles in alternate segment]
∠CAB = 30° [Given]
∠BCD = 30° ...(i)
∠ACB = 90° [Angle in semicircle]
In ∆ABC,
∠A + ∠B + ∠C = 180° [Angle sum property]
⇒ 30° + ∠CBA + 90° = 180°
⇒ ∠CBA = 60°
Also,
∠CBA + ∠CBD = 180° [Linear pair]
⇒ ∠CBD = 180° – 60° = 120°
[as, ∠CBA = 60°]
Now,
In ACBD,
∠CBD + ∠BDC + ∠DCB = 180°
⇒ 120° + ∠BDC + 30° = 180°
⇒ ∠BDC = 30° ...(ii)
From (i) and (ii),
∠BCD = ∠BDC
BC = BD
[Sides opposite to equal angles are equal]
9. Prove that the tangent drawn at the midpoint of an arc of a circle is parallel to the chord joining the end points of the arc.
Solution
Let us take the midpoint of an arc AMB be M and TMT’ be the tangent to the circle.
Join AB, AM and MB.
Since,
arc AM = arc MB =3
Chord AM = Chord MB
In ∆AMB,
AM = MB
∠MAB = ∠MBA ...(i)
[Sides opposite to equal angles are equal]
Since, TMT’ is a tangent line.
Therefore,
∠AMT = ∠MBA [Angles in alternate segments are equal]
= ∠MAB [from equation (i)]
But, ∠AMT and ∠MAB are alternate angles, which is possible only when AB is parallel to
TMT
Hence, the tangent drawn at the midpoint of an arc of a circle is parallel to the chord joining
the end points of the arc.
10. In Fig., the common tangent, AB and CD to two circles with centres O and O' intersect at E. Prove that the points O, E, O' are collinear.
Solution
Join AO, OC and O’D, O’B.
Now,
In ∆EO’D and ∆EO’B,
O’D = O’B
O’E = O’E
ED = EB
[Tangents drawn from an external point to the circle are equal in length]
EO’D ≅ ∆ EO’B [By SSS congruence criterion]
∠O’ED = ∠O’EB ...(i)
Therefore,
O’E is the angle bisector of ∠DEB.
Similarly,
OE is the angle bisector of ∠AEC.
Now, in quadrilateral DEBO’.
∠O’DE = ∠O’BE = 90°
[CED is a tangent to the circle and O’D is the radius, i.e., O’D ⊥ CED]
∠O’DE + ∠O’BE = 180°
∠DEB + ∠DO’B = 180°
[as, DEBO’ is cyclic quadrilateral] ...(ii)
Since,
AB is a straight line.
∠AED + ∠DEB = 180°
⇒ ∠AED + 180° – ∠DO’B = 180° [from (ii)]
⇒ ∠AED = ∠DO’B ...(iii)
Similarly,
∠AED = ∠AOC ...(iv)
Again from eq. (ii),
∠DEB = 180° – ∠DO’B
Dividing by 2 on both sides, we get
(1/2) ∠DEB = 90°  (1/2)∠DO'B
⇒ ∠DEO' = 90°  (1/2)∠DO'B
...(v)Similarly,
∠AEC = 180°  AOC
Dividing 2on both sides,
= ∠AED + 180° – ∠AED
= 180°
So,
∠AED + ∠AEO + ∠DEO = 180°
So,
OEO' is straight line.
Hence, O, E and O' are collinear.
11. In Fig. 9.20. O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.
Solution
OP is perpendicular to PT.
In ∆OPT,
OT^{2} = OP^{2} + PT^{2}
⇒ PT^{2} = OT^{2} – OP^{2}
⇒ PT^{2} = (13)^{2} – (5)^{2}
= 169 – 25
= 144
⇒ PT= 12 cm
Since, the length of pair of tangents from an external point T is equal.
So,
QT= 12 cm
Now,
TA = PT – PA
⇒ TA = 12 PA ...(i)
and
TB = QT – QB
⇒ TB = 12 – QB ...(ii)
Also,
PA = AE and QB = EB ...(iii) [Pair of tangents]
ET = OT – OE [as, OE = 5 cm = radius]
⇒ ET = 13 – 5
⇒ ET = 8 cm
Since, AB is a tangent and OE is the radius.
OE ⊥ AB,
∠OEA = 90°
∠AET = 180° – ∠OEA [Linear pair]
⇒ ∠AET = 90°
Now, in right angled ∆AET,
(AT)^{2} = (AE)^{2} + (ET)^{2} [by Pythagoras theorem]
⇒ (12 – PA)^{2} = (PA)^{2} + (8)^{2}
On solving,
144 + (PA)^{2} – 24 PA = (PA)^{2} + 64
⇒ 24 PA = 80
⇒ PA = 10/3
So,
AE = 10/3
We join OQ,
Similarly,
BE = 10/3
Also,
AB = AE + BE
= 10/3 + 10/3
= 20/3
12. The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, find CBA.
[Hint: Join C with centre O.]
Solution
Join OC. In this, OC is the radius .
We know that, tangent at any point of a circle is – perpendicular to the radius through the
point of contact.
Therefore,
OC ⊥ PC
Now,
∠PCA = 110° [Given]
∠PCO + ∠OCA = 110°
⇒ 90° + ∠OCA = 110°
⇒ ∠OCA = 20°
Also,
OC = OA = radius of circle
∠OCA = ∠OAC = 20° [Sides opposite to equal angles are equal]
Since, PC is a tangent,
∠BCP = ∠CAB = 20° [Angles in alternate segment]
In ∆PAC,
∠P + ∠C + ∠A = 180°
So,
∠P = 180° – (∠C + ∠A)
= 180°(110°+ 20°)
= 180° – 130° = 50°
In ∆PBC,
∠BPC + ∠PCB + ∠CBP = 180°
⇒ 50° + 20° + ∠PBC = 180°
⇒ ∠PBC = 180° – 70°
⇒ ∠PBC = 110°
Since, ABP is a straight line.
Therefore,
∠PBC + ∠CBA = 180°
⇒ ∠CBA = 180° – 110° = 70°
13. If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.
Solution
Join OB, OC and OA.
In ∆ABO and ∆ACO,
AB = AC [Given]
BO = CO [Radii of same circle]
AO = AO [Common side]
∆ABO ≅ ∆ACO [By SSS congruence criterion]
∠1 = ∠2
[CPCT]Now,
In ∆ABM and ∆ACM,
AB = AC [Given]
∠1 = ∠2 [proved above]
AM = AM [Common side]
So,
∆AMB ≅ ∆AMC
[By SAS congruence criterion]∠AMB = ∠AMC
[CPCT]Also,
∠AMB + ∠AMC = 180°
[Linear pair] ⇒ ∠AMB = 90°
We know that a perpendicular from the centre of circle bisects the chord.
So, OA is a perpendicular bisector of BC.
Let AM = x, then OM = 9  x
[as, OA = radius = 9cm]In right angled ∆AMC,
AC^{2} = AM^{2} + MC^{2} [By Pythagoras theorem]
⇒ MC^{2} = 6 – x^{2} ...(i)
In right angle ∆OMC,
OC^{2} = OM^{2} + MC^{2} [By Pythagoras theorem]
⇒ MC^{2} = 9^{2} – (9 – x)^{2}
From equation (i) and (ii),
6^{2} – x^{2} = 9^{2} – (9 – x)^{2}
⇒ 36 – x^{2} = 81 – (81 + x^{2} – 18x)
⇒ 36 = 18x
x = 2
So,
AM = 2 cm
From equation (ii),
MC^{2} = 9^{2} – (9 – 2)^{2}
⇒ MC^{2} = 81 – 49 = 32
⇒ MC = 4√2 cm
So,
BC = 2MC
= 8√2 cm
Area of ABC = (1/2) × base × height
= (1/2) × BC × AM
= (1/2) × 8√2 × 2
= 8√2 cm^{2}
The required area of Î”ABC is 8√2 cm^{2} .
14. A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC .
Solution
We have,
∠OPA = 90°
[Tangent at any point of a circle is perpendicular to the radius through the point of contact]
In ∆OAP,
OA^{2} = OP^{2} + PA^{2}
⇒ 13^{2} = 5^{2} + PA^{2}
⇒ PA = 12 cm
Now,
Perimeter of ∆ABC = AB + BC + CA
= AB + BR + RC + CA
= (AB + BR) + (RC + CA)
= (AB + BP) + (CQ + CA)
[As, BR = BP, RC = CQ i.e., tangents from external point to a circle are equal]
Perimeter of ∆ABC = AP + AQ
= 2AP [as, AP = AQ]
= 2 × 12
= 24 cm
Hence, the perimeter of ∆ABC = 24 cm.