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Page No 102:

Question 1:

Choose the correct answer from the given four options:
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is
(A) 3 cm
(B) 6 cm
(C) 9 cm
(D) 1 cm

Answer:




Let O be the centre of two concentric circles C1 and C, whose radii are r1 = 4 cm and r2 = 5 cm. Now, we draw a chord AC of circle C2 which is tangent to circle C1 at B.
Also, join OB, which is perpendicular to AC.

So OBC is a right-angled triangle
(OB)2 + (BC)2 = (OC)2                            (Pythagoras theorem)
 (4)2 + (BC)2 = (5)2                                 
16 + (BC)2 = 25
(BC)2 = 25 -16 = 9
BC = 3 cm

AB = BC                                               [Perpendicular through the center to a chord in a circle (C2 in this case) bisects the chord]
AC = AB + BC
AC = AB + AB = 2AB = 2(3) = 6 cm
Hence, the correct answer is option B.

Page No 102:

Question 2:

In Fig. 9.3, if ∠AOB = 125°, then ∠COD is equal to

(A) 62.5°
(B) 45°
(C) 35°
(D) 55°

Answer:

As in the given figure ABCD is a quadrilateral circumscribing the circle and we know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
So, we have
∠AOB + ∠COD = 180°
125° + ∠COD = 180°
∠COD = 55°
Hence, the correct answer is option D.

Page No 102:

Question 3:

In Fig. 9.4, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to

(A) 65°
(B) 60°
(C) 50°
(D) 40°

Answer:

We Have
∠CBA = 90°                                            [ As angle in a semicircle is a right angle]
By angle sum property of a triangle
∠ACB + ∠CAB + ∠CBA = 180°
50° + ∠CAB + 90° = 180°
∠CAB = 40°                                        .....(1)


OA ⏊ AT                                             [ As tangent at any point on the circle is perpendicular to the radius through point of contact]
 ∠OAT = 90°
∠OAB + ∠BAT = 90°
∠CAB + ∠BAT = 90°                   [ As ∠OAB =  ∠CAB]
40° + ∠BAT = 90°                               [ using (1) ]
∠BAT = 50°
Hence, the correct answer is option C.



Page No 103:

Question 4:

From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(A) 60 cm2
(B) 65 cm2
(C) 30 cm2
(D) 32.5 cm2

Answer:


Let us draw a circle of radius 5 cm having center O. P is a point at a distance of 13 cm from O. A pair of tangents PQ and PR are drawn.
Also, OQ = OR = radius = 5cm                    .....(1)
And OP = 13 cm

As OQ ⏊ PQ and OR ⏊PR                     [ Since tangent to at any point on the circle is perpendicular to the radius through point of contact]
△POQ and △POR are right-angled triangles.

In △PQO By Pythagoras Theorem
(PQ)2 + (OQ)= (OP)2
(PQ)2 + (5)2 = (13)2
(PQ)2 + 25 = 169
(PQ)2 = 144
PQ = 12 cm

PQ = PR = 12 cm                                          [ tangents through an external point to a circle are equal]           .....(2)
Area of quadrilateral PQRS = area of △POQ + area of △POR
=12×OQ×PQ + 12×OR×PR                     [Area of right angle triangle=12×Base×Perpendicular]
12×5×12+12×5×12
= 30 + 30
= 60 cm2
Hence, the correct answer is option A.

Page No 103:

Question 5:

At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(A) 4 cm
(B) 5 cm
(C) 6 cm
(D) 8 cm

Answer:

Let's make a diagram for the given problem in which we have a circle having AB as diameter and O as center and having radius 5cm and a tangent XAY at point A on the circle.

A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.
∠BAX + ∠AEC = 180°                                 [Sum of co-interior angles is 180°]
∠OAX =∠BAX= 90°                                    [Tangent at any point of a circle is perpendicular to the radius through the point of contact]
90 + ∠AEC = 180
∠AEC = 90°
∠OEC = 90°
So, OE ⏊ CD and △OEC is a right-angled triangle.
Now in △OEC
By Pythagoras theorem 
(OE)2 + (EC)2 = (OC)2
(3)2 + (EC)= (5)2                                  [OE = AE - AO = 8 - 5 = 3 cm]
9 + (EC)2 = 25
(EC)2 = 25 - 9 = 16
EC = 4 cm
Also, CE = ED                                       [since, perpendicular from center to the chord bisects the chord]
 CD = CE + ED = 2CE = 2(4) = 8 cm
Hence, the correct answer is option D.   
 

Page No 103:

Question 6:

In Fig. 9.5, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to


(A) 4 cm

(B) 2 cm

(C) 2 3 cm

(D) 4 3 cm

Answer:

Given, OT = 4 cm and ∠OTA = 30°
Construction : Join OA.

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
So we have
OA ⏊ AT
Therefore, ∠OAT = 90°
In △ OAT
cosθ=basehypotenuse
cos 30°=ATOT
32=AT4
AT=23 cm
Hence, the correct answer is option C.

Page No 103:

Question 7:

In Fig. 9.6, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to

(A) 100°
(B) 80°
(C) 90°
(D) 75°

Answer:

Given : OP is a radius and PR is a tangent in a circle with center O with ∠RPQ = 50°
To find : ∠POQ
Now, OP ⏊ PR                [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]
∠OPR = 90°
∠OPQ + ∠RPQ = 90°
∠OPQ + 50° = 90°
∠OPQ = 40°
In △POQ
OP = OQ                                                               [radii of same circle]
∠OQP = ∠OPQ = 40°                                         [ angles opposite to equal sides are equal]                        .....(1)
In △OPQ By angle sum property of a triangle
∠OPQ + ∠OQP + ∠POQ = 180°
40° + 40° + ∠POQ = 180°
∠POQ = 100°
Hence, the correct answer is option A.

Page No 103:

Question 8:

In Fig. 9.7, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to

(A) 25°
(B) 30°
(C) 40°
(D) 50°

Answer:

Given : A circle with center O and PA and PB are tangents to circle from a common external point P to point A and B respectively and ∠APB = 50°
To find : ∠OAB
OA ⏊ AP and OB ⏊ PB                             [As tangent to at any point on the circle is perpendicular to the radius through point of contact]
∠OBP = ∠OAP = 90°                                                     .....(1)
In Quadrilateral AOBP,
∠OBP + ∠OAP + ∠AOB + ∠APB = 360°                   [By angle sum property of quadrilateral]
90° + 90° + ∠AOB + 50° = 360°
∠AOB = 130°                                                             ..... (2)


Now in △OAB,
OA = OB                                                                        [Radii of same circle]
∠OBA = ∠ OAB                                                          .....(3)
∠OBA + ∠OAB + ∠AOB = 180°                               [By angle sum property of triangle]
∠OAB + ∠OAB + 130 = 180                                       [using (2) and (3)]
2∠OAB = 50°
∠OAB = 25°
Hence, the correct answer is option A.



Page No 104:

Question 9:

If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to

(A) 323 cm

(B) 6 cm

(C) 3 cm

(D) 33 cm

Answer:

Given : A circle with center O and PA and PC are two tangents to the circle at point A and C from an external point P such that ∠APC = 60° [i.e. angle of inclination between two tangents].
To Find : AP and PC


In △OAP and △OCP
AO = OC                                                                                                     [ radii of same circle]
OP = OP                                                                                                      [ common ]
AP = PC                                                                                                      [ tangents through an external point to a circle are equal]
△OAP ≅ △OPC                                                                                         [ By SSS ]
∠APO = ∠OPC                                                                                         [Corresponding parts of congruent triangles are equal]        .....(1)
Now, ∠APC = 60°                                                                                      [Given]
∠APO + ∠OPC = 60°
∠APO + ∠APO = 60°                                                                           [By (1)]
2∠APO = 60°
∠APO = 30°

Now, OA ⏊ AP                                                                                            [ As tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OAP = 90°
So △AOP is a right-angled triangle
And we know that,
tanθ=PerpendicularHeight
tan 30°=3AP
13=3APAP=33 cm
Hence, the correct answer is option D.

Page No 104:

Question 10:

In Fig. 9.8, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to

(A) 20°
(B) 40°
(C) 35°
(D) 45°

Answer:


Given :  PQR is a tangent to a circle Q with center O and AB is a chord parallel to PR and ∠BQR = 70°
To Find :  ∠AQB

Now, ∠OQR = ∠DQR = 90°                      [As tangent at any point on the circle is perpendicular to the radius through point of contact]
∠DQR = ∠DQB + ∠BQR
90° = ∠DQB + 70°
∠DQB = 20°                                              .....(1)

As, AB || PR
∠BDQ + ∠DQR = 180°                               [ sum of co-interior angles is supplementary ]
∠BDQ + 90° = 180°
∠BDQ = 90°

∠ADQ + ∠BDQ = 180°                              [Linear pair]
∠ADQ + 90° = 180°
∠ADQ = 90° = ∠BDQ                             .....(2)

In △AQD and △BQD
QD = QD                                                      [common]
∠ADQ = ∠BDQ                                         [ From(2) ]
AD = BD                                                      [since, perpendicular from center to the chord bisects the chord]
△AQD ≅ △BQD                                         [By Side Angle Side Criterion]
∠AQD = ∠BQD                                     [ Corresponding parts of congruent triangles are equal ]


∠AQD + ∠BQD = ∠AQB
∠AQB = 2∠BQD                                        [As ∠AQD = ∠BQD ]
∠AQB = 2(20) = 40°                                    [From (1)]

Hence, the correct answer is option B.



Page No 105:

Question 1:

Write ‘True’ or ‘False’ and justify your answer
If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is also 60°.

Answer:


Considering the figure in which we have a circle with center O and chord AB with ∠AOB = 60°


Now,
OA ⏊ AC and OB ⏊ CB                         [ As tangent  at any point on the circle is perpendicular to the radius through point of contact]
∠OBC = ∠OAC = 90°                             .....(1)


In Quadrilateral AOBC,
∠OBC + ∠OAC + ∠AOB + ∠ACB = 360°                 [ By angle sum property of quadrilateral]
90° + 90° + 60° + ∠ACB = 360°
∠ACB = 120° 
So the angle between two tangents is 120°. So the above statement is false .

Page No 105:

Question 2:

Write ‘True’ or ‘False’ and justify your answer
The length of tangent from an external point on a circle is always greater than the radius of the circle.

Answer:

Length of tangent from external point may or may not be greater than the radius of circle.
Hence, the given statement is false.

Page No 105:

Question 3:

Write ‘True’ or ‘False’ and justify your answer
The length of tangent from an external point P on a circle with centre O is always less than OP.

Answer:

Consider a figure for the problem in which we have a circle with center O.

PT is a tangent drawn from external point P. Join OT.
OT ⏊ PT                                           [ As tangent at any point on the circle is perpendicular to the radius through point of contact]
So, OPT is a right-angled triangle.

In right angled triangle, hypotenuse is always greater than any of the other two sides of the triangle.
 OP > PT or PT < OP
Hence, length of tangent from an external point P on a circle with center O is always less than OP .
 The given statement is True.

Page No 105:

Question 4:

Write ‘True’ or ‘False’ and justify your answer
The angle between two tangents to a circle may be 0°.

Answer:

True
This may be possible only when both tangent lines coincide or are parallel to each other.

Page No 105:

Question 5:

Write ‘True’ or ‘False’ and justify your answer
If angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°, then OP=a2.

Answer:

True
Let us consider a circle with center O and tangents PT and PR and angle between them is 90° and radius of circle is a.
To show: OP = 2a

Proof :
In △OTP and △ORP
TO = OR                                               [ radii of same circle]
OP = OP                                                [ common ]
TP = PR                                                [ tangents through an external point to a circle are equal]
△OTP ≅ △ORP                                   [  By SSS Criterion ]


∠TPO = ∠OPR                                   [Corresponding parts of congruent triangles are equal ]                  .....(1)

Now, ∠TPR = 90°                                [Given]
∠TPO + ∠OPR = 90°
∠TPO + ∠TPO = 90°                     [From (1)]
∠TPO = 45°

Now, OT ⏊ TP                                     [ As tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OTP = 90°
So △POT is a right-angled triangle
And we know that,

  sin45°=OTOP                       sinθ=PerpendicularHypotenuse12=aOPOP=2a   
Hence Proved.    

Page No 105:

Question 6:

Write ‘True’ or ‘False’ and justify your answer
If angle between two tangents drawn from a point P to a circle of radius a and centre O is 60°, then OP=a3.

Answer:

False

Let us consider a circle with center O and tangents PT and PR and angle between them is 90° and radius of circle is a.
In △OTP and △ORP
TO = OR                                               [ radii of same circle]
OP = OP                                                [ common ]
TP = PR                                                [ tangents through an external point to a circle are equal]
△OTP ≅ △ORP                                   [  By SSS Criterion ]


∠TPO = ∠OPR                                   [Corresponding parts of congruent triangles are equal ]                  .....(1)

Now, ∠TPR = 60°                                [Given]
∠TPO + ∠OPR = 60°
∠TPO + ∠TPO = 60°                     [From (1)]
∠TPO = 30°

Now, OT ⏊ TP                                     [ As tangent at any point on the circle is perpendicular to the radius through point of contact]
∠OTP = 90°
So △POT is a right-angled triangle
And we know that,

  sin30°=OTOP                       sinθ=PerpendicularHypotenuse12=OTOPOP=2a    
​Hence, the given statement is false.

Page No 105:

Question 7:

Write ‘True’ or ‘False’ and justify your answer
The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB = AC, is parallel to BC.

Answer:



To Prove : EF || BC
Construction : Join OA , OB and OC
Proof :
AB = AC                                                  [Given]
∠ACB = ∠ABC                                 [Angles opposite to equal sides are equal]                                         .....(1)
∠EAB = ∠ACB                                      [Alternate Segment Theorem]                                                             .....(2)
∠EAB = ∠ACB                                      [ From (1) and (2) ]
 EF || BC                                               [ two lines are parallel if their alternate interior angles are equal]
Hence, the given statement is True.



Page No 106:

Question 8:

Write ‘True’ or ‘False’ and justify your answer
If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.

Answer:


Let S1, S2, S3 .......S be n circles with centres C1, C2, C3 .......Cn respectively.
And PQ is a common tangent to all the circles at point A which is common to all circles.
As we know, tangent at any point on the circle is perpendicular to the radius at point of contact
We have,
C1APQ                                   [for S1]C2APQ                                   [for S2]C3APQ                                   [for S3]...so on.
So, C1, C2, C3 .......Cn all lie on the perpendicular line to PQ but not on perpendicular bisector as PA may or may not be equal to AQ.
Hence, the given statement is false.
 

Page No 106:

Question 9:

Write ‘True’ or ‘False’ and justify your answer
If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of PQ.

Answer:



Let S1, S2, S3 .......S be n circles with centres C1, C2, C3 .......Cn respectively, passing through P and Q.
Here PQ either will be a chord to a circle 

As, PQ is a chord then we know that perpendicular from the center to the chord of circle bisect the chord, which implies that center lies on the perpendicular bisector of the chord.

So for each circle center lies on the perpendicular bisector of PQ.
So above statement is true .

Page No 106:

Question 10:

Write ‘True’ or ‘False’ and justify your answer
AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.

Answer:




OA = OC                                                                                   [radii of same circle]
∠OCA = ∠OAC = 30°                                                        [Angles opposite to equal sides are equal]
∠ACB = 90°                                                                             [angle in a semicircle is a right angle]
∠OCA + ∠OCB = 90°
30° + ∠OCB = 90°
∠OCB = 60°                                                                          .....(1)

OC = OB                                                                                    [ radii of same circle]
∠OBC = ∠OCB = 60°                                                         [angles opposite to equal sides are equal]
Now, ∠OBC + ∠CBD = 180°                                                  [linear pair]
60° + ∠CBD = 180°
So, ∠CBD = 120°                                                                       .....(2)
Also,
OC ⏊ CD                                                                                    [Tangent at a point on the circle is perpendicular to the radius through point of contact ]
∠OCD = 90°
∠OCB + ∠BCD = 90°
60° + ∠BCD = 90°
∠BCD = 30°                                                                           .....(3)


In △BCD,
∠CBD + ∠BCD + ∠BDC = 180°                                             [Angle sum property of triangle]
120° + 30° + ∠BDC = 180°                                                       [From 2 and 3]
∠BDC = 30°                                                                               .....(4)


From (3) and (4)
∠BCD = ∠BDC = 30°
BC = BD                                                                                     [Sides opposite to equal angles are equal]
Hence Proved.
 



Page No 107:

Question 1:

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

Answer:



Let C1 and Cbe two concentric circles with center O and radius of outer circle be 5 cm.
Given : AC is a chord with length 8 cm that is tangent to inner circle.
To find : Radius of inner circle i.e. OD


OD ⏊ AC                                                    [Tangent at a point on the circle is perpendicular to the radius through point of contact ]
So, OAD is a right-angled triangle at D .
Also it implies that OD is perpendicular to the chord AC in C2
So we have,
AD = DC                                                    [perpendicular from the center to the chord bisects the chord]
AC = AD + DC
8 = AD + AD
AD = 4 cm


In △OAD By Pythagoras Theorem,
(OA)2 = (OD)2 + (AD)2
(5)2 = (OD)2+ (4)2
25 = (OD)2 + 16
(OD)= 9
OD = 3 cm
Hence, the radius of inner circle is 3 cm.

Page No 107:

Question 2:

Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

Answer:




Given : PQ and PR are two tangents drawn from an external point P .
To Prove : QORP is a cyclic Quadrilateral .
Proof :
OR ⏊ PR and OQ ⏊PQ                                         [Tangent at a point on the circle is perpendicular to the radius at point of contact ]
∠ORP = 90°
∠OQP = 90°
∠ORP + ∠OQP = 180°
Hence, QOPR is a cyclic quadrilateral. Since, the sum of the opposite pair of angle is 180°.

Page No 107:

Question 3:

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Answer:



Let PR and PQ be two intersecting lines touching the circle with center O .

To Prove : Center O lies on the angle bisector of PR and PQ i.e. OP is the angle bisector of ∠RPQ

Proof :
Clearly, PQ and PR are tangents to the circle with a common external point P.
In △POR and △POQ
OR = OQ                                                                           [radii of same circle]
OP =OP                                                                             [Common]
PR = PQ                                                                            [Tangents drawn from an external point to a circle are equal ]
△POR ≅ △POQ                                                           [ By SSS criterion ]
∠RPO = ∠OPQ                                                              [ Corresponding parts of congruent triangles are equal ]
This implies that OP is the angle bisector of ∠RPQ .
Thus, O lies on the angle bisector of PR and PQ.

Page No 107:

Question 4:

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC.

Answer:



Given : BC and BD are two tangents drawn from a common external point to a circle with center O such that ∠DBC = 120°
To Prove : BO = 2BC
Proof :
In △OBC and △ODB
OC = OD                                                                              [ Radii of same circle ]
OB = OB                                                                               [Common]
BC = BD                                                                               [Tangents drawn from an external point to a circle are equal ]  .....(1)
△OBC ≅ △OBD                                                                  [ By SSS Criterion ]
∠CBO = ∠DBO                                                              [Corresponding parts of congruent triangles are equal ]
Also,
∠DBC = 120°
∠CBO + ∠DBO = 120°
∠CBO + ∠CBO = 120°
∠CBO = ∠DBO = 60°

OC ⏊BC                                                          [Tangent at a point on the circle is perpendicular to the radius at point of contact ]
Therefore, OBC is a right-angled triangle at C
So we have,
cos 60°=BCOB12=BCOB

BO = 2BC
BO = BC + BC
BO = BC + BD
Hence Proved.
 

Page No 107:

Question 5:

In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii.
Prove that AB = CD.

Answer:

Given : AB and CD are two common tangents to two circles of unequal radii.
To Prove : AB = CD

Proof:
Consider the circle with greater radius.
AP = CP                                                              [Tangents drawn from an external point to a circle are equal] .....(1)
Also,
Consider the circle with smaller radius.
BP = BD                                                               [Tangents drawn from an external point to a circle are equal] .....(2)
Substract (2) from (1). We Get
AP - BP = CP - BD
AB = CD
Hence Proved .
 

Page No 107:

Question 6:

In Question 5 above, if radii of the two circles are equal, prove that AB = CD.

Answer:

Given: AB and CD are two common tangents to two circles of equal radii .
To Prove: AB = CD

Construction: Join OA, OC, O’B and O’D

Proof:
Now, ∠OAB = 90° and ∠OCD = 90° as OA ⏊ AB and OC ⏊ CD       
[tangent at any point of a circle is perpendicular to radius at the point of contact]
Thus, AC is a straight line.
Also,
∠O'BA = ∠O'DC = 90°                                          [Tangent at a point on the circle is perpendicular to the radius through point of contact]
Thus, BD is Also a straight line.
So ABCD is a quadrilateral with four sides as AB, BC, CD and AD
But as
∠A = ∠B = ∠C = ∠D = 90°
So, ABCD is a rectangle.
Hence, AB = CD.                                                    [opposite sides of rectangle are equal]
 

Page No 107:

Question 7:

In Fig. 9.14, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.

Answer:

Given : AB and CD are two tangents to two circles which intersect at E .
To Prove : AB = CD
Proof :
AE = CE                                               [Tangents drawn from an external point to a circle are equal]    .....(1)
EB = ED                                               [Tangents drawn from an external point to a circle are equal]   .....(2)
Adding (1) and (2), we get
AE + EB = CE + ED
AB = CD
Hence Proved.

Page No 107:

Question 8:

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

Answer:

Given : PQ is a chord in a circle with center O and MN is a tangent drawn at point R on the circle such that PQ is parallel to MN

To Prove : R bisects the arc PRQ i.e. arc PR = arc PQ
Proof :
∠1 = ∠2                                                         [Alternate Interior angles]
∠1 = ∠3                                                         [Angle between tangent and chord is equal to angle made by chord in alternate segment]
So, we have
∠2 = ∠3
QR = PR [Sides opposite to equal angles are equal]
As the equal chords cuts equal arcs in a circle.
Arc PR = arc RQ
R bisects the arc PRQ .
Hence Proved.



Page No 108:

Question 9:

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Answer:

Let QR be a chord in a circle with center O and ∠1 and ∠2 are the angles made by tangent at points R and Q with chord respectively.
To Prove : ∠1 = ∠2

Let P be another point on the circle, then, join PQ and PR.
Since, at point Q, there is a tangent.
∠RPQ = ∠2                                                   [angles in alternate segments are equal]    .....(1)
Since, at point R, there is a tangent.
∠RPQ = ∠1                                                   [angles in alternate segments are equal]    .....(2)
From (1) and (2)
∠1 = ∠2
Hence Proved .
 

Page No 108:

Question 10:

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

Answer:


Let AB be diameter in a circle with center O and MN is the tangent at point A .
And CD be any chord parallel to MN intersecting AB at E

To Prove : AB bisects CD .

Proof :
OA ⏊ MN                                        [Tangent at a point on the circle is perpendicular to the radius at point of contact]
∠MAO = 90°
Also, ∠CEO = ∠MAO     [corresponding angles ]

So, we have OE ⏊ CD
CE = ED                                         [ Perpendicular drawn from the center of a circle to chord bisects the chord ]
Implies that AB bisects CD
Hence Proved.



Page No 110:

Question 1:

If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.

Answer:


Given : A Hexagon ABCDEF circumscribe a circle .
To prove : AB + CD + EF = BC + DE + FA
Proof :
As we know, Tangents drawn from an external point to a circle are equal.
We have
AM = RA                                  [tangents from point A]                           .....(1)
BM = BN                                  [tangents from point B]                           .....(2)
CO = NC                                   [tangents from point C]                           .....(3)
OD = DP                                   [tangents from point D]                           .....(4)
EQ = PE                                    [tangents from point E]                           .....(5)
QF = FR                                    [tangents from point F]                           .....(6)

Adding (1), (2), (3), (4), (5) and (6)
AM + BM + CO + OD + EQ + QF = RA + BN + NC + DP + PE + FR
Rearranging
(AM + BM) + (CO + OD) + (EQ + QF) = (BN + NC) + (DP + PE) + (FR + RA)
AB + CD + EF = BC + DE + FA
Hence Proved.

Page No 110:

Question 2:

Let s denote the semi-perimeter 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.

Answer:

Given : 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
Proof :
Semi Perimeter = s
Perimeter = 2s

2s = AB + BC + AC                                                                                         .....(1)
As we know, Tangents drawn from an external point to a circle are equal
So we have
AF = AE                                           [Tangents from point A]                       .....(2)
BF = BD                                          [Tangents From point B]                       .....(3)
CD = CE                                         [Tangents From point C]                        .....(4)
Adding (2), (3) and (4)
AF + BF + CD = AE + BD + CE
AB + CD = AC + BD
Adding BD on both sides
AB + CD + BD = AC + BD + BD
AB + BC - AC = 2BD
AB + BC + AC - AC - AC = 2BD
2s - 2AC = 2BD                                                             [From (1)]
2BD = 2s - 2b                                                                 [as AC = b]
BD = s - b
Hence Proved.

Page No 110:

Question 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 the perimeter of the triangle PCD.

Answer:


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 = 10 cm
To Find : Perimeter of △PCD

As we know, Tangents drawn from an external point to a circle are equal.
So we have
AC = CE                                                                       [Tangents from point C]                      .....(1)
ED = DB                                                                       [Tangents from point D]                      .....(2)
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 = 10 cm                                                         [tangents drawn from an external point to a circle are equal]
So we have
Perimeter = PA + PB = 10 + 10 = 20 cm

Hence, the perimeter of △PCD is 20 cm.

Page No 110:

Question 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. 9.17. Prove that ∠BAT = ∠ACB

Answer:

Given : 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 
∠ABC + ∠ BAC + ∠ACB = 180 °                [By angle sum property of triangle]
∠ACB + 90° = 180° - ∠BAC
∠ACB = 90 - ∠BAC                                                 .....(1)
Now,
OA ⏊ AT                                       [Tangent at a point on the circle is perpendicular to the radius at point of contact ]
∠OAT = ∠CAT = 90°
∠BAC + ∠BAT = 90°
∠BAT = 90° -∠BAC                                                  ..... (2)


From (1) and (2)
∠BAT = ∠ACB
Hence Proved.

Page No 110:

Question 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.

Answer:


Given : 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 at point of contact ]
So, OPO' is a right-angled triangle at P
Using 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
x2 + (PN)2 = (3)2
(PN)2 = 9 - x2                                                  .....(1)


In right angled triangle  O'NP
(O'N)2 + (PN)2 = (O'P)2
(5 - x)2 + (PN)2 = (4)2
25 -10x + x2 + (PN)2 = 16
(PN)2 = -x2 + 10x - 9                                     .....(2)


From (1) and (2)
9 - x2 = -x2 + 10x - 9
10x = 18
x = 1.8
From (1) we have
(PN)2 = 9 - (1.8)2
(PN)2  = 9 - 3.24 = 5.76
PN = 2.4 cm
PQ = 2PN = 2(2.4) = 4.8 cm

Hence, the length of common chord PQ is 4.8 cm.

Page No 110:

Question 6:

In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.

Answer:



Given : In a right angle ΔABC 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 i.e. BQ = QC
Proof :


∠APB = 90°                                                              [Angle in a semicircle is a right-angle]
∠BPC = 90°                                                               [Linear Pair ]
∠3 + ∠4 = 90                                                .....(1)
Now, ∠ABC = 90°
So in △ABC
∠ABC + ∠BAC + ∠ACB = 180°
90 + ∠1 + ∠5 = 180
∠1 + ∠5 = 90                                                .....(2)
Now ,
∠ 1 = ∠ 3                                           [angle between tangent and the chord is equal to angle made by the chord in alternate segment]


Using this in (2) we have
∠3 + ∠5 = 90                                                    .....(3)

From (1) and (3), we have
∠3 + ∠4 = ∠3 + ∠5
∠4 = ∠5                                 
 QC = PQ                                      [Sides opposite to equal angles are equal]
But Also PQ = BQ                           [Tangents drawn from an external point to a circle are equal]
So, BQ = QC, i.e. PQ bisects BC .
Hence Proved.

Page No 110:

Question 7:

In Fig. 9.18, 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.

Answer:

Given : 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]  .....(1)
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 is equal to angle made by the chord in alternate segment]
Now In △RQS
∠RQS + ∠QRS + ∠QSR = 180
∠RQS + 75 + 75 = 180
∠RQS = 30°
 



Page No 111:

Question 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.

Answer:




∠CAB = ∠BCD =30°                           [angle between tangent and chord is equal to angle made by chord in alternate segment]   .....(1) 
Now, ∠ACB = 90°                                         [Angle in a semicircle is a right angle]
∠ACD = ∠ACB + ∠BCD = 90 + 30 = 120°
In triangle ACD, 
∠ACD + ∠CAD + ∠ADC = 180°             [By Angle Sum Property]
120 + ∠CAB + ∠BDC = 180°
120 + 30 + ∠BDC = 180
∠BDC = 30°                                                               .....(2)
From (1) and (2)
∠BCD = ∠BDC = 30°
BD = BC                                                [Sides opposite to equal angles are equal]
Hence Proved.

Page No 111:

Question 9:

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Answer:

Let us draw a circle in which AMB is an arc and M is the mid-point of the arc AMB. Join AM and MB. Also TT' is a tangent at point M on the circle.

To Prove : AB || TT'
Proof :
Arc AM = Arc MB                       [As M is the mid point of Arc AMB]
AM = MB                                 [As equal chords cut equal arcs]
∠ABM = ∠BAM                    [Angles opposite to equal sides are equal]     .....(1)
Now,
∠BMT' = ∠BAM                     [angle between tangent and the chord equals angle made by the chord in alternate segment]     .....(2)
From (1) and (2)
∠ABM = ∠BMT'
So, AB || TT'                              [two lines are parallel if the interior alternate angles are equal]
Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Page No 111:

Question 10:

In Fig. 9.19, 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.

Answer:

Given : AB and CD are tangents to two circles with centers O and O' intersect at E .
To Prove : O, E and O' are collinear.
Construction : Join AO, OC, O'D and O'B

In △AOE and △EOC
OA = OC                                            [radii of same circle]
OE = OE                                             [common]
AE = EC                                             [Tangents drawn from an external point to a circle are equal]
 △AOE ≅ △COE                                [By SSS Criterion]
∠AEO = ∠CEO                            [Corresponding parts of congruent triangles are equal ]
∠AEC = ∠AEO + ∠CEO = ∠AEO + ∠AEO = 2∠AEO                                       .....(1)
Now As CD is a straight line


∠AED + ∠AEC = 180°                                                   [linear pair]
2∠AEO = 180 - ∠AED                                             [From (1)]
∠AEO = 90 - 12∠AED                                                                                         .....(2)


Now, In △O'ED and △O'EB
O'B = O'D                                                              [radii of same circle]
O'E = O'E                                                               [common]
EB = ED                                                                 [Tangents drawn from an external point to a circle are equal]
△O'ED ≅ △O'EB                                                  [By SSS Criterion]
∠O'EB = ∠O'ED                                              [Corresponding parts of congruent triangles are equal ]


∠DEB = ∠O'EB + ∠O'ED = ∠O'ED + ∠O'ED = 2∠O'ED                            .....(3)
Now as AB is a straight line
∠AED + ∠DEB = 180                                            [Linear Pair]
2∠O'ED = 180 -∠AED                                     [From (3)]
∠O'ED = 90 -12∠AED                                                                                     .....(4)

Now,
∠AEO + ∠AED + ∠O'ED = 90 - 12∠AED + ∠AED + 90 - 12∠AED       [ from (2) and (4) ]
∠AEO + ∠AED + ∠O'ED = 180°

So O, E and O' lie on same line                                                                          [By the converse of linear pair]
Hence Proved.

Page No 111:

Question 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.

Answer:

Given : A circle with center O and radius = 5cm, OT = 13 cm.
OP ⏊ PT                                         [Tangent at a point on the circle is perpendicular to the radius at point of contact ]
In △OPT,
(OT)2 = (OP)2 + (PT)2                 [By pythagoras theorem]
(13)2 = (5)2 + (PT)2
(PT)2 = 169 - 25 = 144
PT = 12 cm
PT = TQ = 12 cm                        [Tangents drawn from an external point to a circle are equal]
Now,
OT = OE + ET
ET = OT- OE = 13 - 5 = 8 cm


Now, 
AE = PA                       [Tangents drawn from an external point to a circle are equal]      .....(1)
EB = BQ                      [Tangents drawn from an external point to a circle are equal]      .....(2) 


Also OE ⏊ AB             [Tangent at a point on the circle is perpendicular to the radius at point of contact ]
∠AEO = 90°
∠AEO + ∠AET = 180°                      [By linear Pair]
∠AET = 90°

In △AET By Pythagoras Theorem
(AT)2 = (AE)2 + (ET)2
(PT - PA)2 = (PA)2 + (ET)2             [From (1)]
(12 - PA)2 = (PA)2 + (8)2          
144 + (PA)2 - 24PA = (PA)2 + 64
24PA = 80
PA = 8024 = 103                                                                    .....(3)


∠AET + ∠BET = 180                                                        [Linear Pair]
90 + ∠BET = 180
∠BET = 90


In △BET, By Pythagoras Theorem
(BT)2 = (BE)2 + (ET)2
(TQ - BQ)2 = (BQ)2 + (ET)2                                        [from (2)]
(12 - BQ)2 = (BQ)2 + (8)2
144 + (BQ)2 - 24BQ = (BQ)2 + 64
24BQ = 80
BQ = 8024 = 103                                                             .....(3)

So,
AB = AE + BE
AB = PA + BQ                                                       [From (1) and (2)]
AB = 103103203                                           [From (3) and (4)]

Hence, the required length AB is 203 cm.



Page No 112:

Question 12:

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º, find ∠CBA [see Fig. 9.21].

Answer:


Given : ∠PCA = 110°
To Find : ∠CBA
∠ACB = 90°                                      [Angle in a semicircle is a right angle] .....(1)
Also,
∠PCA = ∠ACB + ∠PCB
110 = 90 + ∠PCB
∠PCB = 20°
Now, ∠PCB = ∠BAC                    [angle between tangent and the chord is equal to angle made by the chord in alternate segment]
∠BAC = 20°                                                                                                .....(2)

Now In △ABC,
∠CBA + ∠BAC + ∠ACB = 180       [ By angle sum property of Triangle]
∠CBA + 20 + 90 = 180
∠CBA = 70°

Page No 112:

Question 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.

Answer:

Given : In a circle, ΔABC is inscribed in which AB = AC = 6 cm and radius of circle is 9 cm
To Find : Area of triangle ABC
Construction : Join OB and OC

In △AOB and △AOC
OB = OC                                                       [Radii of same circle]
AO = AO                                                       [Common]
AB = AC = 6 cm                                           [Given]
△AOB ≅ △AOC                                          [By SSS Criterion]
∠OAB = ∠OAC                                          [Corresponding parts of congruent triangles are equal ]
∠MAB = ∠MAC

Now in △ABM and △AMC
AB = AC = 6 cm                                          [Given]
AM = AM                                                    [Common]
∠MAB = ∠MAC                                        [Proved Above]
△ABM ≅ △ACM                                        [By SAS Criterion]
∠AMB = ∠AMC                                    [Corresponding parts of congruent triangles are equal ]
Now,
∠AMB + ∠ AMC = 180°
∠AMB + ∠AMB = 180°
2 ∠AMB = 180
∠AMB = 90°
We know that a perpendicular from center of circle bisects the chord.
So, OA is perpendicular bisector of BC.
Let OM = x
Then, AM = OA - OM = 9 - x      [ As OA = radius = 9 cm]
In right angled ΔAMC, By Pythagoras theorem
(AM)2 + (MC)2 = (AC)2
(9 - x)2 + (MC)2 = (6)2
81 + x2 - 18x + (MC)2 = 36
(MC)2 = 18x - x2 -45                            ...

In △OMC , By Pythagoras Theorem
(MC)2 + (OM)2 = (OC)2
18x - x2 - 45 + (x)2 = (9)2
18x - 45 = 81
18x = 126
x = 7
AM = 9 - x = 9 - 7 = 2 cm
In △AMC, By Pythagoras Theorem
(AM)2 + (MC)2 = (AC)2
(2)2 + (MC)2 = (6)2
4 + (MC)2 = 36
(MC)2 = 32
MC = 4√2 cm
Now,
As MC = BM
BC = 2MC = 2(4√2) = 8√2 cm
Area ofABC=12×Base×Height=12×BC×AM=12×82×2=82 cm2
Hence, the required area of ABC is 82 cm2.


 

Page No 112:

Question 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.

Answer:

Given : Radius of circle, OP = 5 cm and OA = 13 cm


∠OPA = 90°                               [Tangent at any point of a circle is perpendicular to the radius at the point of contact]
(OA)2 = (OP)2 + (PA)2               [Pythagoras Theorem]
(13)2 = (5)2 + (PA)2
169 - 25 = (PA)2
(PA)2 = 144
PA = 12 cm

PB = BR                     [Tangents from point B]                                                            .....(1)
CR = QC                     [Tangents from point C]                                                            .....(2)


Now, Perimeter of Triangle ABC
= AB + BC + CA
= AB + BR + CR + CA
= AB + PB + QC + CA                                          [From (1) and (2)]
= PA + QA

Now,
PA = QA = 12 cm                                                  [tangents drawn from an external point to a circle are equal]
Therefore, Perimeter of ABC = PA + QA = 12 + 12 = 24 cm.



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