Chapter 10 : Circles– Chapter-Wise Important Questions and Answers | Class 10 Mathematics CBSE Solutions
CBSE Class 10 Mathematics Chapter 10 : Circles Important Questions with Solutions and Answers
Chapter 10 : CIRCLES
SECTION = A
Question: In the circle given in figure, the number of tangents parallel to tangent PQ is : [CBSE 2020 basic]
(a) 0 (b) 1 (c) 2 (d) many
Solution: (a) 1
Question: In figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O . If 
, then length of PQ is : [CBSE 2020 standard]
(a) 3 cm (b) 4 cm (c) 2 cm (d) 
cm
Solution: (b) 4 cm
[ Since OP is the angle bisector of ∠QPR .
In ∆OPQ we have , 
So, OPQ is an isoscele triangle .
cm ]
Question: In figure, AB is a chord of the circle and AOC is its diameter such that 
. If AT is the tangent to the circle at the point A , then 
is equal to :
(a) 65° (b) 60° (c) 50° (d) 40°
Solution: (c) 50°
[ Since AC perpendicular to AT .
In 
we have, 
But , 
Question: From an external point P, tangents PA and PB are drawn to a circle with centre O . If 
, then 
is :
(a) 30° (b) 35° (c) 45° (d) 25°
Solution: (d) 25° .
[ Since APB is an isocele triangle .
So, 
[ 
]
But 
]
Question: In figure, 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) 75° (d) 90°
Solution: (a) 100°
[ Since OP is perpendicular to PR .
POQ is an isosceles triangle .
]
Question: In a right triangle ABC , right-angled at B , 
. The radius of the circle inscribed in the triangle (in cm) is :
(a) 4 (b) 3 (c) 2 (d) 1
Solution: (b) 3
[ Here, 
In 
, we have
A/Q, 
Question: In figure, O is the centre of a circle . PT and PQ are tangents to the circle from an external point P .If 
, then the measure of 
is :
(a) 65° (b) 50° (c) 55° (d) 45°
Solution: (c) 55°
[ We join OT and OQ .
Here, 
Since, OTPQ is quadrilateral .
and 
]
Question: In figure, PQ is a chord of a circle and PT is the tangent at P such that 
.Then ∠PRQ is equal to :
(A) 135° (B) 150° (C) 120° (D) 110°
Solution: (c) 120°
[ Here, ∠OPT=90° , ∠QPT=60° 
Since, POQ is an isosceles triangle .
So, 
In 
, we have
Question: In figure, PQ is a tangent at a point C to a circle with centre O . If AB is a diameter and 
, then 
is : [ CBSE 2016 ]
(a) 30° (b) 50° (c) 60° (d) 45°
Solution: (c) 60°
[ We join OC .
So, OA = OC
AOC is an isosceles triangle .
But, 
]
Question: In figure, PQ is tangent to the circle with centre at O , at the point B .If 
, then 
is equal to : [ CBSE 2020]
(a) 50° (b) 60° (c) 40° (d) 80°
Solution: (c) 40°
[ Since AOB is an isosceles triangle.
In 
, we have 
But 
]
Question: In figure , AOB is a diameter of a circle with centre O and AC is a tangent to the circle at A . If 
, then 
is : [ 2016]
(a) 50° (b) 60° (c) 40° (d) 70°
Solution: (c) 40°
[ Since, AOB is straight angle .
In 
we have , 
]
Question: In figure, O is the centre of the circle , PQ is a chord and PT is tangent to the circle at P . If 
, then 
. [2017]
(a) 40° (b) 45° (c) 35° (d) 65°
Solution: (d) 35°
[ Since OPQ is an isosceles triangle .
But , 
]
Question: In figure, BOA is a diameter of a circle and the tangent at a point P meets BA extended at T . If 
, then ∠PTA is equal to
(a) 35° (b) 25° (c) 30° (d) 55°
Solution: (c) 30°
[ Here, . We join OP .
So, OP = OB [Radius]
POB is an isosceles triangle . So, 
In 
, we have
Since, POA is an isosceles triangle .
In 
, we have 
and 
]
Question: In figure, If PQR is the tangent to a circle at Q whose centre is O , AB is a chord parallel to PR and 
, then 
is equal to : [Examplar]
(a) 20° (b) 40° (c) 35° (d) 45°
Solution: (b) 40°
[ Here,
(Alternative interior angles )
]
Question: In figure, O is the centre of the circle and LN is a diameter . If PQ is a tangent to the circle at K and 
, then 
is : [2017]
(a) 30° (b) 50° (c) 70° (d) 60°
Solution: (d) 60° .
[ Here, . We join OK .
OK = OL (Radius)
OKL is an isosceles triangle .
So,
SECTION = B
Question: The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm .Find the radius of the circle .
Solution: In given figure:
Here, OP = 5 cm and AP = 4 cm
In ∆OAP , We have
cm
Therefore, the radius of the circle is 3 cm .
Question: Two concentric circles are of radii 5cm and 3cm .Find the length of the chord of the larger circle which touches the smaller circle.
Solution: In given figure:
Here, OM = 3 cm and OA = 5 cm .
In ∆OMA, we have 
Since, 
Therefore, the length of the chord is 8 cm .
Question: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the nagle subtended by the line –segment joining the points of contact at the centre.
Solution: Given, AP and BP be two tangents drawn from an external point P to a circle with centre O .
To prove : 
Proof : Since, the tangent to a circle is perpendicular to the radius through the point of contact .
Therefore , 
and 
.
So, 
and 
.
OAPB is cyclic quadrilateral , we have
proved .
Question: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution: Let AB and CD are two tangent touch at X and Y of the diameter XY of the circle with centre O .
To prove : 
.
Proof : Since, the tangent to a circle is perpendicular to the radius through the point of contact .
and 
So, and XY is a transversal .
Proved .
SECTION = C
Question: A quadrilateral ABCD is drawn to circumscribe a circle a circle (see Fig. 10.12). Prove that 
Solution: Let, 
be a quadrilateral and its sides are AB, BC , CD and DA in which the point P, Q , R and S are touch of a circle with centre O respectively .
To Prove : 
.
Proof : Since the length of the two tangents from an external point to a circle are equal .
So , 
(From A)…….(i)
(From B)…….(ii)
(From C)…….(iii)
(From D )…….(iv)
Proved.
Question: In figure , a circle is inscribed in a triangle PQR with PQ = 10 cm , QR = 8 cm and PR = 12 cm . Find the lengths QM , RN and PL .
Solution: Given, PQ = 10 cm , QR = 8 cm and PR = 12 cm
Since the lengths of tangents drawn from an external point to a circle are equal .
, 
and 
Let, 
, 
and 
From 
and 
, we get 
cm
From and 
, we get 
cm
From and 
, we get 
cm
cm , 
cm and 
cm.
Question: A circle is touching the side BC of a 
at P and touching the sides AB and AC when produced at Q and R respectively . Prove that 
.
Solution: Since the lengths of the two tangents from an external point to a circle are equal .
BQ = BP , CP = CR and AQ = AR
Now , 2 AQ = AQ + AQ
2 AQ = AQ + AR
2 AQ = (AB + BQ) + (AC + CR)
2 AQ = AB + AC + (BQ + CR)
2 AQ = AB + AC + (BP + CP)
2 AQ = AB + AC + BC
Question: Two tangents TP and TQ are drawn to a circle with centre O from an external point T . Prove that 
.
Solution: Given, O be a centre of a circle and an external point T and two tangents TP and TQ to the circle , where P and Q are the points of contact .
To prove :
Proof : Since the length of tangents drawn from an external point to a circle are equal .
i.e. , 
In ∆PTQ , we have
and 
So , 
[ 
OP⊥PT ]
In 
we have ,
[ From (i) ]
Proved.
Question: Prove that the lengths of tangents drawn from an external point to a circle are equal .
Solution: Given, a circle with centre , a point P lying outside the circle and two tangents PQ , PR on the circle from P .
To prove : 
Construction : We join 
,
and 
.
Proof : In 
and 
we have ,
( Radius of the same circle)
( Common)
[ OA⊥AP and OB⊥BP]
[ R.H.S rule]
(C.P.C.P.) Proved.
Question: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact .
Solution: Given, a circle with centre O and a tangents XY to the circle at a point P .
To Prove : OP is perpendicular to XY .
Construction : we draw a point Q on XY other than P and join OQ .
Proof : The point Q must lie outside the circle .
Therefore , OQ is longer than the radius OP of the circle .
i.e. , OQ > OP
Every point on the line XY except the point P , OP is the shorter of all the distance of the point O to the points of XY . So, OP is perpendicular to XY .
SECTION = D
Question: Prove that the parallelogram circumscribing a circle is a rhombus.
Solution: Let, be a quadrilateral and its sides are AB, BC , CD and DA in which the point P, Q , R and S are touch of a circle with centre O respectively .
To Prove : .
Proof : Since the length of the two tangents from an external point to a circle are equal .
So , (From A)…….(i)
(From B)…….(ii)
(From C)…….(iii)
(From D )…….(iv)
Since, be a parallelogram . So, 
and 
So , 
Therefore, is a rhombus.
Question: In given figure , 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: Here , 
and OT=13 cm
In 
we have, 
Since, the lengths of the two tangents from an external point to a circle are equal .
Thus , 
cm
. So , 
, 
and 
Let , 
,
and 
In 
We have , 
Therefore , 
Question: From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle .Prove that OT is the right bisector of the line segment PQ .
Solution: Given , O be a centre of a circle whose two tangents are TP and TQ .
To Prove : OT is the right bisector of the line segment PQ .
Proof : In 
, we have
OP = OQ [ Radius of the circle]
∠OPT = ∠OPQ = 90° [ 
]
OT = OT [ common side]
[ R.H.S]
[ C.P.C.T]
In 
, we have
OP =OQ [ Radius of the circle]
[ Given ]
[ common side]
[ S.A.S]
[C.P.C.T ]
[ linear pair of angle ]
OT is the right bisector of the line segment PQ . Proved .
Question: In figure , a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T , are of lengths 12 cm and 9 cm respectively .If the area of 
, then find the lengths of sides PQ and PR .
Solution: Since, the lengths of the two tangents from an external point to a circle are equal.
Let, 
cm , 
and 
Therefore, 
; 
and 
So , 
A/Q , 
Thus, 
and
Question: In Fig. 10.13, XY and 
are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B .Prove that 
.
Solution: Given , XY and 
are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting 
at A and at B .
To Prove
Construction : Join 
and
.
Proof : Since the lengths of the two tangents from an external point to a circle are equal and also the subtend equal angles at the centre .
Therefore , OA is a bisector of 
And 
is a bisect of 
Since , 
and AB is a transversal
[From 
and 
] 
∆AOB we have,
[ From 
]
Proved.
Question: PQ is a chord of length 8 cm of a circle of radius 5 cm . The tangents at P and Q intersect at a point T . Find the length TP.
Solution: Join 
and let OT and PQ intersect at the point R . is the bisector of ∠PTQ .
So, 
then, 
; 
cm
In 
we have , 
Let , 
and
. So, 
In 
we have , 
In we have , 
[ From 
]
Putting 
in we have , 
Therefore, 
.
Question: A triangle ABC is drawn circumscribe a circle of radius 4cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
Solution: Since, the lengths of the two tangents from an external point to a circle are equal.
cm ; 
cm
Let, 
cm
Therefore, 
, 
, 
; 
; 
So, 
A/Q, 
(impossible) or 
Thus, 
cm and 
cm
Question: Prove that the opposite sides of a quadrilateral circumscribing a circle subtended supplementary angles at the centre of the circle.
Solution: Given , ABCD is a quadrilateral circumscribing a circle with centre O and touches the quadrilateral at P , Q , R and S respectively .
To Prove : (i) 
(ii) 
Construction : Join OP , OQ , OR and OS respectively .
Proof : Since the lengths of the two tangents from an external point to a circle are equal .
i.e., 
, 
, 
and 
.
In 
and 
, we have
(Given)
[ Common side]
[ R.H.S rule]
[ C.P.C.T] 
Similarly , ∠BOP=∠BOQ , 
, 
Let, 
, 
, 
, 
, 
, 
, 
and 
We know that the sum of the all angles of subtended at a point is 360° .
Similarly , Proved .
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