CBSE Class 10 Maths Chapter-10 Circles – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 10 – Circles prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-10 Circles Important Questions
CBSE Class 10 Maths Important Questions Chapter 10 – Circles
3 Mark Questions
1. Prove that the angel between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Ans. 
OPA = 
……….(i)
OCA = ……….(ii)
[Tangent at any point of a circle is 
to the radius through the point of contact]
OAPB is quadrilateral.
APB + AOB + OAP + OBP = 
[Angle sum property of a quadrilateral]
APB + AOB + 
+ =
[From eq. (i) & (ii)]
APB + AOB = 
APB and AOB are supplementary.
2. Prove that the parallelogram circumscribing a circle is a rhombus.
Ans. Given: ABCD is a parallelogram circumscribing a circle.
To Prove: ABCD is a rhombus.
Proof: Since, the tangents from an external point to a circle are equal.
AP = AS ……….(i)
BP = BQ ……….(ii)
CR = CQ ……….(iii)
DR = DS ……….(iv)
On adding eq. (i), (ii), (iii) and (iv), we get
(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)
AB + CD = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = AD + AD [Opposite sides of 
gm are equal]
2AB = 2AD
AB = AD
But AB = CD and AD = BC [Opposite sides of gm]
AB = BC = CD = AD
Parallelogram ABCD is a rhombus.
3. Two tangents TP and TQ are drawn from an external point T with centre O as shown in figure. If they are inclined to each other at an angle of 1000, then what is the value of 
Ans. 
and TQ are tangents and O is the centre of the circle
Quadrilateral OPTQ is cyclic.
4. Two concentric circles are of radii 5 cm and 3 cm,find the length of the chord of the larger circle which touches the smaller circle.
Ans. PQ is the chord of the larger circle which touches the smaller circle at the point L. Since PQ is tangent at the point L to the smaller circle with centre O.
is a chord of the bigger circle and 
bisects PQ
In 
=
Chord PQ = 2PL =8 cm
Length of chord PQ = 8 cm
5. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB+CD=AD+BC.
Ans. AP, AS are tangents from a point A (Outside the circle) to the circle.
Similarly, BP = BQ
CQ = CR
DR = DS
Now AB + CD = AP + PB + CR + RD
= AS +BQ + CQ +DS
= (AS +DS) + (BQ + CQ)
= AD + BC
AB + CD = AD + BC
6. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at point T. Find the length TP.
Ans. Join OT.
TP = PQ [tangents from T upon the circle]
And OT bisects PQ
Now 
Now 
[By AA similarity]
7. A circle is touching the side BC of 
at P and touching AB and AC produced at Q and R respectively. Prove that AQ = 
(perimeter of ).
Ans. We know that the two tangents drawn to a circle from an external point are equal.
Perimeter of 
8. If PA and PB are two tangents drawn from a point P to a circle with centreO touching it at A and B. Prove that OP is the perpendicular bisector of AB.
Ans. Let OP intersect AB at a point C, we have to prove that AC = CB and 
are two tangents from a point P to the circle with centre O
[O lies on the bisector of
]
In two 
, ACP and BCP, we have
AP = BP [tangents from P to the circle are equal]
PC = PC [Common]
[Proved]
[By SAS rule]
[CPCT]
And 
But 
Hence, OP is perpendicular bisector of AB.
9. In the given figure, PQ is tangent at point R of the circle with centre O. If 
, find 
Ans. Given PQ is tangent at point R and
[Tangent of a circle is perpendicular to Radius]
10. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Ans. Let the circle touch the sides AB, BC, CD and DA at the points P, Q, R, and S respectively.
Join OP, OQ, OR and OS.
Join OA, OB, OC and OD.
Since the two tangents drawn from an external point subtend equal angles at the centre.
But 
[Sum of all angles around a point = 360
]
And 
And 
and
11. Prove that parallelogram circumscribing a circle is a rhombus.
Ans. Given ABCD is a parallelogram in which all the sides touch a given circle
To prove: ABCD is a rhombus
Proof:
ABCD is a parallelogram
AB=DC and AD = BC
Again AP, AQ are tangents to the circle from the point A
Similarly,
Hence, parallelogram ABCD is a rhombus.
12. If two tangents are drawn to a circle from an external point then
(i) they subtend equal angles at the centre.
(ii) they are equally inclined to the segment joining the centre to that point.
Ans. Given on a circle C (O,r), two tangents AP and AQ are drawn from an external point A.
To prove:
(i) 
(ii) 
Construction: Join AO, PO and QO
Proof: In 
and 
AP = AQ [Length of the tangents drawn from an external point]
PO = QO [Radii of the same circle]
AO = AO [common]
[By SSS theorem of congruence]
(i) [CPCT]
(ii) 
[By CPCT.]
13. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that 
Ans. Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.
To Prove: 
Proof: Let 
Since TP, TQ are tangents drawn from point T to the circle.
TP = TQ
TPQ is an isosceles triangle
Since, TP is a tangent to the circle at point of contact P
Thus,
14. Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Ans. Given: P is an external point to the circle C(O,r).
PQ and PR are two tangents from P to the circle.
To Prove: PQ = PR
Construction: Join OP
Proof:
A tangent to a circle is perpendicular to the radius through the point of contact
Now in right triangles POQ and POR,
OQ = OR [Each radius r]
Hypotenuse. OP = Hypotenuse. OP [common]
[By RHS rule]
15. The circle of touches the sides BC, CA and AB at D,E and F respectively. If AB = AC, prove that BD = CD.
Ans. Tangents drawn from an external point to a circle are equal in length
AF = AE [Tangents from A] …(i)
BF = BD [Tangents from B] …(ii)
CD = CE [Tangents from C] …(iii)
Adding (i), (ii)and (iii), we get
AF + BF + CD = AE + BD + CE
But AB = AC (given)
CD = BD
16. A circle touches the side BC of a at a point P and touches AB and AC when produced at Q and R respectively, show that AQ = [Perimeter of ].
Ans. Since the tangents from an external point to a circle are equal in length,
BP = BQ …(i) [from point B]
CP = CR …(ii) [from point C]
And AQ = AR …(iii) [From point A]
From (iii), we have
AQ = AR
[Using (i) and (ii)]
Now perimeter of
AB + BC +AC = AB +(BP + PC) + AC
= (AB + BP) + (AC +PC)
= 2 (AB + BP) [using (iv)]
=2 (AB + BQ) [using (i)]
= 2 AQ
ÞAQ = (perimeter of )
17. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that
Ans. Given: A circle with centre O 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: Let
In 
we have
TP = TQ
[Length of the tangents drawn from an external point to a circle are equal]
So, TPQ is an isosceles triangle.
In we have
[Sum of three angles of a 
is 
]
But 
[Angle between the tangent and radius of a circle is 90° ]
Now 
18. Prove that the parallelogram circumscribing a circle is a rhombus
Ans. Given: ABCD be the parallelogram circumscribing a circle with centre O such that the sides AB, BC, CD and DA touch a circle at P, Q, R and S respectively.
To prove: ||gm ABCD is a rhombus.
Proof: AP = AS …(i)
BP = BQ …(ii)
CR = CQ …(iii)
DR = DS …(iv)
[Tangents drawn from an external point to a circle are equal]
Adding (i), (ii), (iii) and (iv), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
19. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with chord.
Ans. Let NM be chord of circle with centre C.
Let tangents at M.N meet at the point O.
Since OM is a tangent
is a tangent
Again in 
Thus, tangents make equal angle with the chord.
20. In the given figure, if AB = AC, prove that BE = EC.
Ans. Since tangents from an exterior point A to a circle are equal in length
Similarly, tangents from an exterior point B to a circle are equal in length
Similarly, for C
CE = CF ………….(3)
Now AB = AC
21. Find the locus of centre of circle with two intersecting lines.
Ans. Let 
be two intersection lines.
Let a circle with centre P touch the two lines 
and 
at M and N respectively.
PM = PN [Radii of same circle]
P is equidistance from the lines and
Similarly, centre of any other circle which touch the two intersecting lines will be equidistant from and
P lies on 
a bisector of the angle between and
[The locus of points equidistant from two intersecting lines is the pair of bisectors of the angle between the lines]
Hence, locus of centre of circles which touch two intersecting lines is the pair of bisectors of the angles between the two lines.
22. In the given figure, a circle is inscribed in a quadrilateral ABCD in which 
. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius of the circle.
Ans. In the given figure,
and OQ ^ BA
Also, OP = OQ =r
is a square
BP = BQ =r
But DR = DS = 5 cm …(i)
