09243500460  09243500460

Chapter 10: Circles

Short Answer Questions

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

Sol.    Let the chord FC of the larger circle touch the smaller circle at the point L.

Since FC is tangent at the point L to the smaller circle with the centre O.


∴  OL\(\perp\)FC

Since AC is chord of the bigger circle and OL\(\perp\)FC.

∴ OL bisects FC

∴           FC=2FL

\(\Rightarrow\)            8=2FL

\(\Rightarrow\)          FL=4cm

Now, consider right-angled  \(\Delta\)FLO,we obtain

\(OL^{2}\)= \(FO^{2}\)– \(FL^{2}\)

= \(5^{2}\)–\(4^{2}\)



OL =\(\sqrt{9}\)=3

Hence, the radius of the smaller or inner circle is 3cm


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

Sol.  Let \(l_{1}\) and \(l_{2}\),two intersecting lines, intersect at A, be the tangents from an external point A to a circle with centre X, at B and F respectively.


Join XB and XF

Now, in \(\Delta\)ABX and \(\Delta\)ARX, we have

AX=AX                        (common)

XB=XF                       (radii of same circle)


(tangents from an external point)

∴     \(\Delta\)ABX=\(\Delta\)AFX

(by SSS congruence rule)

\(\Rightarrow\)    \(\angle\)XAB=\(\angle\)XAF

\(\Rightarrow\)   X lies on the bisector of the lines \(l_{1}\) and \(l_{2}\).


3.In the given figure,LM and NP are common tangents to two circles of unequal radii.Prove that LM=NP.


Sol.  Let Chords LM and NP meet at the point R


Since RL and RN are tangents from an external point R to two Circles with centres O and \({O}’\).

∴      RL=RN    …..(i)

And                                            RM=RP     …..(ii)

Subtracting (ii) from (i),we have


∴        LM=NP


4.In the given figure, common tangents PQ and RS to two circles intersect at T. Prove that PQ=RS.


Sol. Clearly, TP and TR are two tangents from an external point T to the circle with centre O.

∴      TP=TR   ……(i)

Also, TQ and TS are two tangents from an external point T to the circle with centre \({O}’\).

∴      TQ=TS   …..(ii)

Adding (i) and (ii),we obtain


\(\Rightarrow\)      PQ=RS


5.A chord XY of a circle is parallel to the tangent drawn at a point Z of the circle. Prove that Z bisects the arc XZY.


Sol. Since XY is parallel to the tangent drawn at the point Z and radius OZ is perpendicular to the tangent.

∴                      OR\(\perp\)XY

∴    OL bisects the chord XY.

∴            XL-LY

∴           arc XZ=arc ZY

i.e., Z bisects arc XZY


Long answer questions

1.If from an external point A of a circle with centre X,two tangents AC and AD are drawn such that \(\angle\)DAC=120° , prove that AC+AD=AX i.e.,AX=2AC.


Sol. Join XC and XD.

since,                XCAC


∴ [tangent to any circle is perpendicular to its radius at point of contact]


∴                \(\angle\)XCA=90°

In \(\Delta XCB\) and \(\Delta XDB\),we have


[tangents from an external point]

XA=XA                       [common side]

XC=XD                     [radii of a circle]

∴   By SSS  congruency , we have

\(\Delta XCA\)\(\cong\)\(\Delta XDA\)

\(\Rightarrow \cong\)   \(\angle\)XAC=\(\angle\)XAD


In  \(\Delta XCA\) ,\(\angle\)XCA=90°

∴  \(\frac{AC}{AX}=\cos 60^{\circ}\)


\(\Rightarrow                \frac{BC}{BX}=\frac{1}{2}\)

\(\Rightarrow\)       AX=2AC

Also, AX=AC + AC

\(\Rightarrow\)    AX=AC + AD     [Since,  AC=AD]


2.If x,y,z are the sides of a right triangle where c is hypotenuse,prove that the radius r of the circle which touches the sides of the triangle is given by


                                            r  = \(\frac{x+y-z}{2}\)


Sol.   Let the circle touches the sides YZ,ZX,XY of the right triangle XYZ at D,E,F respectively,where YZ=x,ZX=y and XY=z.Then XE = XF and YD = YF.


Also,           ZE=ZD=r

I.e.,               y-r=XF,


Or                 XY=z=XF+YF

\(\Rightarrow\)    Z=y-r+x-r

\(\Rightarrow\)    r  = \(\frac{x+y-z}{2}\)


3.In the given figure,from an external poiny P,a tangent PT and a line segment PXY is drawn to a circle with the centre O.ON is perpendicular on the chord XY. Prove that:

                            (i)PX.PY= \(PY^{2}\)–\( XN^{2}\)

                            (ii) \(PN^{2}- XN^{2}\)= \(OP^{2}- OT^{2}\)

                            (iii) \(PX.PY= PT^{2}\)


Sol.                   (i) PX.PY=(PN-XN)(PN+YN)


since, XN=YN,as ON\(\perp\)Chord XY]

= \(PN^{2}- XN^{2}\)

(ii) \(PN^{2}- XN^{2}\)=(\(OP^{2}\)\(ON^{2}\))- \(XN^{2}\)

Since,in rt. \(\triangle\)ONP \(OP^{2}\)= \(ON^{2}\)+ \(PN^{2}\)]

=\(OP^{2}\)– \(ON^{2}\)– \(XN^{2}\)

=\(OP^{2}\)-(\(ON^{2}\))+ \(XN^{2}\))

=\(OP^{2}\)– \(OX^{2}\)

=\(OP^{2}\)– \(OT^{2}\)


(iii)From (i) and(ii),we have

PX.PY= \(OP^{2}\)– \(OT^{2}\)

\(PX.PY= PT^{2}\)

∴ In rt. \(\triangle\)OTP; \(OP^{2}\)= \(OT^{2}\)+ \(PT^{2}\)


4.XY is a diameter and XZ is a chord with centre O such that \(\angle\)YXZ= \(30^{\circ}\).The tangent at Z intersects XY extended at a point D. Prove that YZ=YD.

Sol. Here, XOY is a diameter of the circle, such that \(\angle\)YXZ=\(30^{\circ}\) and ZD be the tangent at Z.


\(\angle\)XZY=\(90^{\circ}\)\(\angle\)in a semi-circle]

Also,          \(\angle\)ZYD=\(\angle\)YXZ+\(\angle\)XZY

[ext. \(\angle\)of a \(\triangle\) is equal to sum of interior opp.angles]

=\(30^{\circ}\)+ \(90^{\circ}\)



In \(\triangle\),by angles sum property,we have





\Rightarrow                 \(\angle\)YDZ=\(\angle\)YDZ


Hence,                           YD=YZ

[sides opp.to equal angles]


5.Two circles with centres O and \({O}’\) of radii 3 cm and 4cm respectively intersect at two points P and Q such that \({O}’\)P are tangents to the two circles.Find the length of the common chord PQ.


Sol.       Clearly, \(\angle\)OP\({O}’\)= \(90^{\circ}\)

O\({O}’\)= \({\sqrt{3^{2}+4^{2}}}’\)




Let                       RO=x

⇒             R\({O}’\)=5-x

Now,                    \(PR^{2}\)= \(PO^{2}\)– \(RO^{2}\)

=9-\(x^{2}\)    ……..(1)

Also,                    \(PR^{2}\)= \(PO^{2}\)– \(RO^{2}\)

=16-\(\left ( 5-x \right )^{_{2}}\)……….(2)

From (1) and (2),we have





From (1),we have






Hence, the required length of the common chord is 2 x PR i.e., 2×2.4 i.e., 4.8cm.


Exemplar problems

1. If V1 , V2 (V2 > V1) be the diameters of two concentric circles and Z be the length of a chord of a circle which is tangent to the other circle, prove that : \({V_{2}}^{2}=Z^{2}+{V_{1}}^{2}\).


Let XY be a chord of a circle which touches the other circle at Z. Then \(\Delta\)OZY is a right triangle.


∴          By Pythagoras Theorem,

\(OZ^{2}\) + \(ZY^{2}\)= \(OY^{2}\)

i.e.,        \(\left ( \frac{V1}{2} \right )^{2}\)+ \(\left ( \frac{Z}{2} \right )^{2}\)= \(\left ( \frac{V2}{2} \right )^{2}\)

\(\Rightarrow\)           \(\frac{{V_{1}}^{2}}{4}\)+ \(\frac{Z^{2}}{4}\)=            \(\frac{{V_{2}}^{2}}{4}\)

\(\Rightarrow\)         \({V_{2}}^{2}+Z^{2}={V_{1}}^{2}\)


2. Two tangents AB and AC are drawn from an external point to a circle with centre 0. Prove that BOCA is a cyclic quadrilateral.

Sol. We know that,  tangents to a circle is perpendicular to its radius at the point of contact.


∴     OC\(\perp\)AC and OB\(\perp\)AB


\(\angle\)OCA+\(\angle\)OBA=\(90^{\circ}\)+ \(90^{\circ}\)


Sum of opposite angles of quadrilateral BOCA is \(180^{\circ}\)

Hence, BOCA is a cyclic quadrilateral.


Practise This Question


  1. Jignesh Jignesh
    April 26, 2018    

    I like it
    It’s very good app to learn

  2. Aryan Aryan
    March 1, 2018    

    I didn’t get the answer what I want…

    • Shankha Samanta Shankha Samanta
      March 8, 2018    

      All the NCERT questions and solutions for Class 9 – Chapter 10: Circles are given here. Please do let us know which question you are referring to and which solution you want.

  3. udit udit
    March 29, 2018    

    this was great but can i know where are the solutions of byjus sample paper for maths

Leave a Reply

Your email address will not be published. Required fields are marked *

Practise This Question

Join BYJU’S NCERT Learning Program

NCERT Solutions

NCERT Solutions Maths