# NCERT Solutions For Class 9 Maths Chapter 10

## NCERT Solutions Class 9 Maths Circles

With the help of our downloadable NCERT Solutions For Class 9 Maths Chapter 10 pdf students can try solving textbook problems by themselves and also correct themselves with the solutions provided with the right steps. Some of the chapters included in the NCERT Solutions For Class 9 Maths Chapter 10 Circles are definitions of circle related concepts, radius, circumference, diameter, chord, arc, secant, sector, segment subtended angle. Test your knowledge in the class 9 maths subject with the help of NCERT Solutions For Class 9 Maths Chapter 10 and step by step solutions.

### NCERT Solutions For Class 9 Maths Chapter 10 Exercises

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  Δ$\Delta$FLO,we obtain

OL2$OL^{2}$= FO2$FO^{2}$FL2$FL^{2}$

= 52$5^{2}$42$4^{2}$

=25-16

=9

OL =9$\sqrt{9}$=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 l1$l_{1}$ and l2$l_{2}$,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 Δ$\Delta$ABX and Δ$\Delta$ARX, we have

AX=AX                        (common)

AB=AF

(tangents from an external point)

∴     Δ$\Delta$ABX=Δ$\Delta$AFX

(by SSS congruence rule)

$\Rightarrow$    $\angle$XAB=$\angle$XAF

$\Rightarrow$   X lies on the bisector of the lines l1$l_{1}$ and l2$l_{2}$.

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${O}’$.

∴      RL=RN    …..(i)

And                                            RM=RP     …..(ii)

Subtracting (ii) from (i),we have

RL-RM=RN-RP

∴        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${O}’$.

∴      TQ=TS   …..(ii)

TP+TQ=TR+TS

$\Rightarrow$      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

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 ΔXCB$\Delta XCB$ and ΔXDB$\Delta XDB$,we have

CA=DA

[tangents from an external point]

XA=XA                       [common side]

∴   By SSS  congruency , we have

ΔXCA$\Delta XCA$$\cong$ΔXDA$\Delta XDA$

$\Rightarrow \cong$   $\angle$XAC=$\angle$XAD

=12$\frac{1}{2}$$\angle$CAD=12$\frac{1}{2}$×$\times$120$120^{\circ}$=60$60^{\circ}$

In  ΔXCA$\Delta XCA$ ,$\angle$XCA=90°

∴  ACAX=cos60$\frac{AC}{AX}=\cos 60^{\circ}$

BCBX=12$\Rightarrow                \frac{BC}{BX}=\frac{1}{2}$

$\Rightarrow$       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  = x+yz2$\frac{x+y-z}{2}$

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,

x-r=yF

Or                 XY=z=XF+YF

$\Rightarrow$    Z=y-r+x-r

$\Rightarrow$    r  = x+yz2$\frac{x+y-z}{2}$

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= PY2$PY^{2}$XN2$XN^{2}$

(ii) PN2XN2$PN^{2}- XN^{2}$= OP2OT2$OP^{2}- OT^{2}$

(iii) PX.PY=PT2$PX.PY= PT^{2}$

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

=(PN-XN)(PN+XN)

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

= PN2XN2$PN^{2}- XN^{2}$

(ii) PN2XN2$PN^{2}- XN^{2}$=(OP2$OP^{2}$ON2$ON^{2}$)- XN2$XN^{2}$

Since,in rt. $\triangle$ONP OP2$OP^{2}$= ON2$ON^{2}$+ PN2$PN^{2}$]

=OP2$OP^{2}$ON2$ON^{2}$XN2$XN^{2}$

=OP2$OP^{2}$-(ON2$ON^{2}$)+ XN2$XN^{2}$)

=OP2$OP^{2}$OX2$OX^{2}$

=OP2$OP^{2}$OT2$OT^{2}$

Since,OX=OT=r]

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

PX.PY= OP2$OP^{2}$OT2$OT^{2}$

PX.PY=PT2$PX.PY= PT^{2}$

∴ In rt. $\triangle$OTP; OP2$OP^{2}$= OT2$OT^{2}$+ PT2$PT^{2}$

4.XY is a diameter and XZ is a chord with centre O such that $\angle$YXZ= 30$30^{\circ}$.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$30^{\circ}$ and ZD be the tangent at Z.

$\angle$XZY=90$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$30^{\circ}$+ 90$90^{\circ}$

=120$120^{\circ}$

$\angle$YZD=$\angle$YXZ=30$30^{\circ}$

In $\triangle$,by angles sum property,we have

$\angle$YDZ+$\angle$ZYD+$\angle$YZD=180$180^{\circ}$

$\angle$YDZ+120$120^{\circ}$+30$30^{\circ}$=180$180^{\circ}$

$\angle$YDZ=180$180^{\circ}$120$120^{\circ}$30$30^{\circ}$

=30$30^{\circ}$

\Rightarrow                 $\angle$YDZ=$\angle$YDZ

=30$30^{\circ}$

Hence,                           YD=YZ

[sides opp.to equal angles]

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

Sol.       Clearly, $\angle$OPO${O}’$= 90$90^{\circ}$

OO${O}’$= 32+42${\sqrt{3^{2}+4^{2}}}’$

=9+16=25$\sqrt{9+16}=\sqrt{25}$

=5cm

Let                       RO=x

⇒             RO${O}’$=5-x

Now,                    PR2$PR^{2}$= PO2$PO^{2}$RO2$RO^{2}$

=9-x2$x^{2}$    ……..(1)

Also,                    PR2$PR^{2}$= PO2$PO^{2}$RO2$RO^{2}$

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

From (1) and (2),we have

9-x2$x^{2}$=16-(25+x2$x^{2}$-10x)

9-x2$x^{2}$=16-25+x2$x^{2}$+10x)

10x=18

X=1.8cm

From (1),we have

PR2$PR^{2}$=9-1.82$1.8^{2}$

=9-3.24

=5.76

PR=5.76$\sqrt{5.76}$

=2.4cm

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 : V22=Z2+V12${V_{2}}^{2}=Z^{2}+{V_{1}}^{2}$.

Solution:-

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

∴          By Pythagoras Theorem,

OZ2$OZ^{2}$ + ZY2$ZY^{2}$= OY2$OY^{2}$

i.e.,        (V12)2$\left ( \frac{V1}{2} \right )^{2}$+ (Z2)2$\left ( \frac{Z}{2} \right )^{2}$= (V22)2$\left ( \frac{V2}{2} \right )^{2}$

$\Rightarrow$           V124$\frac{{V_{1}}^{2}}{4}$+ Z24$\frac{Z^{2}}{4}$=            V224$\frac{{V_{2}}^{2}}{4}$

$\Rightarrow$         V22+Z2=V12${V_{2}}^{2}+Z^{2}={V_{1}}^{2}$

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$90^{\circ}$

$\angle$OCA+$\angle$OBA=90$90^{\circ}$+ 90$90^{\circ}$

=180$180^{\circ}$

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

Hence, BOCA is a cyclic quadrilateral.