**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

\(OL^{2}\)= \(FO^{2}\)– \(FL^{2}\)

= \(5^{2}\)–\(4^{2}\)

=25-16

=9

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)

AB=AF

(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

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 \({O}’\).

∴ TQ=TS …..(ii)

Adding (i) and (ii),we obtain

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

__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

CA=DA

[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

=\(\frac{1}{2}\)\(\angle\)CAD=\(\frac{1}{2}\)\(\times\)\(120^{\circ}\)=\(60^{\circ}\)

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,

x-r=yF

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)

=(PN-XN)(PN+XN)

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 \(OP^{2}\)= \(ON^{2}\)+ \(PN^{2}\)]

=\(OP^{2}\)– \(ON^{2}\)– \(XN^{2}\)

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

=\(OP^{2}\)– \(OX^{2}\)

=\(OP^{2}\)– \(OT^{2}\)

Since,OX=OT=r]

(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]

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

=\(120^{\circ}\)

\(\angle\)YZD=\(\angle\)YXZ=\(30^{\circ}\)

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

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

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

\(\angle\)YDZ=\(180^{\circ}\)–\(120^{\circ}\)–\(30^{\circ}\)

=\(30^{\circ}\)

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

=\(30^{\circ}\)

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\({O}’\)= \({\sqrt{3^{2}+4^{2}}}’\)

=\(\sqrt{9+16}=\sqrt{25}\)

=5cm

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

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

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

10x=18

X=1.8cm

From (1),we have

\(PR^{2}\)=9-\(1.8^{2}\)

=9-3.24

=5.76

PR=\(\sqrt{5.76}\)

=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 : \({V_{2}}^{2}=Z^{2}+{V_{1}}^{2}\).*

Solution:-

Let XY be a chord of a circle which touches the other circle at Z. Then \(\Delta\)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=\(90^{\circ}\)

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

=\(180^{\circ}\)

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

Hence, BOCA is a cyclic quadrilateral.