Question

The circles of radii $5$ cm and $3$ cm intersect at two points and the distance between their centers is $4$ cm. Find the length of the common cord in cm

Answer 1

Given: circle $C_1$ with radii 5$cm$

8$C_2$ with radii 3$cm$

Intersecting at $P\&Q$ .

$OP=5cm,XP=3cm\&OX=4cm$

To find: Length of common chor

i.e., length of PQ

Solution: Let the point where OX intersect PQ be R

$\ln aPOX\&\Delta QOX$

$OP=OQ$

$xP=xQ$

$Ox=Ox$

$\therefore \Delta POX\cong \Delta QOX$

$\angle POX=\angle QOX$

Also

In $\Delta$ POR \& $\Delta QOR$

$OP=OQ$

$\angle POR=\angle QOR$

$OR=OR$

$\Rightarrow \angle PRO=\angle QRO$

$\&PR=RQ$

$\therefore \Delta POR\cong \Delta QOR$

$\Rightarrow \angle PRO=\angle QRO$

$\&PR=RQ$

since PQ is a line

$\angle PRO+\angle QRO={180}^{\circ }$

$\angle PRO+\angle PRO={180}^{\circ }$

$2\angle PRO={180}^{\circ }$

$\angle PRO=\frac{{180}^{\circ }}{2}$

$\angle PRO={90}^{\circ }$

Therefore,

$\angle QRO=\angle PRO={90}^{\circ }$

Also,

$\angle PRX=\angle QRO={90}^{\circ }$

Let $OR=OX-OR=4-x$

In In $\Delta OPR$

$P{R}^2=25-x^2\dots \left(4\right)$

$\ln \Delta XPR$

$P{R}^2=-7-x^2+8x\dots \left(5\right)$

From $\left(4\right)\&\left(5\right)$

$5^2-x^2=-7-x^2+8x$

$25-x^2=-7-x^2+8x$

$25+7-x^2+x^2=8x$

$32=8x$

$8x=32$

$x=\frac{32}{8}$

$x=4$

Putting value

$P{R}^2=25-x^2$

$P{R}^2=25-4^2$

$P{R}^2=25-16$

$P{R}^2=9$

$PR=\sqrt{9}=3$

$\therefore PQ=2PR=2\times 3=6$

Length of chord =6m