Question

If $3,4$ are the radii and $5$ is the distance between the centers of two intersecting circles then the length of the common chord of the circle is

• A. $12/5$
• B. $24/25$
• C. $24/5$
• D. $5/24$

Answer 1

Answer: (C)

$24/5$

From fig, in $△POX$ and $△QOX$

$OP=OQ$ ------ radius of Circle 1

$XP=XQ$ ------ radius of Circle 2

$OX=OX$ ------ common

$\therefore △POX\cong △QOX$ ------ SSS postulate

$\angle POX=\angle QOX$ ------ CPCT ------ (i)

Also in $△POR$ and $△QOR$

$OP=OQ$ ------ radius of Circle 1

$\angle POR=\angle QOR$ ------ from(i)

$OR=OR$ ------ common

$\therefore △POR\cong △QOR$ ------ SAS postulate

$\angle PRO=\angle QRO$ ------ CPCT ------ (ii)

$PR=RQ$ ------ CPCT ------ (iii)

Since $PQ$ is a line,

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

$\angle PRO+\angle PRO={180}^{\circ }$ ------ from(ii)

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

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

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

Also $\angle PRX=\angle QRO={90}^{\circ }$ ----- Vertically opposite angles

Let $OR=x$,

$XR=OX-OR=5-x$

In $△OPR$

$O{P}^2=P{R}^2+O{R}^2$

$4^2=P{R}^2+x^2$

$P{R}^2=16-x^2$ ------ (iv)

In $△XPR$

$X{P}^2=P{R}^2+X{R}^2$

$3^2=P{R}^2+(5-x{)}^2$

$P{R}^2=9-(25+x^2-10x)$

$P{R}^2=-16-x^2+10x$ ------ (v)

From (iv) & (v)

$16-x^2=-16-x^2+10x$

$32=10x$

$x=\frac{16}{5}$

From (iv), $P{R}^2=16-x^2=16-{\left(\frac{16}{5}\right)}^2$

$P{R}^2=\frac{16\times 9}{25}$

$PR=\frac{4\times 3}{5}=\frac{12}{5}$

Common Chord, $PQ=2PR$

$PQ=2\times \frac{12}{5}$

$PQ=\frac{24}{5}$