Question

Two circles of radii $5cm$ and $3cm$ intersect at two points and the distance between their centres is $4cm$. Find the length of the common chord.

Answer 1

Let the common chord be $AB$ and $P$ and $Q$ be the centers of the two circles.

$\therefore AP=5$cm and $AQ=3$cm.

$PQ=4$cm ....given

Now, seg$PQ\perp$chord $AB$

$\therefore AR=RB=\frac{1}{2}AB$ ....perpendicular from center to the chord, bisects the chord

Let $PR=x$cm, so $RQ=(4-x)$cm

In $△ARP$,

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

$A{R}^2=5^2-x^2$ ...(1)

In $△ARQ$,

$A{Q}^2=A{R}^2+Q{R}^2$

$A{R}^2=3^2-(4-x{)}^2$ ...(2)

$\therefore 5^2-x^2=3^2-(4-x{)}^2$ ....from (1) & (2)

$25-x^2=9-(16-8x+x^2)$

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

$32=8x$

$\therefore x=4$

Substitute in eq(1) we get,

$A{R}^2=25-16=9$

$\therefore AR=3cm$.

$\therefore AB=2\times AR=2\times 3$

$\therefore AB=6cm.$

So, length of common chord $AB$ is $6cm.$