Question

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

Answer 1

Let the radius of the circle centered at $O$ and $O^{\prime }$ be $5cm$ and $3cm$ respectively.

$OA=OB=5cm$

$O^{\prime }A=O^{\prime }B=3cm$

$O{O}^{\prime }$ will be the perpendicular bisector of chord $AB.$

$\therefore AC=CB$

It is given that, $O{O}^{\prime }=4cm$

Let $OC$ be $x.$ Therefore, $O^{\prime }C$ will be $x-4$

In $△OAC,$ we have

$O{A}^2=A{C}^2+O{C}^2$

$\Rightarrow 5^2=A{C}^2+x^2$

$\Rightarrow 25-x^2=A{C}^2...\left(1\right)$

In $△O^{\prime }AC,$ we have

$O^{\prime }{A}^2=A{C}^2+O^{\prime }{C}^2$

$\Rightarrow 3^2=A{C}^2+(x-4{)}^2$

$\Rightarrow 9=A{C}^2+x^2+16-8x$

$\Rightarrow A{C}^2=-x^2-7+8x...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right),$ we obtain

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

$8x=32$

$x=4$

Therefore, the common chord will pass through the center of the smaller circle i.e., $O^{\prime }$ and hence, it will be the diameter of the smaller circle.

Length of the common chord $AB=2O^{\prime }A=(2\times 3)cm=6cm$