Question

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

Answer 1

Consider the figure:

Given: $AC=3cm$, $BC=4cm$ and $AB=5cm$

Since, $A{B}^2=B{C}^2+A{C}^2$, $△ABC$ is a right-angled triangle, right angled at $C$.

Let $AD$ be $x$ cm. Then $BD=5-x$ cm

Applying Pythagoras theorem to $△ADC$,

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

$\Rightarrow 3^2=x^2+C{D}^2$

$\Rightarrow C{D}^2=9-x^2$ ... (i)

Applying Pythagoras theorem to $△BDC$,

$B{C}^2=B{D}^2+C{D}^2$

$\Rightarrow 4^2=(5-x{)}^2+C{D}^2$

$\Rightarrow C{D}^2=16-(5-x{)}^2$ ... (ii)

Equating equation (i) and (ii), we get,

$9-x^2=16-(5-x{)}^2$

$\Rightarrow 9-x^2=16-25-x^2+10x$

$\Rightarrow 10x=18$

$\Rightarrow x=1.8$

$\Rightarrow C{D}^2=9-3.24=5.76$

$\Rightarrow CD=2.4cm$

We know that the perpendicular from a centre to a chord bisects the chord.

So, length of the common chord $=2\times CD=2\times 2.4=4.8cm$