If $\theta$  is the angle of intersection of two circle$x^2+y^2=a^2$  and ${\left(x-c\right)}^2+y^2=b^2,$  then the length of common chord of two circle is :

We have the two circles $x^2+y^2=a^2,$  ${\left(x-c\right)}^2+y^2=b^2$  radius of first circle=a

Let $\angle OPM=\alpha ,\angle CPM=\beta ,$

$\therefore \angle OPC=\alpha +\beta =\theta$

Let $PQ=d;$  radius of second circle=b

$\cos \alpha =\frac{PM}{a}=\frac{d}{2a},$ $\cos \beta =\frac{d}{2b}$

Now $\cos \theta =\cos \left(\alpha +\beta \right)$ $=\cos \alpha \cos \beta -\sin \alpha \sin \beta$

squaring on both sides, we get ${\sin }^2\alpha {\sin }^2\beta$ $={\cos }^2\alpha {\cos }^2\beta +{\cos }^2\theta -2\cos \theta \cos \alpha \cos \beta$

                      = $1-{\cos }^2\alpha -{\cos }^2\beta +{\cos }^2\alpha {\cos }^2\beta$

                     $={\cos }^2\alpha {\cos }^2\beta +{\cos }^2\theta -2\cos \theta \cos \alpha \cos \beta$

$\Rightarrow {\sin }^2\theta ={\cos }^2\alpha +{\cos }^2\beta -2\cos \theta \cos \alpha \cos \beta$

$\therefore {\sin }^2\theta =\frac{d^2}{4a^2}+\frac{d^2}{4b^2}-\frac{2d^2}{4ab}\cos \theta$

$\Rightarrow 4a^2{b}^2{\sin }^2\theta =d^2\left(b^2+a^2-2ab\cos \theta \right)$

$\therefore d=\frac{2ab\sin \theta }{\sqrt{a^2+b^2-2ab\cos \theta }}$