Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$ Two circles ,of radii $a$ and $b$,cut each other at an angle $\theta.$Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$
Let the center of two circles be $O$ and $O'$ and the points where they intersect be $P$ and $Q$.Then angle $OPO'=\theta$
$\cos \theta=\frac{a^2+b^2-OO'^2}{2ab}$
$OO'^2=a^2+b^2-2ab\cos\theta$
In triangle $PO'Q$,angle $PO'Q=\pi-\theta$
$\cos(\pi-\theta)=\frac{b^2+b^2-l^2}{2b^2}$ Then i am stuck.Please help me to reach upto proof.

The above method is too long, a faster method would be
Let A & B be the centers of the circles with radii a & b respectively which intersect each other at an angle θ such that they have a common chord PQ.
Angle between AP and BP is $180-θ$
Using cosine rule in $\triangle APB$,
$$AB=\sqrt {AM^2+BM^2-2(AM)(BM)\cos (\angle APB)}$$
$$AB=\sqrt {a^2+b^2-2(a)(b)\cos (180-θ)}$$
We have,
Area of quadilateral $APBQ = \frac{(PQ*AB)}{2}$
Also, area of quadilateral $APBQ = AP*BP*sin(\angle APB)$
So,$$\frac{(PQ*AB)}{2} = AP*BP*sin(\angle APB)$$
$$\frac{(PQ*\sqrt {a^2+b^2-2(a)(b)\cos (180-θ)})}{2} = a*b*sin(180-θ)$$
$$ PQ=\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$
Sorry for the late answer.
