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.

Let $A$ & $B$ be the centers of the circles with radii $a$ & $b$ respectively such that they have a common chord $MN=2x$ & intersecting each other at an angle $\theta$. 
Let $O$ be the mid-point of common chord $MN$. Point $O$ lies on the line AB joining the centers of circle then we have $$MO=ON=x$$
In right $\triangle AOM$ $$AO=\sqrt{AM^2-MO^2}=\sqrt{a^2-x^2}$$
Similarly, in right $\triangle BOM$ $$OB=\sqrt{BM^2-MO^2}=\sqrt{b^2-x^2}$$
$$AB=AO+OB=\sqrt{a^2-x^2}+\sqrt{b^2-x^2}$$
Now, applying cosine rule in $\triangle AMB$ as follows $$AB^2=AM^2+BM^2-2(AM)(BM)\cos (\angle AMB)$$
Setting the corresponding values, we get
$$(\sqrt{a^2-x^2}+\sqrt{b^2-x^2})^2=a^2+b^2-2(a)(b)\cos (180^\circ-\theta)$$
$$a^2-x^2+b^2-x^2+2\sqrt{(a^2-x^2)(b^2-x^2)}=a^2+b^2+2ab\cos\theta$$
$$\sqrt{(a^2-x^2)(b^2-x^2)}=x^2+ab\cos\theta$$
$$(a^2-x^2)(b^2-x^2)=(x^2+ab\cos\theta)^2$$
$$a^2b^2-b^2x^2-a^2x^2+x^4=x^4+a^2b^2\cos^2\theta+2abx^2\cos\theta$$
$$(a^2+b^2+2ab\cos\theta)x^2=a^2b^2-a^2b^2\cos^2\theta$$ $$x^2=\frac{a^2b^2\sin^2\theta}{a^2+b^2+2ab\cos\theta}$$
$$x=\frac{ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$
Hence, the length of the common chord MN,
$$MN=2x=\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$ 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Length of common chord}=\color{blue}{\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}}}}$$
