Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$

Question

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.

Answer 1

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, $$MN=2x=\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Length of common chord}=\color{blue}{\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}}}}$$

Answer 2

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.