IIT-JEE 1996Question 1
A circle passes through three points $A$, $B$ and $C$ with the line segment $AC$ as its diameter. A line passing through $A$ intersects the chord $BC$ at a point $D$ inside the circle. If angles $DAB$ and $CAB$ are $\alpha$ and $\beta$ respectively and the distance between the point $A$ and the mid point of the line segment $DC$ is $d$, prove that the area of the circle is $\dfrac{\pi d^{2}\cos^{2}\alpha}{\cos^{2}\alpha+\cos^{2}\beta+2\cos\alpha\cos\beta\cos(\beta-\alpha)}$.
Circle
Answer$proved: pi d^2 cos^2(a)/(cos^2 a+cos^2 b+2cos a cos b cos(b-a))$
Solution
Let $R$ be the radius, $A$ the origin, $AB$ the $x$-axis: $B=(2R\cos\beta,0)$, $C=(2R\cos\beta,2R\cos\beta\tan\beta)$, $D=(2R\cos\beta,2R\cos\beta\tan\alpha)$. Then $d^2=\left|\tfrac{D+C}{2}\right|^2$ simplifies to $R^2\,\frac{\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta\cos(\beta-\alpha)}{\cos^2\alpha}$, so area $\pi R^2$ is the stated expression.