$A$ 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 $α$ and $β$ respectively and the distance between the point $A$ and the mid-point of the line segment $DC$ is $d$ .
Then the area of the Circle is

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$\frac{π d \times \cos^{2} α}{\cos^{2} α + \cos^{2} β + 2 \cos α \cos β \cos (β - α)}$

$\frac{π d \times \cos^{2} α}{\cos^{2} α - \cos^{2} β + 2 \cos α \cos β \cos (β - α)}$

$\frac{2 π d \times \cos^{2} α}{\cos^{2} α + \cos^{2} β + 2 \cos α \cos β \cos (β - α)}$

$\frac{π d \times \cos^{2} α}{\cos^{2} α + \cos^{2} β - 2 \cos α \cos β \cos (β - α)}$

answer is A.

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Detailed Solution

Let r be the radius of a circle, then $A C = 2 r$

Since, $AC$ is the diameter $∠ A B C = 90^{\circ}$

In $△ A B C, B C = 2 r \sin β, A B = 2 r \cos β$

$\begin{array}{l} B D = A B \tan α = 2 r \cos β \tan α \\ A D = A B \sec α = 2 r \cos β \sec α \\ D C = B C - B D = 2 r \sin β - 2 r \cos β \tan α \end{array}$

Since E is the mid-point of $DC$ , so

$D E = \frac{D C}{2} = r \sin β - r \cos β \tan α$

Now, in $△ A D C, A E$ is the median

$\begin{array}{l} 2 (A E^{2} + D E^{2}) = A D^{2} + A C^{2} \\ 2 (d^{2} + r^{2} (\sin β - \cos β \tan α)^{2}) \\ = 4 r^{2} \cos^{2} β \sec^{2} α + 4 r^{2} \\ r^{2} = \frac{d^{2} \cos^{2} α}{\cos^{2} α + \cos^{2} β + 2 \cos α \cos β \cos (β - α)} \end{array}$