Finding the area of a portion of circle between two chords I recently solved a problem with the following scenario. A circle has equation $x^2 + y^2 - ax - by = 0$ with $b \neq 0$. Two chords from the point $A(a,b)$ are such that they are bisected by the $x$-axis (i.e. $AX = XB$ and $AY = YC$ in the diagram). 

As an extension to the problem, I wondered if it would be possible from this information alone to determine the area of the shaded portion of the diagram in terms of $a$ and $b$ (this wasn't related to the original problem that I solved, but I was curious). Somehow I feel as though there wouldn't be enough information, but if this is indeed the case could anyone suggest a piece of extra information that would allow me to determine the area?
 A: Equation of chord of circle , with midpoint as a point $(x_1,y_1)$ is given by the equation $$T = S_1$$ $$\implies xx_1 + yy_1-\frac{a}{2}(x+x_1)-\frac{b}{2}(y+y_1) = x_1^2+y_1^2-ax_1-by_1$$ Now, in the above equation, for $(x_1,y_1)$, we substitute $(x,0)$, and since we know that the chords pass through point $(a,b)$, we substitute that to get a quadratic in $x$. Hence, we have at most two solutions. Therefore, we have enough data to solve
A: Elaborating on Dhanvi's answer, the abscissae of the mid-points of $AB$ and $AC$ are given by the quadratic:
$$
2x^2 - 3ax + (a^2 + b^2) = 0.
$$
Let the roots of the equation be $x_1$ and $x_2$. Then,
$$
|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1x_2} = \frac{\sqrt{a^2 - 8b^2}}{2}.
$$
Let the centre of the circle be $C_1(a/2, b/2)$. Then, $C_1C = C_1B = r = \frac{\sqrt{a^2 + b^2}}{2}$. Now, the points B and C happen to be $B(2x_1 - a, -b)$ and $C(2x_2 - a, -b)$. Notice that both of their ordinates are equal. Hence, 
$$
BC = 2|x_1-x_2| = \sqrt{a^2 - 8b^2}.
$$
Let the angle between $C_1C$ and $C_1B$ be $\theta$. Then, $r \sin(\theta/2) = BC/2$. Thus,
$$
\theta = 2 \sin^{-1} \left ( \frac{BC}{2r} \right ).
$$
Now we are ready to find the area of the shaded region as follows:
$$
\Delta = [\Delta ABC] + [\textrm{sector } C_1BC] - [\Delta C_1BC] \\
\implies \Delta = {1 \over 2} BC \times 2|b| + r^2 \theta - {1 \over 2} BC \sqrt{r^2 - (BC/2)^2}.
$$
Where $[\textrm{fig}]$ is the area of a figure. The final answer is:
$$
\Delta = {|b|\sqrt{a^2 - 8b^2} \over 4} + {a^2 + b^2 \over 2} \sin^{-1} \sqrt{ a^2 - 8b^2 \over {a^2+b^2}}
$$
