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(A) Let the centre of the circle be $O = (c, 0)$.Let the points on the circle be $A = (0, a)$ and $B = (b, h)$.Since $O$ is the centre,the distance fr

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$\frac{b^2-a^2+h^2}{2 b}$

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(A) Let the centre of the circle be $O = (c, 0)$.Let the points on the circle be $A = (0, a)$ and $B = (b, h)$.Since $O$ is the centre,the distance from $O$ to $A$ must equal the distance from $O$ to $B$,so $OA^2 = OB^2$.$OA^2 = (c - 0)^2 + (0 - a)^2 = c^2 + a^2$.$OB^2 = (c - b)^2 + (0 - h)^2 = (c - b)^2 + h^2$.Equating the two distances:$c^2 + a^2 = (c - b)^2 + h^2$.$c^2 + a^2 = c^2 - 2bc + b^2 + h^2$.$a^2 = -2bc + b^2 + h^2$.$2bc = b^2 + h^2 - a^2$.$c = \frac{b^2 - a^2 + h^2}{2b}$.

Suppose a circle passes through $(0, a)$ and $(b, h)$ having its centre at $(c, 0)$. Then the value of $c$ is

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