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

Suppose a circle passes through $(0, a)$ and $(b, h)$ having its centre at $(c, 0)$. Then the value of $c$ is
  1. $\frac{b^2-a^2+h^2}{2 b}$
  2. $\frac{b^2+a^2-h^2}{2 b}$
  3. $\frac{b^2-a^2+h^2}{2 a}$
  4. $\frac{b^2+a^2-h^2}{2 a}$

Solution

Let $m$ be mid point of $\mathrm{AB}$ Coordinates of $\mathrm{M}$ are $\left(\frac{0+b}{2}, \frac{a+h}{2}\right) \equiv\left(\frac{b}{2}, \frac{a+h}{2}\right)$ $\mathrm{OA}^2=\mathrm{AM}^2+\mathrm{OM}^2$ $(\mathrm{C}-0)^2+(0-a)^2=$ $\left(\frac{b}{2}-0\right)^2+\left(\frac{a+h}{2}-a\right)^2+\left(\frac{c-b}{2}\right)^2+\left(0-\frac{a+h}{2}\right)^2$ $\begin{aligned} C^2+a^2=\frac{b^2}{4}+\frac{h^2}{4}+\frac{a^2}{4}-\frac{a h}{2}+ & c^2+\frac{b^2}{4}-b c \\ + & \frac{a^2}{4}+\frac{h^2}{4}+\frac{a h}{2}\end{aligned}$ $C^2+a^2=\frac{a^2}{2}+\frac{b^2}{2}+\frac{h^2}{2}+c^2-b c$ $\begin{aligned} & \frac{a^2}{2}=\frac{b^2+h^2}{2}-b c \\ & -b c=\frac{a^2}{2}-\frac{b^2+h^2}{2} \\ & c=\frac{a^2-b^2-h^2}{-2 b} \\ & c=\frac{b^2-a^2+h^2}{2 b}\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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