Question

Find the equation to the circle passing through the points $(0,a)$ and $(b,h)$, and having its centre on the axis of $x$.

Answer 1

Let the centre of circle be (k,0), then the equation of circle become $x^2+y^2-2kx+c=0$ ..... $\left(i\right)$

$(0,a),(b,h)$ satisfies the equation as the circle passes through these points.

From equation $\left(i\right)$, we get

$a^2+0+c=0$

$\Rightarrow c=-a^2$

Also, for $(b,h)$ in $\left(i\right)$, we get

$b^2+h^2-2kb-a^2=0$

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

Thus, the equation of circle is $x^2+y^2-\frac{b^2+h^2-a^2}{b}x-a^2=0$

i.e. $b(x^2+y^2-a^2)=x(b^2+h^2-a^2).$