Find the equation of the circle which passes through the points $\left(4,2\right)$, $\left(-6,-2\right)$ and has its center on the Y axis.
The center of the circle lies on the Y axis.
Let center $= C = \left(0,b\right)$
The required circle passes through the points $A \left(4,2\right)$ and $B \left(-6,-2\right)$.
Radius of circle $= CA = CB$ $\;\;\; \cdots \; (1)$
Now, $CA = \sqrt{\left(4 - 0\right)^2 + \left(2 - b\right)^2} = \sqrt{20 - 4b + b^2}$ $\;\;\; \cdots \; (2a)$
and $CB = \sqrt{\left(0 + 6\right)^2 + \left(b + 2\right)^2} = \sqrt{40 + 4b + b^2}$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes
$\sqrt{20 - 4b + b^2} = \sqrt{40 + 4b + b^2}$
i.e. $20 - 4b + b^2 = 40 + 4b + b^2$
i.e. $8b = -20$ $\implies$ $b = \dfrac{-5}{2}$
$\therefore$ $\;$ Center of required circle $= C = \left(0, \dfrac{-5}{2}\right)$
Substituting the value of $b$ in equation $(2a)$ gives
radius of required circle $= CA = \sqrt{20 - 4 \times \left(\dfrac{-5}{2}\right) + \left(\dfrac{-5}{2}\right)^2} = \sqrt{\dfrac{145}{4}}$
$\therefore$ $\;$ Equation of required circle is
$\left(x - 0\right)^2 + \left(y + \dfrac{5}{2}\right)^2 = \dfrac{145}{4}$
i.e. $x^2 + y^2 + 5y + \dfrac{25}{4} = \dfrac{145}{4}$
i.e. $x^2 + y^2 + 5y - 30 = 0$