Find the equation of the circle passing through $\left(h, 0\right)$, $\left(k, 0\right)$, $\left(0, h\right)$ and $\left(0, k\right)$. Also find the center of the circle.
Let the equation of the required circle be
$x^2 + y^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (1)$
Since the required circle passes through the points $\left(h, 0\right)$, $\left(k, 0\right)$ and $\left(0, h\right)$, we have
$h^2 + 2gh + c = 0$ $\;\;\ \cdots \; (2)$
$k^2 + 2gk + c = 0$ $\;\;\; \cdots \; (3)$
$h^2 + 2 fh + c = 0$ $\;\;\; \cdots \; (4)$
We have from equations $(2)$ and $(4)$,
$2h \left(g - f\right) = 0$
i.e. $\;$ $g - f = 0 \;\; \because \; h \neq 0$ as $ \left(h, 0\right)$ and $\left(0, h\right)$ are finite points
$\implies$ $g = f$ $\;\;\; \cdots \; (5)$
We have from equations $(2)$ and $(3)$,
$\left(h^2 - k^2\right) + 2g \left(h - k\right) = 0$
i.e. $\;$ $\left(h + k\right) \left(h - k\right) + 2g \left(h - k\right) = 0$
i.e. $\;$ $\left(h - k\right) \left(h + k + 2g\right) = 0$
Now,
$h - k = 0$ $\implies$ $h = k$ $\;$ which is not possible as it would make the four given points trivial in nature.
$\therefore \;$ $h + k + 2g = 0$ $\implies$ $g = \dfrac{-h - k}{2}$ $\;\;\; \cdots \; (6a)$
$\therefore \;$ We have from equation $(5)$, $\;$ $f = g = \dfrac{-h - k}{2}$ $\;\;\; \cdots \; (6b)$
Substituting the value of $g$ from equation $(6a)$ in equation $(2)$ gives
$h^2 - 2 \times \left(\dfrac{h + k}{2}\right) \times h + c = 0$
i.e. $\;$ $h^2 - h^2 - kh + c = 0$ $\implies$ $c = kh$ $\;\;\; \cdots \; (6c)$
Substituting the values of $f$, $g$ and $c$ from equations $(6a)$, $(6b)$ and $(6c)$ in equation $(1)$ gives the equation of the required circle as
$x^2 + y^2 + 2 \times \left(\dfrac{-h - k}{2}\right) \times x + 2 \times \left(\dfrac{-h - k}{2}\right) \times 2 + kh = 0$
i.e. $\;$ $x^2 + y^2 - \left(h + k\right)x - \left(h + k\right) y + kh = 0$ $\;\;\; \cdots \; (7)$
Substituting the point $\left(0, k\right)$ in equation $(7)$ gives
$0^2 + k^2 - \left(h + k\right) \times 0 - \left(h + k\right) k + kh = 0$
i.e. $\;$ $k^2 - hk - k^2 + kh = 0$
i.e. $\;$ $0 = 0$ $\;$ which is true.
$\implies$ The required circle, given by equation $(7)$, also passes through the point $\left(0,k\right)$.
The center of the required circle is
$\left(-g, -f\right) = \left(\dfrac{h + k}{2}, \dfrac{h + k}{2}\right)$