Find the equation of circles that touch both the axes and pass through the point $\left(-4,-2\right)$, in general form.
The required circle passes through the point $P \left(-4,-2\right)$.
$\because$ $\;$ Point $P$ is in the third quadrant, the required circle will touch the negative X and Y axes.
Let the required circle touch the X axis at the point $A \left(-a,0\right)$ and the Y axis at the point $B \left(0,-b\right)$.
Then, the center of the circle is $C \left(-a,-b\right)$.
Now, $AC = BC = PC =$ radius of the circle
$\implies$ $AC^2 = BC^2 = PC^2$ $\;\;\; \cdots \; (1)$
Now, $AC^2 = BC^2$ $\implies$ $b^2 = a^2$ $\implies$ $b = a$
$\therefore$ $\;$ Center of circle is $C \left(-a, -a\right)$
$\therefore$ $\;$ $PC^2 = AC^2$ $\implies$ $\left(-4 + a\right)^2 + \left(-2 + a\right)^2 = a^2$
i.e. $16 + a^2 - 8a + 4 + a^2 - 4a = a^2$
i.e. $a^2 - 12 a + 20 = 0$
i.e. $\left(a - 10\right) \left(a - 2\right) = 0$
i.e. $a = 10$ $\;$ or $\;$ $a = 2$
$\because$ $\;$ $b = a$, $b = 10$ $\;$ or $\;$ $b = 2$
$\therefore$ $\;$ Center of the circle is $C \left(-10,-10\right)$ or $C \left(-2,-2\right)$
and radius of circle $= a = 10$ or $a = 2$
$\therefore$ $\;$ The equations of the circles are
for center $\left(-10,-10\right)$ and radius $=10$: $\;\;$ $\left(x + 10\right)^2 + \left(y + 10\right)^2 = 100$
i.e. $x^2 + y^2 + 20 x + 20 y + 100 = 0$
and for center $\left(-2,-2\right)$ and radius $=2$: $\;\;$ $\left(x + 2\right)^2 + \left(y + 2\right)^2 = 4$
i.e. $x^2 + y^2 + 4x + 4y + 4 = 0$