The equation of a circle that touches both the x-axis and y-axis must have its center at $(h, k)$, where $h = k$, because the distance from the center to each axis (the radius) must be the same. So, we assume the center of the circle is $(a, a)$, and its equation is:

$(x - a)^2 + (y - a)^2 = r^2$

Since the circle touches both axes, the radius $r$ must be equal to $a$, meaning:

$r = a$

Thus, the equation becomes:

$(x - a)^2 + (y - a)^2 = a^2$

The circle passes through the point $(-3, -2)$. Substituting $x = -3$ and $y = -2$ into the equation:

$(-3 - a)^2 + (-2 - a)^2 = a^2$

Expanding this:

$(9 + 6a + a^2) + (4 + 4a + a^2) = a^2$

Simplifying:

$9 + 6a + a^2 + 4 + 4a + a^2 = a^2$ $13 + 10a + 2a^2 = a^2$ $2a^2 - a^2 + 10a + 13 = 0$ $a^2 + 10a + 13 = 0$

This is a quadratic equation. Solving for $a$ using the quadratic formula:

$a = \frac{-10 \pm \sqrt{10^2 - 4(1)(13)}}{2(1)}$ $a = \frac{-10 \pm \sqrt{100 - 52}}{2}$ $a = \frac{-10 \pm \sqrt{48}}{2}$ $a = \frac{-10 \pm 4\sqrt{3}}{2}$ $a = -5 \pm 2\sqrt{3}$

Thus, the two possible values for $a$ are:

$a = -5 + 2\sqrt{3} \quad \text{or} \quad a = -5 - 2\sqrt{3}$

Therefore, the equation of the circle is either:

$(x - (-5 + 2\sqrt{3}))^2 + (y - (-5 + 2\sqrt{3}))^2 = (-5 + 2\sqrt{3})^2$

or

$(x - (-5 - 2\sqrt{3}))^2 + (y - (-5 - 2\sqrt{3}))^2 = (-5 - 2\sqrt{3})^2$

Would you like further details on solving quadratic equations or anything else?

Related Questions:

1. What is the geometric interpretation of the circle touching both axes?
2. How can we solve the quadratic equation using completing the square?
3. Can we find the distance from the center to the point (-3, -2)?
4. How does the discriminant affect the nature of the solutions?
5. What happens if the circle doesn't touch both axes?

Tip: Always double-check your work when expanding and simplifying equations.