Solution:
Let the vertices of the triangle be $A(-2,1)$, $B(2,5)$, $C(5,2)$.

Let the equation of the inscribed circle be $(x-h)^2 + (y-k)^2 = r^2$.

The center $(h,k)$ of the incircle is the intersection point of the angle bisectors, i.e., the incenter.

Let $a$, $b$, $c$ be the lengths of the sides opposite $A$, $B$, $C$ respectively.

First, compute the side lengths:

$a = BC = \sqrt{(2-5)^2 + (5-2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

$b = AC = \sqrt{(-2-5)^2 + (1-2)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$

$c = AB = \sqrt{(-2-2)^2 + (1-5)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

The incenter $(h,k)$ is given by:

$$h = \frac{a x_A + b x_B + c x_C}{a + b + c}$$
$$k = \frac{a y_A + b y_B + c y_C}{a + b + c}$$

Plug in the values:

$a = 3\sqrt{2}$, $b = 5\sqrt{2}$, $c = 4\sqrt{2}$

$a + b + c = (3 + 5 + 4)\sqrt{2} = 12\sqrt{2}$

$x_A = -2$, $x_B = 2$, $x_C = 5$

$h = \frac{3\sqrt{2} \times (-2) + 5\sqrt{2} \times 2 + 4\sqrt{2} \times 5}{12\sqrt{2}}$

$= \frac{-6\sqrt{2} + 10\sqrt{2} + 20\sqrt{2}}{12\sqrt{2}}$

$= \frac{24\sqrt{2}}{12\sqrt{2}} = 2$

$y_A = 1$, $y_B = 5$, $y_C = 2$

$k = \frac{3\sqrt{2} \times 1 + 5\sqrt{2} \times 5 + 4\sqrt{2} \times 2}{12\sqrt{2}}$

$= \frac{3\sqrt{2} + 25\sqrt{2} + 8\sqrt{2}}{12\sqrt{2}}$

$= \frac{36\sqrt{2}}{12\sqrt{2}} = 3$

So, the incenter is $(2,3)$.

Now, the radius $r$ is given by:

$$r = \frac{\text{Area}}{s}$$
where $s = \frac{a + b + c}{2} = \frac{12\sqrt{2}}{2} = 6\sqrt{2}$

Compute the area of the triangle using the determinant formula:

$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Plug in the values:

$x_1 = -2$, $y_1 = 1$

$x_2 = 2$, $y_2 = 5$

$x_3 = 5$, $y_3 = 2$

$= \frac{1}{2} | -2(5-2) + 2(2-1) + 5(1-5) |$

$= \frac{1}{2} | -2 \times 3 + 2 \times 1 + 5 \times (-4) |$

$= \frac{1}{2} | -6 + 2 - 20 |$

$= \frac{1}{2} | -24 | = 12$

So, $\text{Area} = 12$

Therefore,

$r = \frac{12}{6\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Thus, the equation of the inscribed circle is:

$$(x-2)^2 + (y-3)^2 = 2$$