We are given three points $A(5, 2)$, $B(-1, 0)$, and $C(3, -2)$. (1) We need to find the equation of the circle passing through points A, B, and C. (2) We need to find the coordinates of the circumcenter of triangle ABC and the radius of the circumcircle.

1. Problem Description

We are given three points $A(5, 2)$, $B(-1, 0)$, and $C(3, -2)$.

(1) We need to find the equation of the circle passing through points A, B, and C.

(2) We need to find the coordinates of the circumcenter of triangle ABC and the radius of the circumcircle.

2. Solution Steps

(1) Let the equation of the circle be $x^2 + y^2 + ax + by + c = 0$.

Since the points A, B, and C lie on the circle, we can substitute their coordinates into the equation:

For A(5, 2):

$5^2 + 2^2 + 5a + 2b + c = 0 \implies 25 + 4 + 5a + 2b + c = 0 \implies 5a + 2b + c = -29$ (1)

For B(-1, 0):

$(-1)^2 + 0^2 - a + 0b + c = 0 \implies 1 - a + c = 0 \implies -a + c = -1$ (2)

For C(3, -2):

$3^2 + (-2)^2 + 3a - 2b + c = 0 \implies 9 + 4 + 3a - 2b + c = 0 \implies 3a - 2b + c = -13$ (3)

From (1) and (3), adding the two equations we get:

$5a + 2b + c = -29$

$3a - 2b + c = -13$

$8a + 2c = -42 \implies 4a + c = -21$ (4)

From (2), $c = a - 1$. Substituting this into (4),

$4a + a - 1 = -21 \implies 5a = -20 \implies a = -4$

Then, $c = a - 1 = -4 - 1 = -5$.

Substituting $a = -4$ and $c = -5$ into (1),

$5(-4) + 2b + (-5) = -29 \implies -20 + 2b - 5 = -29 \implies 2b = -4 \implies b = -2$

Therefore, the equation of the circle is $x^2 + y^2 - 4x - 2y - 5 = 0$.

(2) Completing the square:

$(x^2 - 4x) + (y^2 - 2y) - 5 = 0$

$(x^2 - 4x + 4) + (y^2 - 2y + 1) - 5 - 4 - 1 = 0$

$(x - 2)^2 + (y - 1)^2 = 10$

The center of the circle is $(2, 1)$ and the radius is $\sqrt{10}$.

Therefore, the coordinates of the circumcenter of triangle ABC are $(2, 1)$ and the radius of the circumcircle is $\sqrt{10}$.

3. Final Answer

(1) The equation of the circle is $x^2 + y^2 - 4x - 2y - 5 = 0$.

(2) The circumcenter is $(2, 1)$ and the radius is $\sqrt{10}$.