Find an equation of the circle that has center −2, 3 and passes through −−2, 6.

The equation of a circle is given by (x - h)² + (y - k)² = r², where (h, k) is the center of the circle and r is the radius.

Given that the center of the circle is (-2, 3), we can substitute h = -2 and k = 3 into the equation to get (x - (-2))² + (y - 3)² = r², which simplifies to (x + 2)² + (y - 3)² = r².

We also know that the circle passes through the point (-2, 6). We can substitute x = -2 and y = 6 into the equation to find the radius:

((-2 + 2)² + (6 - 3)² = r²) simplifies to (0² + 3² = r²), which further simplifies to (0 + 9 = r²), so r² = 9.

Substituting r² = 9 back into the equation of the circle gives us the final equation of the circle:

(x + 2)² + (y - 3)² = 9.