Find the general form of the equation of a circle with center (2, -1) passing through point (-6, 5).

The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Given the center D is $(2, -1)$, so $h = 2$ and $k = -1$.
Substituting these values, the equation becomes:

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

The circle passes through point P $(-6, 5)$. The distance between the center D and point P is the radius $r$. We can use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ or directly calculate $r^2$ using the points.

$$r^2 = (-6 - 2)^2 + (5 - (-1))^2$$ $$r^2 = (-8)^2 + (5 + 1)^2$$ $$r^2 = (-8)^2 + (6)^2$$ $$r^2 = 64 + 36$$ $$r^2 = 100$$

Expand $(x - 2)^2$:

$$(x - 2)^2 = x^2 - 2(x)(2) + 2^2 = x^2 - 4x + 4$$

Expand $(y + 1)^2$:

$$(y + 1)^2 = y^2 + 2(y)(1) + 1^2 = y^2 + 2y + 1$$

Substitute these expanded forms back into the equation:

$$(x^2 - 4x + 4) + (y^2 + 2y + 1) = 100$$

Combine like terms and move the constant to the left side:

$$x^2 + y^2 - 4x + 2y + 4 + 1 - 100 = 0$$ $$x^2 + y^2 - 4x + 2y - 95 = 0$$