1. **State the problem:** Find the general form of the equation of the circle with diameter endpoints at (5, -6) and (-3, -6). 2. **Formula and rules:** The general form of a circle's equation is $$x^2 + y^2 + Dx + Ey + F = 0$$ where $(h, k)$ is the center and $r$ is the radius. 3. **Find the center:** The center is the midpoint of the diameter endpoints. $$h = \frac{5 + (-3)}{2} = \frac{2}{2} = 1$$ $$k = \frac{-6 + (-6)}{2} = \frac{-12}{2} = -6$$ So, center is $(1, -6)$. 4. **Find the radius:** Radius is half the distance between the endpoints. Distance between points: $$d = \sqrt{(5 - (-3))^2 + (-6 - (-6))^2} = \sqrt{8^2 + 0^2} = 8$$ Radius: $$r = \frac{d}{2} = 4$$ 5. **Write the standard form:** $$(x - 1)^2 + (y + 6)^2 = 4^2 = 16$$ 6. **Expand to general form:** $$(x - 1)^2 + (y + 6)^2 = 16$$ $$x^2 - 2x + 1 + y^2 + 12y + 36 = 16$$ $$x^2 + y^2 - 2x + 12y + 37 = 0$$ **Final answer:** The general form of the circle is $$x^2 + y^2 - 2x + 12y + 37 = 0$$. This matches the circle centered at (1, -6) with radius 4 passing through (5, -6) and (-3, -6).