1. **State the problem:** Find the area of a square whose vertices lie on the circle given by the equation $$x^2 + y^2 - 4x + 6y + 4 = 0$$. 2. **Rewrite the circle equation in standard form:** Complete the square for both $x$ and $y$ terms. For $x$: $$x^2 - 4x = (x^2 - 4x + 4) - 4 = (x - 2)^2 - 4$$ For $y$: $$y^2 + 6y = (y^2 + 6y + 9) - 9 = (y + 3)^2 - 9$$ Substitute back: $$(x - 2)^2 - 4 + (y + 3)^2 - 9 + 4 = 0$$ Simplify constants: $$(x - 2)^2 + (y + 3)^2 - 9 = 0$$ So, $$(x - 2)^2 + (y + 3)^2 = 9$$ This is a circle with center at $(2, -3)$ and radius $r = 3$. 3. **Relate the square to the circle:** The square is inscribed in the circle, so its vertices lie on the circle. The diagonal of the square equals the diameter of the circle: $$d = 2r = 6$$ 4. **Find the side length of the square:** If $s$ is the side length, then the diagonal $d$ relates to $s$ by: $$d = s\sqrt{2}$$ So, $$s = \frac{d}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$$ 5. **Calculate the area of the square:** $$\text{Area} = s^2 = (3\sqrt{2})^2 = 9 \times 2 = 18$$ **Final answer:** The area of the square is $18$ square units.