1. **State the problem:** Convert the general form of the circle equation into center-radius form for the first equation: $$3x^2 + 3y^2 + 12x - 18y - 33 = 0$$ 2. **Divide entire equation by 3** to simplify coefficients of $x^2$ and $y^2$ to 1: $$\cancel{3}x^2 + \cancel{3}y^2 + \cancel{3} \cdot 4x - \cancel{3} \cdot 6y - \cancel{3} \cdot 11 = 0 \implies x^2 + y^2 + 4x - 6y - 11 = 0$$ 3. **Group $x$ and $y$ terms:** $$x^2 + 4x + y^2 - 6y = 11$$ 4. **Complete the square for $x$ and $y$ terms:** - For $x^2 + 4x$, half of 4 is 2, square it: $2^2 = 4$ - For $y^2 - 6y$, half of -6 is -3, square it: $(-3)^2 = 9$ Add 4 and 9 to both sides: $$x^2 + 4x + 4 + y^2 - 6y + 9 = 11 + 4 + 9$$ 5. **Rewrite as perfect squares:** $$(x + 2)^2 + (y - 3)^2 = 24$$ 6. **Final center-radius form:** Center: $(-2, 3)$ Radius: $\sqrt{24} = 2\sqrt{6}$ **Answer:** $$ (x + 2)^2 + (y - 3)^2 = 24 $$