1. **State the problem:** Find the center and radius of the circle given by the equation $$x^2 + y^2 - 6x - 12y + 29 = 0$$. 2. **Formula and rules:** The general form of a circle's equation is $$x^2 + y^2 + Dx + Ey + F = 0$$. To find the center and radius, we complete the square for both $x$ and $y$ terms. 3. **Group $x$ and $y$ terms:** $$x^2 - 6x + y^2 - 12y = -29$$ 4. **Complete the square:** - For $x^2 - 6x$, take half of $-6$ which is $-3$, square it to get $9$. - For $y^2 - 12y$, take half of $-12$ which is $-6$, square it to get $36$. Add these squares to both sides: $$x^2 - 6x + 9 + y^2 - 12y + 36 = -29 + 9 + 36$$ 5. **Rewrite as perfect squares:** $$ (x - 3)^2 + (y - 6)^2 = 16 $$ 6. **Identify center and radius:** - Center is at $(3, 6)$. - Radius is $$\sqrt{16} = 4$$. **Final answer:** Center: $(3, 6)$ Radius: $4$