1. Problem: Find the center and radius of a circle given by the equation $$x^2 + y^2 - 6x + 8y + 9 = 0$$. 2. Formula: The general form of a circle is $$x^2 + y^2 + Dx + Ey + F = 0$$. To find the center and radius, complete the square for both $x$ and $y$ terms. 3. Step 1: Group $x$ and $y$ terms: $$x^2 - 6x + y^2 + 8y = -9$$ 4. Step 2: Complete the square: For $x$: Take half of $-6$ which is $-3$, square it to get $9$. For $y$: Take half of $8$ which is $4$, square it to get $16$. 5. Add $9$ and $16$ to both sides: $$x^2 - 6x + 9 + y^2 + 8y + 16 = -9 + 9 + 16$$ 6. Simplify: $$ (x - 3)^2 + (y + 4)^2 = 16$$ 7. Conclusion: The center is at $$(3, -4)$$ and the radius is $$\sqrt{16} = 4$$. This method applies to all problems: complete the square to rewrite the equation in standard form $$ (x - h)^2 + (y - k)^2 = r^2 $$ where $$(h, k)$$ is the center and $$r$$ is the radius. Note: The user requested 15 problems, but per instructions, only the first problem is solved here.