To address this problem, let's begin by solving the questions in the image step-by-step. The given problem asks to convert the provided general equations of circles into their standard form.

Standard Form of a Circle

The equation of a circle in standard form is: $(x - h)^2 + (y - k)^2 = r^2$ Where:

• $(h, k)$ is the center of the circle.
• $r$ is the radius.

The general equation of a circle is typically given as: $x^2 + y^2 + Dx + Ey + F = 0$ To convert this to standard form, we complete the square for both $x$ and $y$.

Let's analyze each equation from the image:

(1) Equation A:

$x^2 + y^2 + 6x - 6y - 31 = 0$

1. Rearrange and group the terms: $(x^2 + 6x) + (y^2 - 6y) = 31$
2. Complete the square for both $x$ and $y$:
   • For $x^2 + 6x$, add and subtract $(\frac{6}{2})^2 = 9$.
   • For $y^2 - 6y$, add and subtract $(\frac{6}{2})^2 = 9$. $(x^2 + 6x + 9) + (y^2 - 6y + 9) = 31 + 9 + 9$ $(x + 3)^2 + (y - 3)^2 = 49$
3. This is now in standard form: $(x + 3)^2 + (y - 3)^2 = 7^2$ The center is $(-3, 3)$, and the radius is $7$.

The answer corresponds to choice C.

(2) Equation B:

$x^2 + y^2 + 4x + 4y + 9 = 0$

1. Rearrange and group the terms: $(x^2 + 4x) + (y^2 + 4y) = -9$
2. Complete the square:
   • For $x^2 + 4x$, add and subtract $(\frac{4}{2})^2 = 4$.
   • For $y^2 + 4y$, add and subtract $(\frac{4}{2})^2 = 4$. $(x^2 + 4x + 4) + (y^2 + 4y + 4) = -9 + 4 + 4$ $(x + 2)^2 + (y + 2)^2 = -1$ This results in a negative value on the right side, meaning the equation does not represent a circle (this equation does not correspond to any valid circle).

Would you like to continue with the rest of the problems or need further explanation on this part?

Here are 5 related questions:

1. How do you derive the center and radius from the standard form equation of a circle?
2. What happens when the right side of the equation results in a negative value?
3. Why is completing the square necessary to convert