The question asks to find the radius and center of the circle from the equation provided, using the method of completing the square.

The equation given is:

$x^2 + y^2 + 4x - 6y + 12 = 0$

Steps to find the center and radius:

1. Rearrange the equation: $x^2 + 4x + y^2 - 6y = -12$ Move the constant to the other side of the equation.

2. Complete the square:

   • For the x terms: $x^2 + 4x$
     • Take half of 4 (which is 2), square it to get $2^2 = 4$. Add 4 to both sides.
   • For the y terms: $y^2 - 6y$
     • Take half of -6 (which is -3), square it to get $(-3)^2 = 9$. Add 9 to both sides.

   Adding 4 and 9 to both sides gives: $(x^2 + 4x + 4) + (y^2 - 6y + 9) = -12 + 4 + 9$ Simplifies to: $(x + 2)^2 + (y - 3)^2 = 1$

3. Identify the center and radius:

   • The equation is now in the standard form of a circle: $(x - h)^2 + (y - k)^2 = r^2$
   • The center is $(-2, 3)$ and the radius is $\sqrt{1} = 1$.

Final answer:

• Center: $(-2, 3)$
• Radius: $1$

Would you like further details on completing the square, or have any questions?

Here are 5 related questions to expand your understanding:

1. How do you identify the general form of a circle equation?
2. Why do we add and subtract the same value when completing the square?
3. What is the significance of the center and radius in the geometry of a circle?
4. How would the equation change if the radius was larger, say 5?
5. Can completing the square be used for any quadratic equation, and how?

Tip: Always verify your final equation by expanding the completed square form to check if it matches the original equation.