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The general equation of a circle is given by:

\[
x^2 + y^2 + 2Ax + 2By + C = 0
\]

where \((A, B)\) represents the coordinates of the center of the circle \((-A, -B)\), and the radius \(r\) is related to \(C\) by the formula:

\[
r = \sqrt{A^2 + B^2 - C}
\]

Now, the equation in the image is:

\[
x^2 + y^2 + 4x - 6y + q = 0
\]

We need to rewrite it to match the standard form. First, rewrite the coefficients:

- \(2A = 4 \implies A = 2\)
- \(2B = -6 \implies B = -3\)

So, the center of the circle is \((-2, 3)\). The radius \(r = 5\), as given in the problem.

Using the formula for the radius:

\[
r = \sqrt{A^2 + B^2 - C}
\]

Substitute the values:

\[
5 = \sqrt{2^2 + (-3)^2 - q}
\]
\[
5 = \sqrt{4 + 9 - q}
\]
\[
5 = \sqrt{13 - q}
\]

Now square both sides:

\[
25 = 13 - q
\]

Solve for \(q\):

\[
q = 13 - 25 = -12
\]

So, the value of \(q\) is \(-12\).

Would you like more details on this? Here are five related questions to explore this further:

1. How do we complete the square to derive the center and radius from a circle equation?
2. What is the geometric significance of the equation of a circle in different forms?
3. How do we handle cases where the radius of the circle is not provided?
4. Can we generalize the process of finding the center and radius for any conic sections?
5. How does this method differ for equations of ellipses or hyperbolas?

**Tip:** Always ensure to match the equation to the standard form of the circle before attempting to solve for parameters like radius or center.

Jika persamaan umum lingkaran x^2 + y^2 + 4x - 6y + q = 0 memiliki jari-jari 5, nilai dari q adalah ....

Solve the circle equation x^2 + y^2 + 4x - 6y + q = 0 to find the value of q, given that the radius is 5. This problem involves the equation of a circle, algebraic manipulation, and the use of the formula for the radius. Suitable for Grades 10-12.

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