Sure! Let's tackle the problem of finding the center and radius of the circle given the equation [tex]\(x^2 + y^2 + 4x - 6y + 4 = 0\)[/tex]. Here is the step-by-step solution:
1. Given equation:
[tex]\[
x^2 + y^2 + 4x - 6y + 4 = 0
\][/tex]
2. Rearrange the terms (grouping [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms together):
[tex]\[
x^2 + 4x + y^2 - 6y = -4
\][/tex]
3. Complete the square for the [tex]\(x\)[/tex]-terms.
Take the [tex]\(x\)[/tex]-terms: [tex]\(x^2 + 4x\)[/tex]
To complete the square, add and subtract [tex]\((\frac{4}{2})^2 = 4\)[/tex]:
[tex]\[
x^2 + 4x = (x + 2)^2 - 4
\][/tex]
4. Complete the square for the [tex]\(y\)[/tex]-terms.
Take the [tex]\(y\)[/tex]-terms: [tex]\(y^2 - 6y\)[/tex]
To complete the square, add and subtract [tex]\((\frac{-6}{2})^2 = 9\)[/tex]:
[tex]\[
y^2 - 6y = (y - 3)^2 - 9
\][/tex]
5. Substitute back the completed squares into the rearranged equation:
[tex]\[
(x + 2)^2 - 4 + (y - 3)^2 - 9 = -4
\][/tex]
6. Combine constant terms:
[tex]\[
(x + 2)^2 + (y - 3)^2 - 13 = -4
\][/tex]
[tex]\[
(x + 2)^2 + (y - 3)^2 = 9
\][/tex]
7. Equation of a circle:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
Comparing the resulting equation [tex]\((x + 2)^2 + (y - 3)^2 = 9\)[/tex] to the general form of a circle's equation, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we can find [tex]\(h\)[/tex], [tex]\(k\)[/tex], and [tex]\(r\)[/tex]:
- Center: [tex]\((h, k) = (-2, 3)\)[/tex]
- Radius: [tex]\(r = \sqrt{9} = 3\)[/tex]
Therefore, the center of the circle is [tex]\((-2, 3)\)[/tex] and the radius is [tex]\(3\)[/tex].
