To find the center and radius of the circle given by the equation [tex]\( x^2 + y^2 + 3x - 12y = 0 \)[/tex], we follow these steps:
1. Rewrite the equation in the standard form of a circle's equation:
The general form of the equation of a circle is:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
2. Complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
The given equation is:
[tex]\[
x^2 + y^2 + 3x - 12y = 0
\][/tex]
Rearrange the equation to group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[
x^2 + 3x + y^2 - 12y = 0
\][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms:
For the [tex]\(x\)[/tex] terms [tex]\(x^2 + 3x\)[/tex]:
- Take half the coefficient of [tex]\(x\)[/tex] (which is 3), square it, and add and subtract that value inside the equation. Half of 3 is [tex]\(\frac{3}{2}\)[/tex], and [tex]\(\left(\frac{3}{2}\right)^2 = \frac{9}{4}\)[/tex]:
[tex]\[
x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 = x^2 + 3x + \frac{9}{4} - \frac{9}{4}
\][/tex]
- This results in:
[tex]\[
\left(x + \frac{3}{2}\right)^2 - \frac{9}{4}
\][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms:
For the [tex]\(y\)[/tex] terms [tex]\(y^2 - 12y\)[/tex]:
- Take half the coefficient of [tex]\(y\)[/tex] (which is -12), square it, and add and subtract that value inside the equation. Half of -12 is -6, and [tex]\((-6)^2 = 36\)[/tex]:
[tex]\[
y^2 - 12y + 36 - 36 = (y - 6)^2 - 36
\][/tex]
5. Combine the completed squares back into the equation:
Substitute the completed squares back into the original equation:
[tex]\[
\left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + (y - 6)^2 - 36 = 0
\][/tex]
Simplify and combine constants:
[tex]\[
\left(x + \frac{3}{2}\right)^2 + (y - 6)^2 = \frac{9}{4} + 36 = \frac{9 + 144}{4} = \frac{153}{4} = 38.25
\][/tex]
6. Determine the center and the radius:
The standard form is now:
[tex]\[
(x + \frac{3}{2})^2 + (y - 6)^2 = 38.25
\][/tex]
From this equation:
- The center [tex]\((h, k)\)[/tex] of the circle is:
[tex]\[
h = -\frac{3}{2} = -1.5, \quad k = 6
\][/tex]
- The radius [tex]\(r\)[/tex] is:
[tex]\[
r = \sqrt{38.25} \approx 6.18
\][/tex]
Therefore, the center of the circle is [tex]\((-1.50, 6)\)[/tex] and the radius of the circle is approximately [tex]\(6.18\)[/tex].
