Question

give the center and radius of the circle represented by the equation. x² + y² + 12x + 2y + 33 = 0 the center of the circle is □. (type an ordered pair, using integers or fractions.) the radius of the circle is □. (type an integer or a simplified fraction.)

Explanation:

Step1: Rearrange and complete the square for \(x\) terms

Group \(x\) - terms and \(y\) - terms: \((x^{2}+12x)+(y^{2}+2y)= - 33\).
For \(x^{2}+12x\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = x\) and \(2ab=12x\) (so \(b = 6\)). Then \(x^{2}+12x=(x + 6)^{2}-36\).

Step2: Complete the square for \(y\) terms

For \(y^{2}+2y\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = y\) and \(2ab = 2y\) (so \(b = 1\)). Then \(y^{2}+2y=(y + 1)^{2}-1\).

Step3: Substitute back into the equation

Substitute into \((x^{2}+12x)+(y^{2}+2y)=-33\):
\((x + 6)^{2}-36+(y + 1)^{2}-1=-33\).
Simplify the equation: \((x + 6)^{2}+(y + 1)^{2}-37=-33\).
Then \((x + 6)^{2}+(y + 1)^{2}=4\).

Answer:

The center of the circle is \((-6,-1)\).
The radius of the circle is \(2\).