Question

write the equation in standard form for the circle $x^{2}+y^{2}+12x + 3y+32 = 0$.

Explanation:

Step1: Group \(x\) and \(y\) terms

Group \(x\) - terms and \(y\) - terms: \((x^{2}+12x)+(y^{2}+3y)= - 32\)

Step2: Complete the square for \(x\) - terms

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 \(x^{2}+12x=(x + 6)^{2}-36\)

Step3: Complete the square for \(y\) - terms

For \(y^{2}+3y\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = y\) and \(2ab = 3y\) (so \(b=\frac{3}{2}\)). Then \(y^{2}+3y=(y+\frac{3}{2})^{2}-\frac{9}{4}\)

Step4: Substitute back into the equation

Substitute \((x + 6)^{2}-36+(y+\frac{3}{2})^{2}-\frac{9}{4}=-32\)

Step5: Simplify the equation

\((x + 6)^{2}+(y+\frac{3}{2})^{2}=-32 + 36+\frac{9}{4}\)
\((x + 6)^{2}+(y+\frac{3}{2})^{2}=4+\frac{9}{4}\)
\((x + 6)^{2}+(y+\frac{3}{2})^{2}=\frac{16 + 9}{4}\)
\((x + 6)^{2}+(y+\frac{3}{2})^{2}=\frac{25}{4}\)

Answer:

\((x + 6)^{2}+(y+\frac{3}{2})^{2}=\frac{25}{4}\)