Question

the equation for a circle is ( x^{2}+10x + y^{2}+12y + 52 = 0 ).
what is the equation of the circle in standard form?
( circ(x + 5)^{2}+(y + 6)^{2}=9 )
( circ(x + 5)^{2}+(y + 6)^{2}=36 )
( circ(x + 25)^{2}+(y + 36)^{2}=9 )
( circ(x + 25)^{2}+(y + 36)^{2}=36 )

Explanation:

Step1: Group x and y terms

$$x^{2}+10x + y^{2}+12y+52 = 0$$
$$(x^{2}+10x)+(y^{2}+12y)= - 52$$

Step2: Complete the square for x - terms

For \(x^{2}+10x\), using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2ab = 10x\), so \(b = 5\). Then \(x^{2}+10x=(x + 5)^{2}-25\)

Step3: Complete the square for y - terms

For \(y^{2}+12y\), using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a = y\), \(2ab=12y\), so \(b = 6\). Then \(y^{2}+12y=(y + 6)^{2}-36\)

Step4: Substitute back into the equation

\((x + 5)^{2}-25+(y + 6)^{2}-36=-52\)
\((x + 5)^{2}+(y + 6)^{2}-61=-52\)
\((x + 5)^{2}+(y + 6)^{2}=-52 + 61\)
\((x + 5)^{2}+(y + 6)^{2}=9\)

Answer:

\((x + 5)^{2}+(y + 6)^{2}=9\) (the first option)