Question

write the equation in standard form for the circle ( y^{2}+10y + 34=-x^{2}-10x ).

Explanation:

Step1: Rearrange the equation

First, move all terms to one side of the equation:
$$y^{2}+10y + 34+x^{2}+10x = 0$$
Then, group the \(x\) - terms and \(y\) - terms:
$$(x^{2}+10x)+(y^{2}+10y)+34 = 0$$

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

For the \(x\) - terms \(x^{2}+10x\), using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a = x\) and \(2b=10\) (so \(b = 5\)).
\(x^{2}+10x=(x + 5)^{2}-25\)

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

For the \(y\) - terms \(y^{2}+10y\), using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a = y\) and \(2b = 10\) (so \(b=5\)).
\(y^{2}+10y=(y + 5)^{2}-25\)

Step4: Substitute back into the equation

Substitute \((x + 5)^{2}-25\) for \(x^{2}+10x\) and \((y + 5)^{2}-25\) for \(y^{2}+10y\) in the equation \((x^{2}+10x)+(y^{2}+10y)+34 = 0\):

$$
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Answer:

\((x + 5)^{2}+(y + 5)^{2}=16\)