Question

write the equation in standard form for the circle $-10x=-x^{2}-y^{2}-24$.

Explanation:

Step1: Rearrange the equation

Move all terms to one side:
$$x^{2}-10x + y^{2}+24 = 0$$

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

For the \(x\) - terms \(x^{2}-10x\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), where \(a = x\) and \(2ab=10x\) (so \(b = 5\)).
\(x^{2}-10x=(x - 5)^{2}-25\).
The equation becomes \((x - 5)^{2}-25+y^{2}+24 = 0\).

Step3: Simplify the equation

Simplify \((x - 5)^{2}-25+y^{2}+24 = 0\) to \((x - 5)^{2}+y^{2}-1 = 0\).
Then \((x - 5)^{2}+y^{2}=1\).

Answer:

\((x - 5)^{2}+y^{2}=1\)