Question

write the equation in standard form for the circle ( x^{2}=-y^{2}+8x - 12 ).

Explanation:

Step1: Rearrange the equation

$$x^{2}+y^{2}-8x + 12=0$$

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

For \(x^{2}-8x\), we have \((x - 4)^{2}=x^{2}-8x + 16\).
So, \(x^{2}-8x=(x - 4)^{2}-16\).
The equation becomes \((x - 4)^{2}-16+y^{2}+12 = 0\).

Step3: Simplify the equation

$$(x - 4)^{2}+y^{2}-4=0$$
$$(x - 4)^{2}+y^{2}=4$$

Answer:

\((x - 4)^{2}+y^{2}=4\)