Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation.
$x^{2}+y^{2}+4x - 4y-17 = 0$
the equation of the circle in standard form is
(simplify your answer.)

Explanation:

Step1: Group x and y terms

$(x^{2}+4x)+(y^{2}-4y)-17 = 0$

Step2: Complete the square for x - terms

For $x^{2}+4x$, add $(\frac{4}{2})^{2}=4$. So we have $(x^{2}+4x + 4)+(y^{2}-4y)-17=4$.

Step3: Complete the square for y - terms

For $y^{2}-4y$, add $(\frac{-4}{2})^{2}=4$. So $(x^{2}+4x + 4)+(y^{2}-4y + 4)-17=4 + 4$.

Step4: Rewrite in standard form

$(x + 2)^{2}+(y - 2)^{2}=25$

Answer:

$(x + 2)^{2}+(y - 2)^{2}=25$