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}+6x + 6y+17 = 0) the equation in standard form is (simplify your answer.)

Explanation:

Step1: Group x and y terms

$(x^{2}+6x)+(y^{2}+6y)+17 = 0$

Step2: Complete the square for x

For $x^{2}+6x$, add $(\frac{6}{2})^{2}=9$. So we have $(x^{2}+6x + 9)-9+(y^{2}+6y)+17 = 0$

Step3: Complete the square for y

For $y^{2}+6y$, add $(\frac{6}{2})^{2}=9$. So we get $(x^{2}+6x + 9)-9+(y^{2}+6y + 9)-9 + 17=0$

Step4: Rewrite in standard form

$(x + 3)^{2}+(y + 3)^{2}-1=0$, which simplifies to $(x + 3)^{2}+(y + 3)^{2}=1$

Answer:

$(x + 3)^{2}+(y + 3)^{2}=1$