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.

$x^{2}+y^{2}+4x - 2y-31 = 0$

the equation of the circle in standard form is (square).
(simplify your answer.)

Explanation:

Step1: Group x - terms and y - terms

$(x^{2}+4x)+(y^{2}-2y)=31$

Step2: Complete the square for x - terms

For the expression $x^{2}+4x$, add $(\frac{4}{2})^{2}=4$ to both sides of the equation.
$(x^{2}+4x + 4)+(y^{2}-2y)=31 + 4$

Step3: Complete the square for y - terms

For the expression $y^{2}-2y$, add $(\frac{-2}{2})^{2}=1$ to both sides of the equation.
$(x^{2}+4x + 4)+(y^{2}-2y+1)=31 + 4+1$

Step4: Write in standard form

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$.
$(x + 2)^{2}+(y - 1)^{2}=36$

Answer:

$(x + 2)^{2}+(y - 1)^{2}=36$