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 (x + 2)^2 + (y - 1)^2 = 36. (simplify your answer.) the center of the circle is (-2,1). (type an ordered pair.) the radius of the circle is r = 6. use the graphing tool to graph the circle. click to enlarge graph

Explanation:

Step1: Group x - terms and y - terms

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

Step2: Complete the square for x - terms

For $x^{2}+4x$, add $(\frac{4}{2})^{2}=4$ inside the first parenthesis.
$(x^{2}+4x + 4)+(y^{2}-2y)-31=4$

Step3: Complete the square for y - terms

For $y^{2}-2y$, add $(\frac{-2}{2})^{2}=1$ inside the second parenthesis.
$(x^{2}+4x + 4)+(y^{2}-2y + 1)-31=4 + 1$

Step4: Rewrite in standard form

$(x + 2)^{2}+(y - 1)^{2}=36$
The standard - form of a circle is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $(h,k)$ is the center and $r$ is the radius.
Comparing $(x + 2)^{2}+(y - 1)^{2}=36$ with $(x - h)^{2}+(y - k)^{2}=r^{2}$, we have:

Step5: Find the center

$h=-2,k = 1$, so the center is $(-2,1)$.

Step6: Find the radius

$r^{2}=36$, so $r = 6$.

Answer:

The equation of the circle in standard form is $(x + 2)^{2}+(y - 1)^{2}=36$.
The center of the circle is $(-2,1)$.
The radius of the circle is $r = 6$.