Question

given a circle with the equation $x^{2}+y^{2}+8x + 2y+1 = 0$, what is the coordinate of its center and the length of its radius? enter your answers in the boxes to complete the equation. be sure to add an addition or subtraction symbol where needed. the coordinate of the circles center is ( , ). the radius of the circle is units.

Explanation:

Step1: Rewrite the circle equation in standard form

The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius. Given $x^{2}+y^{2}+8x + 2y+1 = 0$. Complete the square for $x$ - terms and $y$ - terms.
For the $x$ - terms: $x^{2}+8x=(x + 4)^{2}-16$. For the $y$ - terms: $y^{2}+2y=(y + 1)^{2}-1$.
So the equation becomes $(x + 4)^{2}-16+(y + 1)^{2}-1+1=0$, which simplifies to $(x + 4)^{2}+(y + 1)^{2}=16$.

Step2: Identify the center and radius

From the standard - form equation $(x + 4)^{2}+(y + 1)^{2}=16=(x-(-4))^{2}+(y - (-1))^{2}=4^{2}$, the center of the circle is $(-4,-1)$ and the radius $r = 4$.

Answer:

The coordinate of the circle's center is $(-4,-1)$.
The radius of the circle is $4$ units.