Question

question 8 of 10
what is the value of a for the following circle in general form?
$x^{2}+y^{2}+ax + by + c = 0$

Explanation:

Step1: Find the center - radius form

First, find the center and radius of the circle from the graph. The center of the circle is at $(- 4,2)$ and the radius $r = 2$. 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. Substituting $h=-4,k = 2,r = 2$ gives $(x + 4)^2+(y - 2)^2=4$.

Step2: Expand the standard - form

Expand $(x + 4)^2+(y - 2)^2=4$.
$(x + 4)^2=x^{2}+8x + 16$ and $(y - 2)^2=y^{2}-4y + 4$.
So $x^{2}+8x + 16+y^{2}-4y + 4=4$.

Step3: Convert to general form

Rearrange the equation $x^{2}+8x + 16+y^{2}-4y + 4=4$ to the general form $x^{2}+y^{2}+ax+by + c = 0$.
$x^{2}+y^{2}+8x-4y+16 = 0$.

Answer:

$8$