Question

on a coordinate grid, a circle has a center at (-2, -4) and passes through (0, -3). what is the equation of the circle?

a) (x^{2}+y^{2}-4x - 8y + 15=0)
b) (x^{2}+y^{2}+4x + 8y + 15=0)
c) (x^{2}+y^{2}-6y + 4=0)
d) (x^{2}+y^{2}-6y + 4=0)

Explanation:

Step1: Recall the standard - form of the circle equation

The standard - form of a circle equation is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center of the circle and $r$ is the radius. Given the center $(a,b)=(-2,-4)$, the equation is $(x + 2)^2+(y + 4)^2=r^2$.

Step2: Find the radius

Since the circle passes through the point $(0,-3)$, we substitute $x = 0$ and $y=-3$ into the equation $(x + 2)^2+(y + 4)^2=r^2$. So, $(0 + 2)^2+(-3 + 4)^2=r^2$. Then $4 + 1=r^2$, and $r^2 = 5$.

Step3: Expand the circle equation

Expand $(x + 2)^2+(y + 4)^2=5$. Using the formula $(m + n)^2=m^2+2mn + n^2$, we have $x^2+4x + 4+y^2+8y+16 = 5$.

Step4: Simplify the equation

Combine like - terms: $x^2+y^2+4x + 8y+4 + 16-5=0$, which simplifies to $x^2+y^2+4x + 8y+15 = 0$.

Answer:

B. $x^{2}+y^{2}+4x + 8y+15 = 0$