Question

which is the equation for a circle with center at $(-2,-4)$ that passes through the point $(3,8)$?
$(x + 2)^2+(y + 4)^2 = 169$
$(x - 2)^2+(y + 4)^2 = 169$
$(x - 2)^2+(y + 4)^2 = 144$
$(x + 2)^2+(y + 4)^2 = 144$

Explanation:

Step1: Recall circle equation form

The standard circle equation is $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center. Here, center is $(-2,-4)$, so $h=-2$, $k=-4$. Substitute into the equation: $(x - (-2))^2 + (y - (-4))^2 = r^2$ → $(x+2)^2 + (y+4)^2 = r^2$.

Step2: Calculate radius squared

Use the point $(3,8)$ on the circle. Substitute $x=3$, $y=8$ into the equation to find $r^2$:
$(3+2)^2 + (8+4)^2 = r^2$ → $5^2 + 12^2 = r^2$ → $25 + 144 = r^2$ → $r^2=169$.

Step3: Form the final equation

Combine the center and $r^2$: $(x+2)^2 + (y+4)^2 = 169$.

Answer:

A. $(x + 2)^2 + (y + 4)^2 = 169$