Find the equation of the circle with center at $(-7,-2)$ and passing through the point $(-3,6)$.
A) $x^{2}+y^{2}-6 x+12 y-35=0$
B) $x^{2}+y^{2}+6 x-12 y-35=0$
C) $x^{2}+y^{2}+14 x+4 y-27=0$
D) $x^{2}+y^{2}-14 x-4 y-27=0$

Question

Quizplus · Accepted Answer

The equation of a circle with center $(h, k)$ is $(x-h)^2 + (y-k)^2 = r^2$. Here, $h = -7$ and $k = -2$. The radius $r$ can be found using the distance formula between the center and the point $(-3, 6)$: $r = \sqrt{(-3 + 7)^2 + (6 + 2)^2} = \sqrt{16 + 64} = \sqrt{80}$. Thus, the equation is $(x+7)^2 + (y+2)^2 = 80$, which expands to $x^2 + y^2 + 14x + 4y - 27 = 0$.

Find the Equation of the Circle with Center At (−7,−2)(-7,-2)(−7,−2) And Passing Through the Point

Find the Equation of the Circle with Center At $(-7,-2)$ And Passing Through the Point