Find the equation of the circle that passes through the origin and has its center at (0, 9). A) x ² + y ² + 18 x = 0 B) x 2 + y 2 $-$ 18 x $-$ 81 = 0 C) x ² + y ² + 18 y = 0 D) x 2 $-$ 18 x + y 2 = 0 E) x 2 + y 2 $-$ 18 y = 0

Question

Quizplus · Accepted Answer

The general equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where (h, k) is the center and r is the radius. Here, the center is (0, 9), so $h = 0$ and $k = 9$. Since the circle passes through the origin (0,0), we can use this point to find the radius: $r^2 = (0 - 0)^2 + (0 - 9)^2 = 81$. Substituting $h = 0$, $k = 9$, and $r^2 = 81$ into the general equation gives $x^2 + (y - 9)^2 = 81$, which simplifies to $x^2 + y^2 - 18y + 81 = 81$. Subtracting 81 from both sides gives the equation $x^2 + y^2 - 18y = 0$.

Find the Equation of the Circle That Passes Through the Origin