SAT Question 9d159400: Circles answer and explanation with Desmos solution

Choice C is correct. The graph of the equation ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ in the xy-plane is a circle with center $\left(h,k\right)$ and a radius of length $r$. The radius of a circle is the distance from the center of the circle to any point on the circle. If a circle in the xy-plane intersects the y-axis at exactly one point, then the perpendicular distance from the center of the circle to this point on the y-axis must be equal to the length of the circle's radius. It follows that the x-coordinate of the circle's center must be equivalent to the length of the circle's radius. In other words, if the graph of ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ is a circle that intersects the y-axis at exactly one point, then $r=\left|h\right|$ must be true. The equation in choice C is ${\left(x-4\right)}^2+{\left(y-9\right)}^2=16$, or ${\left(x-4\right)}^2+{\left(y-9\right)}^2=4^2$. This equation is in the form ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where $h=4$, $k=9$, and $r=4$, and represents a circle in the xy-plane with center $\left(4,9\right)$ and radius of length $4$. Substituting $4$ for $r$ and $4$ for $h$ in the equation $r=\left|h\right|$ yields $4=\left|4\right|$, or $4=4$, which is true. Therefore, the equation in choice C represents a circle in the xy-plane that intersects the y-axis at exactly one point.