Circles SAT question

Choice C is correct. The graph of the equation ${(x-h)}^2+{(y-k)}^2=r^2$ in the xy-plane is a circle with center $(h,k)$ 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 ${(x-h)}^2+{(y-k)}^2=r^2$ is a circle that intersects the y-axis at exactly one point, then $r=|h|$ must be true. The equation in choice C is ${(x-4)}^2+{(y-9)}^2=16$, or ${(x-4)}^2+{(y-9)}^2=4^2$. This equation is in the form ${(x-h)}^2+{(y-k)}^2=r^2$, where $h=4$, $k=9$, and $r=4$, and represents a circle in the xy-plane with center $(4,9)$ and radius of length $4$. Substituting $4$ for $r$ and $4$ for $h$ in the equation $r=|h|$ yields $4=|4|$, 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. 

Choice A is incorrect. This is the equation of a circle that does not intersect the y-axis at any point.

Choice B is incorrect. This is an equation of a circle that intersects the x-axis, not the y-axis, at exactly one point.

Choice D is incorrect. This is the equation of a circle with the center located on the y-axis and thus intersects the y-axis at exactly two points, not exactly one point.