Question 736·Hard·Math

A tangent line touches a circle once. The segment from the center to that touchpoint is perpendicular to the line and has length $r$, the radius. The $y$-axis is vertical, so the center has height $2$ and is $r$ units horizontally from $(0,2)$. It is either $(r,2)$ or $(-r,2)$. Tangency fixes the center’s height and its distance from the axis.

The center cannot be $(-r,2)$. From there, the horizontal distance alone to $(3,6)$ would be $3+r$, greater than the radius $r$. That point couldn’t lie on the circle, so the center is $(r,2)$.

From the center $(r,2)$ to $(3,6)$, the horizontal change is $3-r$ and the vertical change is $6-2=4$. The Pythagorean theorem says their squares add to the squared distance, which is $r^2$ because $(3,6)$ is on the circle: $(3-r)^2+4^2=r^2$. The segment between the two given points does not start at the center, so its length is not the radius.

Type $(3-r_1)^2+4^2\sim r_1^2$ in Desmos. The subscript tells Desmos to fit a number rather than graph a variable. Under PARAMETERS, it shows $r_1\approx4.1666667$. Type $r_1$ on the next line and use its fraction button to see $\frac{25}{6}$. Since $r_1$ is the radius, the circle’s radius is $\frac{25}{6}$. Choice C.