Question 9·Hard·Circles

The radius is the vertical distance from $(t,k)$ to the $x$-axis, which is $|k|$. In the distance formula for a point on the circle, the $k^2$ terms will cancel.

Write one equation using each given point. One equation has $4k$ and the other has $32k$, so multiply the first equation by $8$ before setting the left sides equal.

Algebra is the fastest method because tangency gives a direct relationship between the center's $y$-coordinate and the radius. Desmos can quickly verify the value of $t$ after eliminating $k$.

Enter $y=(x-13)^2+256$ and $y=8((x+1)^2+4)$. Their intersections represent the possible values of $t$.

Choose the intersection with positive $x$-coordinate. Then enter $((5+1)^2+4)/4$ to evaluate $k$, whose absolute value is the radius.

Let the center be $(t,k)$. Because the circle is tangent to the $x$-axis, its radius is $|k|$, and its point of tangency has $x$-coordinate $t$.

For any point $(x,y)$ on the circle,

$(x-t)^2+(y-k)^2=k^2$.

After expanding and canceling $k^2$, this becomes $(x-t)^2+y^2=2yk$.

Substitute $(-1,2)$:

$(t+1)^2+4=4k$.

Substitute $(13,16)$:

$(t-13)^2+256=32k$.

Since $32k=8(4k)$, set

$(t-13)^2+256=8((t+1)^2+4)$.

Simplifying gives $t^2+6t-55=0$, so $(t-5)(t+11)=0$. Thus, $t=5$ or $t=-11$.

The tangency point has a positive $x$-coordinate, so $t=5$.

Substitute $t=5$ into $(t+1)^2+4=4k$:

$6^2+4=4k$.

Thus, $k=10$. The radius is $|k|$, so it is $10$.