Question 636·Hard·Math

A tangent line touches the circle once. Both given points are above the $x$-axis, so the circle’s center must be above it too. The center’s height equals the radius when the circle touches the $x$-axis from above. Call the radius $r$ and the center $(h,r)$. The touchpoint is $(h,0)$, so its positive $x$-coordinate gives $h>0$; also $r>0$.

Type $\operatorname{midpoint}((-1,2),(13,16))$ in Desmos and click the plotted point. It shows $(6,9)$, the midpoint of the segment joining the given points. They aren’t stated to be diameter endpoints, so this point isn’t necessarily the circle’s center.

Keep the midpoint line and type $13-(-1)$ and $16-2$. Desmos prints $14$ for each. The segment goes the same distance right and up, so its slope, or rise per unit right, is $1$.

The perpendicular bisector runs through $(6,9)$ at a right angle to that segment. The center lies on it because it’s equally far from both given points. Its slope is $-1$: every move right is matched by a move down, so $x+y$ stays constant. At the midpoint, that sum is $6+9$. For the center $(h,r)$, the same sum gives:

Add the midpoint coordinates in $h+r=6+9$:

Subtract $r$ from both sides of $h+r=15$, leaving one unknown for the circle equation:

The radius is positive, and $h=15-r$ must also be positive. So the allowed range is:

Without this condition, a circle touching the negative side of the $x$-axis could fit the two points.

The radius is the distance from the center to $(-1,2)$. From $(h,r)$, the horizontal change is $h+1$ and the vertical change is $r-2$. Square those changes and add them to get the radius squared:

Substitute $h=15-r$:

Type $(16-x_1)^2+(x_1-2)^2\sim x_1^2\{0<x_1<15\}$. Here $x_1$ stands for $r$; $\sim$ asks Desmos to find a value that fits, and the braces keep it in the allowed range. Under PARAMETERS, Desmos shows $x_1=10$. Then $h=15-10=5>0$, so the touchpoint meets the condition. The circle’s radius is $10$. Grid in 10.