Question 680·Hard·Math

A tangent line touches the circle once. Since the center is in the first quadrant, above the $x$-axis, the radius $r$ equals the center’s $y$-coordinate. Call its $x$-coordinate $h$. The center lies on $y=-x+10$, so: $r=-h+10$. Solve for $h$: $h=10-r$. So the center is $(10-r,r)$. Its two coordinates must be positive, giving $0<r<10$.

The circle passes through $(6,4)$, so its distance from the center is exactly $r$. The horizontal and vertical changes form a right triangle; the Pythagorean theorem says their squares add to the radius squared: $r^2=(6-(10-r))^2+(4-r)^2$. Simplify the horizontal change: $r^2=(r-4)^2+(4-r)^2$. The two squares are equal, so:

Type $2(x_1-4)^2\sim x_1^2\{0<x_1<10\}$, using $x_1$ for the radius. The $\sim$ asks Desmos to fit it, and the restriction keeps the center in the first quadrant. Under PARAMETERS, Desmos gives $x_1\approx2.343$. Type the four radical choices on separate lines; they display approximately $5.657$, $13.657$, $3.172$, and $2.343$, in that order. The matching exact radius is $8-4\sqrt2$. Choice D.