Question 141·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

A tangent axis touches the circle at one point. Since the circle is in the first quadrant and touches both axes, its center is $r$ units from each axis, where $r$ is the radius. So the center is $(r,r)$. It is inside the square, so $0<r<20$.

The center $(r,r)$ lies on $y=x$. Type $x+y=14$ and $y=x$, then click their crossing. Desmos shows $(7,7)$. These lines are perpendicular, meaning they meet at a right angle: their slopes are $-1$ and $1$. A line from a circle’s center perpendicular to a chord cuts the chord in half, so $(7,7)$ is halfway between $A$ and $B$.

Call that distance $d$. From $(r,r)$ to $(7,7)$, both coordinate changes are $r-7$. By the Pythagorean theorem, square and add them:

Combine the equal squares:

Using only $|r-7|$ as the distance would miss one coordinate change.

The midpoint splits the chord of length $8$ into two halves of length $4$. A radius from the center to $A$ is the long side of a right triangle; its shorter sides have lengths $d$ and $4$. Apply the Pythagorean theorem:

Replace $d^2$ with the squared distance from the center to the midpoint:

Type $r_1^2\sim 2(r_1-7)^2+4^2\{0<r_1<20\}$. In this regression, $\sim$ asks Desmos to find $r_1$; the restriction keeps the center inside the square. Under PARAMETERS, Desmos shows $r_1\approx4.944615$. Type $14-\sqrt{82}$ on the next line; it evaluates to the same value. The other solution, $14+\sqrt{82}$, exceeds $20$. So circle $C$ has radius $14-\sqrt{82}$. Choice D.