Question 147·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

A circle tangent to an axis touches it once, so its center is one radius from that axis. Call the radius $r$. The center is in the first quadrant, so both coordinates are positive. Two axis tangencies make the center $(r,r)$.

The point $(8,2)$ is on the circle, so its squared distance from the center equals $r^2$. The horizontal change from $(r,r)$ is $8-r$, and the vertical change is $2-r$, so:

Type $(8-r_1)^2+(2-r_1)^2\sim r_1^2\{0<r_1<10\}$. Here $r_1$ stands for the radius, and $\sim$ tells Desmos to fit it. The restriction uses the problem’s less than 10 condition. Under PARAMETERS, Desmos gives $r_1\approx4.343$; without the restriction, it could find the larger radius instead.

The circle’s center-radius form uses center $(r,r)$ and radius $r$:

Expand the squares:

Subtract $r^2$ from both sides and combine the constants:

Compared with the given form, the constant term is $c=r^2$.

Type $r_1^2$, then $132-80\sqrt2$ on the next Desmos line. Both print about $18.863$, so the exact expression that matches the required value of $c$ is $132-80\sqrt2$. Choice A.

Start with $(8-r)^2+(2-r)^2=r^2$ to find an exact radius. Expand the squared differences:

Combine like terms:

Subtract $r^2$ from both sides:

A perfect square will make the two possible radii visible. Move $68$ to the right:

Add $100$ to both sides because half of $-20$ is $-10$, and $(-10)^2=100$:

Rewrite the left side as a square:

Take square roots, keeping both signs:

Simplify the square root:

Add $10$:

The less than 10 condition selects $r=10-4\sqrt2$, not $10+4\sqrt2$. From the circle expansion, $c=r^2$. Substitute the allowed radius:

Expand, including the square of $4\sqrt2$:

Combine the whole numbers:

So the circle’s constant term is $132-80\sqrt2$. Choice A.