Question 197·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

The points $(1,2)$ and $(5,2)$ are at the same height, with $x=3$ halfway between them. The center must be equally far from both, so it lies on their perpendicular bisector, the vertical line through $x=3$. Call the center $(3,k)$. Two points on a circle do not necessarily form a diameter. Their midpoint gives the center’s $x$-coordinate, not its height.

From the center $(3,k)$ to $(1,2)$, the horizontal change is $1-3$ and the vertical change is $2-k$. The Pythagorean theorem says their squares add to the distance squared. That distance is the radius, $5$, so $(1-3)^2+(2-k)^2=5^2$.

Type $(1-3)^2+(2-k_1)^2\sim5^2\{k_1<0\}$. The $\sim$ asks Desmos to fit $k_1$, and the braces keep only values below the $x$-axis. Under PARAMETERS, Desmos shows $k_1\approx-2.5826$. That identifies which square-root branch to use for an exact answer.

Start with the distance equation. Square the horizontal change:

Square the radius:

Subtract $4$:

Since $k<0$, $2-k$ is positive. Take the positive square root:

Subtract $2$:

Multiply by $-1$:

At the origin, $x=y=0$, so the given expanded equation’s left side is $c$. In center-radius form, the circle is $(x-3)^2+(y-k)^2=5^2$. Subtract $5^2$ from the center-radius form’s left side to match the given equation’s left side, then put in the origin to get $c$. Type $3^2+(2-\sqrt{21})^2-5^2$ in Desmos; it shows about $-9.3303$. For the exact value, expand the square:

Evaluate the integer squares:

Combine the constants:

That is the value of $c$. Choice A.