Question 3·25 Super-Hard SAT Advanced Math Questions·Advanced Math

In center-radius form, the circle has center $(1,-2)$ and radius $\sqrt{4}=2$. Exactly one shared point means the line is tangent, so its perpendicular distance from the center equals the radius.

The center-to-line distance formula uses a line written with $0$ on one side. Move $4-4k$ to the left to put the given line in that form:

For $Ax+By+C=0$, the perpendicular distance from $(u,v)$ is $\frac{|Au+Bv+C|}{\sqrt{A^2+B^2}}$. Substitute the center $(1,-2)$ and set that distance equal to $2$: $\frac{|k(1)+3(-2)+4k-4|}{\sqrt{k^2+3^2}}=2$. Combine terms:

Square both sides to remove the absolute value and square root: $25(k-2)^2=4(k^2+9)$. Type $25(k_1-2)^2\sim4(k_1^2+9)\{k_1\ge1\}$ in Desmos. Here $k_1$ tells Desmos what to find, $\sim$ starts a regression, and the restriction keeps $k_1\ge1$ because $k$ is a positive integer. Desmos shows $k_1=4$ under PARAMETERS.

Type the given circle, then type the line with $k=4$ as $4x+3y=4-4(4)$. Click their one intersection. Desmos shows approximately $(-0.6,-3.2)$. Use that point as a guide; the next move finds its coordinates exactly.

Type $4-4(4)$ and $4(1)+3(-2)$ in Desmos; they show $-12$ and $-2$. The line now requires $4x+3y=-12$, but at the center, $4(1)+3(-2)=-2$. So $4x+3y$ must decrease by $10$. Type $4^2+3^2$; Desmos shows $25$. The direction $(4,3)$ is perpendicular to the line; moving $t(4,3)$ changes $4x+3y$ by $t(4^2+3^2)=25t$. Set that change equal to $-10$: $25t=-10$. Divide by $25$:

Move $-\frac25(4,3)$ from the center $(1,-2)$. Type each coordinate calculation in Desmos: $1-\frac25\cdot4$ shows $-0.6$, whose fraction button gives $-\frac35$; $-2-\frac25\cdot3$ shows $-3.2$, or $-\frac{16}{5}$. So the touch point is $\left(-\frac35,-\frac{16}{5}\right)$.

The question asks for $x+y$, not either coordinate alone. Type $-\frac35-\frac{16}{5}$ as the last Desmos line. It shows $-3.8$; the fraction button gives $-\frac{19}{5}$. So $x+y=-\frac{19}{5}$. Choice B.