Question 25·200 Super-Hard SAT Math Questions·Advanced Math

Swap the coordinates of a solution $(x,y)$. The point $(y,x)$ still satisfies both equations: the squares still add to $50$, and the sum is still $p$. Since there is exactly one ordered-pair solution, the swapped point cannot be different. One solution means swapping the coordinates must leave the point unchanged. So $x=y$.

Replace $y$ with $x$ in the circle equation, because the two coordinates are equal: $x^2+x^2=50$. Combine the matching terms: $2x^2=50$.

The circle is centered at $(0,0)$, so the radius to a point $(x,x)$ follows $y=x$ and has slope $1$. Rewrite the line to see its slope: $y=-x+p$. Its slope is $-1$, making it perpendicular to that radius. A line perpendicular to a circle's radius at a point on the circle is a tangent, so it touches only once.

Type $2x_1^2\sim50\{x_1>0\}$ in Desmos. The $\sim$ makes this a regression that finds $x_1$, and the positive restriction selects the positive root; the problem does not require $x$ to be positive. Desmos shows $x_1=5$ under PARAMETERS. Squaring also allows $-5$, but that value isn't listed. So $x$ could be $5$ when the system has exactly one solution. Choice C.