Question 180·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Type the circle equation as written in Desmos, without adding $y=$. Select the circle. Desmos marks its highest and lowest points at $(2,5)$ and $(2,1)$.

Those points are the ends of a vertical diameter, a segment across the circle through its center. Type $\frac{5+1}{2}$ and $\frac{5-1}{2}$; Desmos shows $3$ and $2$. So the center is $(2,3)$, halfway between the points, and the radius is $2$, half their distance.

Put the center $(2,3)$ into $y=x+k$: $3=2+k$, so $k=1$. Type $k=1$ and $y=x+k$ in Desmos. The line crosses the circle twice, as a line through its center should; it isn’t the one-solution line.

Changing $k$ moves the line vertically by $|k-1|$ from the line through the center. A slope-$1$ line makes a $45^\circ$ angle with a vertical line, so its perpendicular distance from the center is that vertical gap divided by $\sqrt2$. For exactly one intersection, the line must be tangent, or one radius from the center:

Square both sides to remove the absolute value and the square root:

Keep the division by $2$: the vertical gap isn’t the perpendicular distance.

Call the requested $(k-1)^2$ by the name $q$. Type $q/2\sim4$; $\sim$ tells Desmos to find $q$ rather than make a slider. Under PARAMETERS, it shows $q=8$. Both tangent lines give that same squared value, so $(k-1)^2=8$. Choice C.