Question 138·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Since $b$ can be any real number, choose $b=1$ as a test value. Type $b=1$, $x^2+y^2=20$, and $y=\frac12x+b$ in Desmos. With $b=1$, the line crosses the circle at two points, so that value doesn't give exactly one solution. For one solution, the line must be a tangent, touching the circle once.

The circle is centered at $(0,0)$ because its equation is $x^2+y^2=20$. A radius to a tangent point is perpendicular to the tangent. Perpendicular slopes multiply to $-1$, and $\frac12(-2)=-1$, so the radius line through the center is $y=-2x$.

Add $y=-2x$ in Desmos and click where it meets the circle. The points are $(-2,4)$ and $(2,-4)$. These are the only places a line with slope $\frac12$ can be tangent: at each point, its slope is perpendicular to the radius.

Subtract $\frac12x$ from the given line equation: $b=y-\frac12x$. So each touching point gives the $y$-intercept $b$ of its own tangent line. Don't stop after finding only one point.

For the touching point $(-2,4)$, type $4-\frac{-2}{2}$ in Desmos. It shows $5$, so one tangent has intercept $b=5$.

For $(2,-4)$, add $-4-\frac{2}{2}$ in Desmos. It shows $-5$, the intercept of the tangent on the other side. Both values give exactly one solution, not just the first one you found.

The two values are opposites, so their sum is $5+(-5)=0$. The question asks for the sum of all possible values of $b$, not either intercept alone. Choice C.