Question 200·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Try the positive value $c=7$ in Desmos: type $c=7$, then type both given equations. The line misses the circle, so that value won't work. For exactly one shared point, the line must be a tangent: it touches the circle without cutting through it.

At a shared point, the line gives $y=2x+c$. Substitute that entire expression for $y$ in the circle equation:

Each solution for $x$ gives one point, because the line then fixes $y$.

Expand the squares, keeping the middle term from $(2x+c-1)^2$:

Subtract $20$ and collect like terms so the equation has quadratic form, $Ax^2+Bx+C=0$:

Here $c$ is still the line's constant; the coefficient of $x$ is $4c-12$.

The discriminant is $B^2-4AC$. Exactly one $x$-value means it equals zero. Use $A=5$, $B=4c-12$, and $C=(c-1)^2-4$:

Expand the squares:

Combine like terms:

Divide by $-4$:

Type $c_1^2+14c_1-51\sim0\{c_1>0\}$ in Desmos. The $\sim$ tells Desmos to solve for $c_1$ rather than make a slider, and the restriction keeps the positive solution. Under PARAMETERS, Desmos shows $c_1=3$. So the positive value of the line's constant is $3$. Choice B.

No Desmos needed. The circle's center and squared radius are visible in its equation. In center-radius form, $(x-h)^2+(y-k)^2=r^2$, the center is $(h,k)$. So this circle has center $(4,1)$ and squared radius $r^2=20$.

The line's slope is $2$: it rises $2$ for every $1$ to the right. A radius to a tangent point is perpendicular to the line, so its slope is $-\frac12$. Moving $4$ right and $2$ down fits that slope, and $4^2+2^2=20$ matches the squared radius. The opposite move, $4$ left and $2$ up, works too.

Apply those opposite moves to the center $(4,1)$:

At each point, the radius is perpendicular to a line of slope $2$, so each point gives a possible tangent.

In $y=2x+c$, $c=y-2x$ is the line's $y$-intercept. Find it for $(0,3)$:

Find it for $(8,-1)$:

The second tangent has a negative constant, so the required positive value is $3$. Choice B.