Question 210·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

A solution must satisfy both equations. The line says $y=x+t$, so replace $y$ in the circle equation:

For each $x$ that works, the line gives exactly one $y$. So counting solutions to this equation counts points in the system.

Expand $(x+t)^2$ and combine the $x^2$ terms:

Keep the $t^2$: it comes from squaring the whole expression $x+t$.

Subtract $10t$ from both sides to get the quadratic form needed for the discriminant:

Here the coefficients of $x^2$, $x$, and the constant term are $a=2$, $b=2t$, and $c=t^2-10t$.

Because each $x$ gives one $y$, exactly one point in the system means exactly one real root of this quadratic. The discriminant $D=b^2-4ac$ is zero in that case: the plus and minus versions of the quadratic formula become the same root. Substitute the coefficients:

Square $2t$ and multiply $4$ by $2$:

Distribute the $-8$:

Combine the $t^2$ terms:

Factor out $4t$ to expose both ways $D$ can be zero:

The factor $4t$ makes $D=0$ at $t=0$. Check it in the original system: $x^2+y^2=0$ forces both squares to be zero, so $(x,y)=(0,0)$. That point satisfies $y=x+0$, making it the system's one solution. A zero-radius circle still contributes this value of $t$.

For the other case, $t>0$, so divide $4t(20-t)=0$ by $4t$ to get $20-t=0$. Type $20-t_1\sim0\left\{t_1>0\right\}$; the restriction tells Desmos to find the positive value. Under PARAMETERS, Desmos shows $t_1=20$. The possible values of $t$ are $0$ and $20$. Add them, rather than averaging them: type $t_1+0$ beneath the regression, and Desmos prints $20$. So the sum of all possible values of $t$ is $20$. Choice C.