After substituting, you should get a quadratic equation in $x$. For a quadratic $ax^2 + bx + c = 0$, what condition on the discriminant $b^2 - 4ac$ gives exactly one real solution?

Write $a$, $b$, and $c$ in terms of $t$ and compute $b^2 - 4ac$. Set that equal to $0$ and solve for $t$. Make sure you include all $t$ values you find before taking their sum.

Translate the discriminant you derived algebraically into a function Desmos can graph. In Desmos, type something like D(x) = x^2 - 4*(x^2/2 - 5x). Here, x is playing the role of $t$.

Look at the graph of $y = D(x)$. Use Desmos’s intercept tool or trace along the graph to find the $x$-values where the graph crosses the $x$-axis (where $D(x) = 0$). These $x$-values correspond to the $t$-values that make the system have exactly one solution.

Once you have identified the two $x$-values from the previous step, enter an expression like a + b in Desmos, replacing a and b with those values. The resulting number is the sum of all possible values of $t$ for the original system.

The equations represent:

• A line: $y = x + t$ (slope $1$, vertical shift $t$).
• A circle: $x^2 + y^2 = 10t$ (centered at the origin, with radius $\sqrt{10t}$ if $t \ge 0$).

In the $xy$-plane, a line and a circle have exactly one intersection point if the line is tangent to the circle, or if the circle collapses to a single point and the line passes through that point. Algebraically, this means the system leads to a quadratic in $x$ with exactly one real solution.

Use $y = x + t$ in the circle equation $x^2 + y^2 = 10t$.

Expand and simplify:

Divide everything by $2$ to get a simpler quadratic in $x$:

This has the form $ax^2 + bx + c = 0$ with $a = 1$, $b = t$, and $c = \dfrac{t^2}{2} - 5t$.

For a quadratic $ax^2 + bx + c = 0$, the discriminant is $D = b^2 - 4ac$.

• Two real solutions if $D > 0$.
• One real solution if $D = 0$.
• No real solutions if $D < 0$.

We need exactly one solution in $x$, so set $D = 0$.

Compute $D$ for $x^2 + tx + \left(\dfrac{t^2}{2} - 5t\right) = 0$:

Now simplify this expression.

Simplify the discriminant step by step:

Set $D = 0$ for exactly one real solution:

So the possible values of $t$ are the solutions to this equation: $t = 0$ or $t = 20$.

Both values are valid:

• When $t = 0$, the circle becomes $x^2 + y^2 = 0$, which is the single point $(0,0)$, and the line $y = x$ passes through $(0,0)$, giving exactly one intersection.
• When $t = 20$, the circle has positive radius and the line is tangent to it, again giving exactly one intersection.

The question asks for the sum of all possible values of $t$.

From the discriminant condition, $t$ can be $0$ or $20$.

Add them:

So, the sum of all possible values of $t$ is $20$.