Substitute your expression for $y$ from the first equation into $x^2 + y^2 = 2k$. You should get a quadratic equation involving only $x$ and $k$.

For a quadratic equation in $x$, what condition on the discriminant $b^2 - 4ac$ gives exactly one real solution? Apply that condition to the quadratic you found to get an equation in $k$ and solve it.

In Desmos, enter the equation x + y = 12. This will display the straight line representing all $(x, y)$ pairs that satisfy the first equation.

Type x^2 + y^2 = 2k. Desmos will prompt you to add a slider for k; create the slider so that you can vary the radius of the circle $x^2 + y^2 = 2k$.

Move the k slider and watch how the circle moves relative to the line. Find the value of k at which the line just touches the circle at exactly one point (they are tangent). At that setting, there should be a single visible intersection point.

Once you have adjusted the slider so that there is exactly one intersection point between the line and the circle, note the corresponding value of k shown on the slider. That is the required value of $k$ for the system to have exactly one real solution pair.

From the first equation

you can solve for $y$ in terms of $x$:

This lets you write everything using only $x$.

Substitute $y = 12 - x$ into the second equation $x^2 + y^2 = 2k$:

Now expand $(12 - x)^2$:

So the equation becomes

Combine like terms:

This is a quadratic equation in $x$ of the form $ax^2 + bx + c = 0$.

For a quadratic equation $ax^2 + bx + c = 0$:

• There is exactly one real solution when the discriminant $b^2 - 4ac = 0$.

In our quadratic

we have $a = 2$, $b = -24$, and $c = 144 - 2k$.

Set the discriminant equal to $0$:

Now simplify step by step:

So the condition for exactly one real solution reduces to

Solve the equation

by dividing both sides by $16$:

So, the value of $k$ that makes the system have exactly one real solution pair $(x, y)$ is 36.