After substitution you will get a quadratic equation in $x$. Think about what condition on a quadratic equation makes it have exactly one real solution instead of two or zero.

Write the discriminant $b^2 - 4ac$ of your quadratic in terms of $k$, set it equal to $0$, and solve for $k$. Then use the fact that $k$ is positive to choose the correct value.

In Desmos, type k = 1 and Desmos will create a slider for $k$. Adjust the slider range so it covers several positive values (for example, from 0 to 20).

Enter the equations x^2 + y^2 = 4k and y = x + k. As you move the $k$ slider, you will see how the circle and the line move relative to each other.

Drag the $k$ slider and watch the intersection points of the line and the circle. Look for the specific $k$ where the line just touches the circle at exactly one point (the line is tangent to the circle). The corresponding value of $k$ on the slider is the solution.

From the system,

• Circle: $x^2 + y^2 = 4k$
• Line: $y = x + k$

Substitute $y = x + k$ into the circle equation:

Expand and combine like terms:

Set this equal to $4k$:

Move all terms to one side:

so the quadratic in $x$ is

For a quadratic $ax^2 + bx + c = 0$ to have exactly one real solution, its discriminant must be $0$.

Here, $a = 2$, $b = 2k$, and $c = k^2 - 4k$.

The discriminant is

Simplify:

For exactly one real solution, set $\Delta = 0$:

From

we get two possible values:

• $k = 0$ or
• $k = 8$.

But the problem states that $k$ is positive, so $k = 0$ is not allowed.

Therefore, the value of $k$ is 8.