After substitution, you will get a quadratic equation in $x$. What condition on the discriminant $b^2 - 4ac$ makes a quadratic have exactly one real solution?

Write $a$, $b$, and $c$ in terms of $k$ from your quadratic in $x$, then set $b^2 - 4ac = 0$ and simplify to get an equation involving $k^2$.

Once you find $k^2$, you will have two possible values for $k$. Use the condition that $k$ is a positive real constant to decide which value to keep and then match that value to the answer choices.

In Desmos, enter the circle equation: x^2 + y^2 = 10. This shows the circle centered at the origin with radius $\sqrt{10}$.

For each option, type the corresponding line into Desmos (for example, y = x + sqrt(5), then y = x + 2, etc.). After entering each line, observe how many intersection points it has with the circle: two points means two solutions, one point means exactly one solution, and no points means no real solutions.

Compare the four lines you tested and determine which one touches the circle at exactly one point (is tangent). The value of $k$ used in that line is the correct choice.

Use the second equation $y = x + k$ in the first equation $x^2 + y^2 = 10$.

Substitute:

Expand $(x + k)^2$:

Combine like terms:

Move $10$ to the left side so the equation is equal to $0$:

Now you have a quadratic equation in $x$ with parameter $k$.

For a quadratic $ax^2 + bx + c = 0$, there is exactly one real solution when the discriminant $b^2 - 4ac$ equals $0$.

Here:

• $a = 2$
• $b = 2k$
• $c = k^2 - 10$

So the discriminant is:

Set this equal to $0$ because we want exactly one real $x$:

Simplify the equation from the previous step:

Compute each part:

• $(2k)^2 = 4k^2$
• $4(2)(k^2 - 10) = 8(k^2 - 10) = 8k^2 - 80$

So the equation becomes:

Distribute the minus sign:

Combine like terms:

Add $4k^2$ to both sides:

Divide both sides by $4$:

Now solve this equation for $k$, remembering that $k$ is a positive real constant.

From the previous step, we have:

Take the square root of both sides:

Simplify $\sqrt{20}$:

So $k = \pm 2\sqrt{5}$. The problem states that $k$ is positive, so we take

Among the answer choices, this corresponds to choice C) $2\sqrt{5}$.