Use the fact that any point that lies on both graphs must satisfy $y = 2x + c$. Substitute this expression for $y$ into the circle equation.

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?

Once you set the discriminant equal to zero and simplify, you will get an equation involving $c$. Solve it and then use the fact that $c$ must be positive to choose the correct value.

Type (x-4)^2 + (y-1)^2 = 20 into Desmos. This will display the circle with center $(4,1)$ and radius $\sqrt{20}$.

For each choice of $c$ (−17, 3, 7, 10), type y = 2x + c with that value substituted in (for example, y = 2x + 7). Check how many intersection points the line has with the circle in each case.

Among the four lines you graphed, identify which one intersects the circle at exactly one point (is tangent) and has a positive value of $c$. That line corresponds to the correct answer choice.

The equation $(x-4)^2 + (y-1)^2 = 20$ is a circle with center $(4,1)$ and radius $\sqrt{20} = 2\sqrt{5}$. The equation $y = 2x + c$ is a straight line with slope $2$ and vertical intercept $c$.

"Intersect at exactly one point" means the line is tangent to the circle.

Since every point on the line satisfies $y = 2x + c$, plug this into the circle equation:

becomes

This gives an equation in terms of $x$ and $c$ only.

Expand each squared term:

• $(x-4)^2 = x^2 - 8x + 16$
• $(2x + c - 1)^2 = (2x + (c-1))^2 = 4x^2 + 4x(c-1) + (c-1)^2$

Add them and set equal to $20$:

Combine like terms and move $20$ to the left:

This is a quadratic in $x$ of the form $ax^2 + bx + d = 0$ with

• $a = 5$
• $b = 4(c-3)$
• $d = (c-1)^2 - 4$.

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

Here that means:

Simplify this equation (carefully expanding and combining like terms) to get a quadratic equation in $c$:

Now solve this equation for $c$.

Solve $c^2 + 14c - 51 = 0$ using the quadratic formula:

So the two possible values are

• $c = \dfrac{-14 + 20}{2} = 3$
• $c = \dfrac{-14 - 20}{2} = -17$

The problem states that $c$ is positive, so the valid value is $\boxed{3}$.