Try substituting the expression for $y$ from the line, $y = 2x + 1$, into the circle equation to get a single equation in $x$. What type of equation will that be?

Once you have a quadratic in $x$, recall how the discriminant $b^2 - 4ac$ tells you whether there are 0, 1, or 2 real solutions. Which inequality should you use for "no real solution"?

After you write and simplify the discriminant inequality in terms of $k$, solve it and then think about which integer values satisfy it. Remember the question asks for the least positive integer.

In Desmos, type (x-k)^2 + y^2 = 4 and y = 2x + 1. When you type k, Desmos will prompt you to create a slider for k; accept this so you can vary k continuously.

Move the k slider and watch how the circle (whose center moves along the x-axis) intersects the fixed line. Notice for which values of k there are two intersection points, one point (tangent), or no intersection points.

Gradually increase the slider and test positive integer values of k (for example, k = 1, 2, 3, ...). Look for the smallest positive integer at which the circle and the line do not intersect at all; that integer is the solution.

The system is

Substitute $y = 2x + 1$ into the circle equation:

Now expand both squares:

Add them and move 4 to the left side:

Combine like terms:

So we have a quadratic in $x$:

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

• Two real solutions if $b^2 - 4ac > 0$.
• One real solution (tangent) if $b^2 - 4ac = 0$.
• No real solutions if $b^2 - 4ac < 0$.

Here,

• $a = 5$,
• $b = -2k + 4$,
• $c = k^2 - 3$.

We want no real solutions, so we need:

First expand $(-2k+4)^2$ and $4(5)(k^2 - 3)$:

• $(-2k+4)^2 = (4-2k)^2 = 16 - 16k + 4k^2$,
• $4(5)(k^2 - 3) = 20k^2 - 60$.

Substitute into the inequality:

Distribute the minus sign and combine like terms:

Multiply by $-1$ (and flip the inequality sign):

Divide everything by 4 to simplify:

So $k$ must satisfy the inequality

First find where $4k^2 + 4k - 19 = 0$ using the quadratic formula:

Compute the discriminant:

So

Approximate these:

• $\sqrt{5} \approx 2.236$, so $2\sqrt{5} \approx 4.472$.
• $k_1 \approx \dfrac{-1 - 4.472}{2} \approx -2.736$,
• $k_2 \approx \dfrac{-1 + 4.472}{2} \approx 1.736$.

Since the coefficient of $k^2$ in $4k^2 + 4k - 19$ is positive, the inequality $4k^2 + 4k - 19 > 0$ holds for

That is, for all $k$ less than about $-2.736$ or greater than about $1.736$. Among integers, this means

The question asks for the least positive integer $k$ that makes the system have no real solution, so the required value is $\boxed{2}$.