Try substituting $y = mx + 1$ into the circle’s equation $(x - 3)^2 + (y + 2)^2 = 13$ to get an equation in $x$ only. What type of equation in $x$ do you get?

Once you have a quadratic in $x$, recall how the discriminant $b^2 - 4ac$ tells you the number of real solutions. What must $b^2 - 4ac$ equal if you want exactly one solution?

After setting the discriminant equal to 0, you’ll get a quadratic equation in $m$. Solve it, then see which of the resulting values of $m$ match the answer options.

In Desmos, type (x-3)^2 + (y+2)^2 = 13 to graph the circle with center $(3,-2)$ and radius $\sqrt{13}$.

For each option, replace $m$ in y = m x + 1 with that value and enter the line into Desmos (for example, for choice C, type y = 2x + 1). Observe how many intersection points each line has with the circle.

Among the four lines, look for the one that just touches the circle at a single point (tangent) rather than crossing it at two points or missing it entirely. The slope used in that line matches the correct answer choice for $m$.

A line that intersects a circle at exactly one point is tangent to the circle.

Algebraically, after you substitute the line equation into the circle equation, you get a quadratic in $x$:

• If it has 2 real solutions, the line cuts the circle in 2 points.
• If it has 0 real solutions, the line does not meet the circle.
• If it has exactly 1 real solution, the line is tangent.

So we want the quadratic in $x$ to have discriminant equal to 0.

The circle is

The line is $y = mx + 1$, so

Substitute this into the circle:

Now expand:

• $(x-3)^2 = x^2 - 6x + 9$
• $(mx+3)^2 = m^2x^2 + 6mx + 9$

So

Combine like terms:

Move 13 to the left:

This is our quadratic in $x$.

For a quadratic $ax^2 + bx + c = 0$ to have exactly one real solution, its discriminant $b^2 - 4ac$ must be 0.

Here,

• $a = m^2 + 1$
• $b = 6(m - 1)$
• $c = 5$

Set the discriminant equal to 0:

Simplify step by step:

Divide everything by 4:

Expand:

Finally, divide by 2:

Now solve this quadratic for $m$.

Use the quadratic formula on $2m^2 - 9m + 2 = 0$:

So both $\dfrac{9 + \sqrt{65}}{4}$ and $\dfrac{9 - \sqrt{65}}{4}$ are mathematically possible slopes.

Looking at the answer choices, only

appears, so that is the value of $m$ that makes the line tangent to the circle and intersect it at exactly one point.