A line can intersect a circle in 0, 1, or 2 points. What special situation gives exactly one point of intersection between a line and a circle?

Rewrite $x^2 + y^2 - 4x - 6y + 9 = 0$ by completing the square for $x$ and $y$ so that it looks like $(x - h)^2 + (y - k)^2 = r^2$. Identify the center and radius.

Write the line $y = x + k$ in the form $ax + by + c = 0$ and express the distance from the circle’s center to this line. For tangency, set this distance equal to the radius and solve for $(k - 1)^2$.

In Desmos, enter the circle equation exactly as: x^2 + y^2 - 4x - 6y + 9 = 0. Desmos will display the circle so you can see its center and size visually.

Type y = x + k. Desmos will create a slider for k, giving you a whole family of parallel lines with slope 1 that slide up and down as you move the slider.

Move the k slider slowly and watch how the line intersects the circle. Look for the value(s) of k where the line just touches the circle at exactly one point (the line is tangent). Note the approximate value of k at tangency.

In an empty expression line, type (k - 1)^2 using the tangency value of k from the slider. Compare this result to the answer choices to determine which one matches.

The first equation is a circle and the second is a line.

• A system of a circle and a line has:
  • 2 real solutions if the line cuts through the circle in two points,
  • 1 real solution if the line is tangent to the circle (just touches it),
  • 0 real solutions if the line misses the circle.

Since the system has exactly one real solution, the line must be tangent to the circle.

Start with the circle equation:

Group $x$-terms and $y$-terms and complete the square:

• For $x^2 - 4x$:

• For $y^2 - 6y$:

Substitute back:

Combine constants:

So:

This is a circle with center $(2, 3)$ and radius $2$ (since $r^2 = 4$).

The line is $y = x + k$. Rewrite it in standard form $ax + by + c = 0$:

So $a = 1$, $b = -1$, and $c = k$.

The distance from a point $(x_0, y_0)$ to the line $ax + by + c = 0$ is

Here, the center is $(2, 3)$, so the distance from $(2, 3)$ to the line $x - y + k = 0$ is

For the line to be tangent to the circle, the distance from the center to the line must equal the radius of the circle.

• Radius $r = 2$.
• Distance from center to line $= \dfrac{|k - 1|}{\sqrt{2}}$.

Set them equal:

Multiply both sides by $\sqrt{2}$:

Now square both sides to get $(k - 1)^2$:

So the value of $(k - 1)^2$ is 8, which corresponds to choice C.