What special geometric relationship between a line and a circle occurs when they intersect at exactly one point instead of two?

For tangency, the distance from the circle's center $(0,0)$ to the line $y = kx + 7$ must equal the radius $5$. Rewrite the line in the form $Ax + By + C = 0$ and apply the distance formula $d = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

After setting the distance equal to $5$, carefully solve for $k^2$, then take the square root, and remember to choose the positive value of $k$.

Type x^2 + y^2 = 25 into Desmos. You will see a circle centered at the origin with radius 5.

Type y = kx + 7. Desmos will prompt you to add a slider for k; create the slider so you can adjust the slope of the line.

Move the k slider and watch how the line intersects the circle. For most values of k, the line will cut the circle in two points; adjust k until the line just touches the circle at a single point (the line is tangent).

When the line is tangent (only one intersection point with the circle), look at the value of k on the slider. That positive value should match one of the answer choices; that is the correct choice.

The first equation is

This is a circle centered at $(0,0)$ with radius $5$ (because $25 = 5^2$).

The second equation is

which is a line with slope $k$ and $y$-intercept $7$.

A line and a circle can intersect in $0$, $1$, or $2$ points.

• Two solutions: the line cuts through the circle.
• No solutions: the line misses the circle.
• Exactly one solution: the line just touches the circle at one point; it is tangent to the circle.

So we want the value of $k$ that makes the line $y = kx + 7$ tangent to the circle $x^2 + y^2 = 25$.

A line is tangent to a circle if the distance from the center of the circle to the line equals the radius.

• Center of the circle: $(0,0)$
• Radius: $5$
• Line: $y = kx + 7$

First, rewrite the line in standard form $Ax + By + C = 0$:

So $A = k$, $B = -1$, and $C = 7$.

The distance from $(0,0)$ to the line $Ax + By + C = 0$ is

Here,

For tangency, this distance must equal the radius $5$:

Start from

1. Multiply both sides by $\sqrt{k^2 + 1}$:

2. Divide both sides by $5$:

3. Square both sides:

4. Subtract $1$ (which is $\tfrac{25}{25}$) from both sides:

5. Take the square root of both sides. This gives two possible values, $k = \pm \sqrt{\tfrac{24}{25}}$. Since the problem asks for the positive value of $k$, we take the positive root and simplify:

So the positive value of $k$ is $\frac{2\sqrt{6}}{5}$, which corresponds to choice D.