Substitute $y = kx + 7$ into $x^2 + y^2 = 25$ and simplify. You should end up with a quadratic equation in $x$ whose coefficients involve $k$.

For the quadratic in $x$ that you found, recall that the discriminant is $b^2 - 4ac$. What must this discriminant equal if the quadratic is to have exactly one real solution?

After setting the discriminant to zero, solve the resulting equation for $k^2$, then take the square root. Be careful when simplifying square roots like $\sqrt{24}$ and when reducing fractions.

In Desmos, type x^2 + y^2 = 25 to graph the circle with center at the origin and radius 5.

For each option (A–D), enter a line of the form y = (value)x + 7, for example y = (2*sqrt(6)/5)x + 7 for one of the choices. Observe how many intersection points the line has with the circle.

For each value of $k$ you try, look carefully at the graph: if the line cuts the circle in two points, that $k$ gives two solutions; if it misses the circle entirely, it gives no real solutions. The correct $k$ is the one where the line just touches the circle at exactly one point (is tangent). Choose the answer option that corresponds to that value of $k$.

Instead of plugging in the choices one by one, you can type y = m x + 7 and let Desmos create a slider for m. Move the slider until the line just barely touches the circle at one point. Then compare that m value to the answer choices to see which one matches.

The equations represent:

• A circle: $x^2 + y^2 = 25$, which has center $(0,0)$ and radius $5$.
• A line: $y = kx + 7$.

A line and a circle can intersect in $0$, $1$, or $2$ points. Having exactly one real solution means the line just touches the circle at a single point, i.e., the line is tangent to the circle.

Algebraically, when you substitute the line into the circle, you get a quadratic equation in $x$. A tangent line corresponds to that quadratic having exactly one real solution, which means its discriminant is zero.

Substitute $y = kx + 7$ into $x^2 + y^2 = 25$:

Expand the square:

So the equation becomes

Combine like terms:

This is a quadratic in $x$ with

• $a = 1 + k^2$
• $b = 14k$
• $c = 24$.

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

Here,

Compute and simplify:

Set this equal to zero for tangency:

Solve the equation

Add $96$ to both sides:

Divide by $100$:

Take the square root of both sides. Since $k$ is given to be positive:

So the value of $k$ that makes the line tangent to the circle, and therefore gives exactly one real solution, is $\dfrac{2\sqrt{6}}{5}$, which corresponds to choice A.