Use substitution: replace $y$ in the circle equation with the expression from $y=kx+7$. What kind of equation in $x$ do you get?

Once you have a quadratic in $x$, recall the discriminant $b^2-4ac$. What must this value be for the quadratic to have exactly one real solution?

After setting the discriminant equal to $0$, carefully simplify and isolate $k^2$. Then think about how the condition that $k$ is positive affects your final choice.

In Desmos, enter the circle equation as x^2 + y^2 = 25. This will display the circle centered at the origin with radius 5.

Type y = kx + 7. Desmos will automatically create a slider for k. This line will move (change its slope) as you adjust the slider.

Slowly drag the k slider and watch how the line intersects the circle. Look for the value of k where the line just touches the circle at exactly one point (it is tangent), instead of crossing it at two points or missing it entirely.

To check algebraically, add a new expression: (14k)^2 - 4(1+k^2)*24. This is the discriminant of the quadratic in $x$. Adjust the k slider and observe when this expression equals 0; the corresponding positive value of k is the one that makes the system have exactly one real solution.

The equation $x^2+y^2=25$ represents a circle centered at the origin with radius $5$.

The equation $y=kx+7$ is a line with slope $k$ and $y$-intercept $7$.

The system has exactly one real solution $(x,y)$, which means the line intersects the circle at exactly one point. Geometrically, that means the line is tangent to the circle.

Use substitution: replace $y$ in the circle equation with $kx+7$ from the line.

Starting with

substitute $y=kx+7$:

Now expand $(kx+7)^2$:

So the equation becomes

Combine like terms and move $25$ to the left:

This is a quadratic in $x$.

A quadratic $ax^2+bx+c=0$ has:

• two real solutions if $b^2-4ac>0$,
• one real solution (a double root) if $b^2-4ac=0$,
• no real solutions if $b^2-4ac<0$.

Here, $a=1+k^2$, $b=14k$, and $c=24$.

Set the discriminant equal to $0$ for exactly one real solution:

Now simplify the left-hand side:

Distribute $-96$:

Combine like terms:

Solve

for $k$:

Now take the square root of both sides:

The problem states that $k$ is positive, so

This matches answer choice D.