Question 71·Hard·Advanced Math

Set $k=1$ as a trial and type all three lines into Desmos. Click the intersections: $(-4,3)$ and $(-3,4)$. Those are two shared points. For exactly one, the line must be tangent: it must touch the circle at one point without cutting through it.

At a shared point, the line says $y=kx+7$. Use substitution: replace the circle’s $y$ with the whole expression $kx+7$:

Expand the squared parentheses:

Subtract $25$:

Combine the $x^2$ terms:

Each $x$ in this quadratic gives one $y$ on the line, so count its real $x$-values. In $ax^2+bx+c=0$, the discriminant is $b^2-4ac$. One shared point requires a zero discriminant: the two roots merge when the square root in the quadratic formula is zero. Here $a=k^2+1$, $b=14k$, and $c=24$, so set it to zero:

Square and multiply the coefficients:

Distribute:

Combine like terms:

Keep the graph and add $100k_1^2-96\sim0\{k_1>0\}$. Use $k_1$ because $k=1$ is only the graph’s trial value; $\sim$ tells Desmos to find a fit. The positive root appears under PARAMETERS as $k_1\approx0.9798$. The restriction excludes the negative root.

The choices are exact, so solve $100k^2-96=0$ without rounding. Add $96$:

Divide by $100$:

Reduce the fraction:

Take the positive square root, as the question requires:

Since $24=4\times6$, simplify:

This is the positive slope that makes the line touch the circle at exactly one point. Choice D.