Question 1033·Medium·Math

Type $k=1$ as a temporary positive trial, then graph $y=kx+7$ and $x^2+y^2=25$. Desmos shows two shared points, $(-4,3)$ and $(-3,4)$. Each shared point solves both equations. Exactly one solution means the line is tangent to the circle: it touches at one point instead of crossing twice.

At a shared point, the line gives $y=kx+7$. Use substitution, replacing the whole $y$ in the circle equation: $x^2+(kx+7)^2=25$. Each solution for $x$ gives one point, because the line then determines its $y$.

Square $kx+7$; the middle term is $2(kx)(7)=14kx$: $x^2+k^2x^2+14kx+49=25$. Keep that middle term. Leaving it out changes how many solutions the equation has.

Subtract $25$ from both sides so the equation equals zero: $x^2+k^2x^2+14kx+24=0$.

Combine $x^2$ and $k^2x^2$ to put the quadratic in $ax^2+bx+c=0$ form: $(1+k^2)x^2+14kx+24=0$. Its coefficients are $a=1+k^2$, $b=14k$, and $c=24$.

The discriminant is $b^2-4ac$, the quantity under the square root in the quadratic formula. Exactly one real root means it equals zero: $D=(14k)^2-4(1+k^2)(24)=0$. With $D=0$, the formula’s plus and minus versions give the same $x$, so the line and circle share one point.

Type $(14k_1)^2-4(1+k_1^2)(24)\sim0\{k_1>0\}$. The $\sim$ asks Desmos to find a fit; $k_1$ is the value it solves for, and the restriction keeps it positive. Under PARAMETERS, Desmos reports $k_1\approx0.9797958971$.

Type each answer value on a new Desmos line. They display about $0.9798$, $0.4899$, $6.1237$, and $1.0206$, respectively. Only $\frac{2\sqrt6}{5}$ matches the positive value Desmos found. So the slope that makes the line touch the circle once is $k=\frac{2\sqrt6}{5}$. Choice A.