Question 688·Hard·Math

No Desmos needed. A sum-of-roots fact will give the requested total without finding either slope. Let $m$ be the slope, the change in $y$ divided by the change in $x$. A line with that slope through the given outside point $(7,6)$ is $y-6=m(x-7)$.

Distribute $m$ so the line’s terms can be collected: $y-6=mx-7m$. The slope is still unknown because we haven’t imposed the tangent condition yet.

Move every term to one side: $mx-y+6-7m=0$. This is the form $Ax+By+C=0$ needed for the distance from a point to a line.

A radius to the point where a tangent touches is perpendicular to that tangent. So the center’s perpendicular distance, its shortest distance to the line, equals the radius. For $Ax+By+C=0$, the distance from $(u,v)$ is $|Au+Bv+C|/\sqrt{A^2+B^2}$. Substitute the center $(1,-2)$, the line’s coefficients, and the given radius: $\frac{|m(1)+(-1)(-2)+(6-7m)|}{\sqrt{m^2+(-1)^2}}=\sqrt{13}$.

Combine the terms inside the absolute value and square root: $\frac{|8-6m|}{\sqrt{m^2+1}}=\sqrt{13}$. Keep the absolute value: distance cannot be negative.

Square both sides to remove the absolute value and square roots: $\frac{(8-6m)^2}{m^2+1}=13$. The denominator $m^2+1$ is always positive, so the next multiplication is safe.

Multiply both sides by $m^2+1$: $(8-6m)^2=13(m^2+1)$. Now this is an equation whose two solutions are the tangent slopes.

Expand the square and distribute $13$: $64-96m+36m^2=13m^2+13$. The middle term comes from twice the product $8(-6m)$.

Subtract $13m^2+13$ from both sides and collect like terms: $23m^2-96m+51=0$. Here $23$ is the coefficient of $m^2$, not the constant term.

For $am^2+bm+c=0$, the sum of its two roots is the opposite of the middle coefficient divided by the leading coefficient. Use $a=23$ and $b=-96$: $m_1+m_2=-\frac{-96}{23}=\frac{96}{23}$. Don’t use $51/23$: that’s the product, $c/a$. So the sum of the two tangent slopes is $\frac{96}{23}$. Choice C.