Question 75·200 Super-Hard SAT Math Questions·Advanced Math

Type $(x-1)^2+(y+2)^2=25$ and $(7,-2)$ in Desmos. It draws the circle and plots $P$ to its right. In standard form $(x-h)^2+(y-k)^2=r^2$, the center is $C=(1,-2)$ and the radius is $5$. The $+2$ means the center’s $y$-coordinate is $-2$, not $2$.

Since $C$ and $P$ have the same $y$-coordinate, their distance is the difference between their $x$-coordinates: $CP=7-1=6$. This is the center-to-point distance, not the radius.

Call either touch point $T$. A tangent meets the circle at one point, and the radius to that point is perpendicular to the tangent. So $CT\perp PT$: triangle $CPT$ is right at $T$, with $CP$ as its hypotenuse.

Apply the Pythagorean theorem, using $CP$ as the hypotenuse:

Type $6^2-5^2$ in Desmos; it shows $11$. Since $PT$ is a length, $PT=\sqrt{11}$.

Because $CP$ is horizontal, the angle at $P$ is the tangent line’s angle with a horizontal line. The tangent ratio, opposite leg over adjacent leg, gives the size of the line’s slope. The opposite leg is $CT$, not $PT$:

Reflect one tangent across the horizontal line through $C$ and $P$ to get the other. Their slopes have the same size but opposite signs. So $m_1=\frac{5}{\sqrt{11}}$ and $m_2=-\frac{5}{\sqrt{11}}$, in either order.

Multiply the opposite slopes:

Since $\sqrt{11}\cdot\sqrt{11}=11$, type $-\frac{5^2}{11}$ in Desmos. It shows about $-2.273$; the fraction button gives $-\frac{25}{11}$. That’s the product of the tangent slopes. Choice C.