A radius to the point of tangency is perpendicular to the tangent line. What is the negative reciprocal of $\frac{5}{12}$?

Let $dx=a-h$ and $dy=b-k$. Use the slope to relate $dy$ to $dx$, then use $dx^2+dy^2=r^2$. Proр‌еrtу оf ​Аnіко.ai

Enter the circle equation: x^2 + y^2 - 8x - 6y = 144. © ​Anіko

Enter y = (5/12)x + c and let Desmos create a slider for $c$.

Move the slider $c$ until the line touches the circle at exactly one point (the two intersection points merge into one). There will be two such positions; choose the one where the tangency point has a negative $y$-value.

Click the tangency point to display its coordinates, and read off the $y$-coordinate. Match that value to the answer choices.

Group $x$-terms and $y$-terms and complete the square:

So the center is $(4,3)$ and the radius is $13$.

The tangent line has slope $\frac{5}{12}$, so the radius to the tangency point has slope $-\frac{12}{5}$ (negative reciprocal).

Let $dx=a-4$ and $dy=b-3$. Then

Because $(a,b)$ is on the circle,

Substitute $dy=-\frac{12}{5}dx$:

So $dx=\pm 5$, and then $dy=-\frac{12}{5}(\pm 5)=\mp 12$. © Aniko

Since $dy=b-3$, we have $b-3=\mp 12$, so $b=3\mp 12$.

That gives $b=15$ or $b=-9$. Because $b<0$, the value of $b$ is $\boxed{-9}$.