Question

a circle in the xy - plane has its center at (-4,-6). line k is tangent to this circle at the point (-7,-7). what is the slope of line k?
a. -3
b. -\frac{1}{3}
c. \frac{1}{3}
d. 3
think back to what we said about perpendicular slopes heres a problem to

Explanation:

Step1: Find slope of radius

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. The center of the circle is $(x_1,y_1)=(-4,-6)$ and the point of tangency is $(x_2,y_2)=(-7,-7)$. So the slope of the radius $m_{radius}=\frac{-7-(-6)}{-7 - (-4)}=\frac{-7 + 6}{-7+4}=\frac{-1}{-3}=\frac{1}{3}$.

Step2: Use perpendicular - slope relationship

A tangent line to a circle is perpendicular to the radius at the point of tangency. If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1\times m_2=-1$. Let the slope of the tangent line $k$ be $m_{tangent}$. We know $m_{radius}=\frac{1}{3}$ and $m_{radius}\times m_{tangent}=-1$. So $m_{tangent}=- 3$.

Answer:

A. $-3$