Question

Line L is tangent to a circle whose center is at (3,2). If the point of tangency is (6,6) what is the slope of line L?

Answer 1

Answer:

Slope of the line L will be $-\frac{3}{4}$

Step-by-step explanation:

We will use the theorem that tangent to a circle is always perpendicular to the radius joining center of the circle and point of tangency.

Let $m_{1}$ is the slope of line joining center of the circle (3, 2) and point of tangency (6, 6).

And $m_{2}$ is the slope of line L.

Then by the fact given above $m_{1}\times m_{2}=-1$-----(1)

Now $m_{1}=\frac{y-y'}{x-x'}$

$m_{1}=\frac{6-2}{6-3}$

$m_{1}=\frac{4}{3}$

Now from the equation 1

$\frac{4}{3}\times m_{2}=-1$

$m_{2}=-\frac{3}{4}$

Therefore, slope of the line L will be $-\frac{3}{4}$