Answer:
Slope of the line L will be [tex]-\frac{3}{4}[/tex]
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 [tex]m_{1}[/tex] is the slope of line joining center of the circle (3, 2) and point of tangency (6, 6).
And [tex]m_{2}[/tex] is the slope of line L.
Then by the fact given above [tex]m_{1}\times m_{2}=-1[/tex]-----(1)
Now [tex]m_{1}=\frac{y-y'}{x-x'}[/tex]
[tex]m_{1}=\frac{6-2}{6-3}[/tex]
[tex]m_{1}=\frac{4}{3}[/tex]
Now from the equation 1
[tex]\frac{4}{3}\times m_{2}=-1[/tex]
[tex]m_{2}=-\frac{3}{4}[/tex]
Therefore, slope of the line L will be [tex]-\frac{3}{4}[/tex]
