Hey,
I must be an idiot because the answer to this is probably simple.
Anywho.
Circle has equation X^2+y^2+4x-8y+10=0
So Center is (-2,4) and Radius is Route 10.
Now L has an equation X-3y+4=0
All I have to do is show that L is a tangent to C.
Hey,
I must be an idiot because the answer to this is probably simple.
Anywho.
Circle has equation X^2+y^2+4x-8y+10=0
So Center is (-2,4) and Radius is Route 10.
Now L has an equation X-3y+4=0
All I have to do is show that L is a tangent to C.
If L is a tangent to C then they must have a common point, say $\displaystyle (x_0,y_0)$.
Common point means:
$\displaystyle x_0^2+y_0^2+4x_0-8y_0+10=0, \, x_0-3y_0+4=0$
Now solve these two equations simultaneously. Then $\displaystyle (x_0,y_0)$ will satisfy both and be the required common point.