# Thread: L is Tangent to C?

1. ## L is 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.

2. Originally Posted by StephenPoco
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 $(x_0,y_0)$.

Common point means:
$x_0^2+y_0^2+4x_0-8y_0+10=0, \, x_0-3y_0+4=0$

Now solve these two equations simultaneously. Then $(x_0,y_0)$ will satisfy both and be the required common point.