L is Tangent to C?

Feb 2010
18
0
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. (Crying)
 

Isomorphism

MHF Hall of Honor
Dec 2007
1,465
674
IISc, Bangalore
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. (Crying)
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.