Find the equation of the circle which passes through the points (2, 3) and (-1, 1) and has its center on the line x – 3y – 11 = 0.

Question

anonymous · Answer

lord i bet this is going to be a pain 
lets solve for $x$

anonymous · Answer

get $x=3y+11$ so any point on the line will look like $(3y+11,y)$

anonymous · Answer

the square of the distance between $(3y+11,y)$ and $(2,3)$ must equal the square of the distance between $(3y+11,y)$ and $(-1,1)$

anonymous · Answer

in other words 
$$(3y+11-2)^2+(y-3)^2=(3y+11+1)^2+(y-1)^2$$ simplify this mess and solve for $y$
then solve for $x$ and you will have the center

anonymous · Answer

if it was me, i would cheat

anonymous · Answer

@Loser66  look good to you?

loser66 · Answer

XD yup.