Question

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.

Answer 1

lord i bet this is going to be a pain
lets solve for \(x\)

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

if it was me, i would cheat

Answer 6

@Loser66 look good to you?

Answer 7

XD yup.