Originally Posted by

**wintersoltice** Find the equation of the circle which passes through the points P(-2,0), Q(4,6) and has its centre on the line 2x+y=6.

what i tried to do is:

$\displaystyle y=-2x+6 -----(1)$

let centre of circle be (x,y)

$\displaystyle (-2+x)^2+(0+y)^2=r^2$

$\displaystyle 4-4x+x^2+y^2=r^2 ------(2)$

$\displaystyle (4+x)^2+(6+y)^2=r^2$

$\displaystyle 16+8x+x^2+36+12y+y^2=r^2 -----(3)$

Subst. (1) and (2) into (3),

$\displaystyle x=10$

Subst. x=10 into (1),

$\displaystyle y=-14$

centre of circle is (10,-14)..

but the answer is (2,2).....where did i go wrong? is my approach to the question even correct in the first place??