The line $y = (x - 2)/2$ intersects the circle $x^2 + y^2 = 8$ at $A$ and $B$. Find the midpoint of $\overline{AB}$. Express your answer in the form "$(x,y)$."
y = (x - 2) / 2 (1)
x^2 + y^2 = 8 (2)
Sub (1) into (2)
x^2 + [ (x - 2) / 2 ]^2 = 8
x^2 + (1/4) (x^2 - 4x + 4 ) = 8
4x^2 + x^2 - 4x + 4 = 32
5x^2 - 4x - 28 = 0
(5x - 14) (x + 2) = 0
Setting each factor to 0 and solving for x we have that
x = 14/5 and x = -2
The corresponding y coordinates are
y = (1/2) (14/5 - 2) = (1/2)(4/5) = 4/10 = 2/5
y = (1/2) (-2 - 2) = -2
So....the intersection points are ( 14/5, 2/5) and ( -2, -2)
And the midpoint of this is ( [ 14/5 -2] / 2 , [2/5 - 2] / 2 ) =
( [4/5]/2 , [-8/5]/2) = ( 4/10, -8/10) = (2/5, -4/5)