The line y = (x - 2)/2 intersects the circle x^2 + y^2 = 8 at A and B. Find the midpoint of AB. Express your answer in the form (x,y).

Guest May 1, 2017

#1**0 **

First, let's find where the line y = (x - 2)/2 intersects the circle x^{2} + y^{2} = 8 by substituting the first value for y into the second equation:

---> x^{2} + [ (x-2)/2 ]^{2} = 8

Expand the second term:

---> x^{2} + ( x^{2} - 4x + 4 ) / 4 = 8

Multiply all term by 4:

---> 4x^{2} + x^{2} - 4x + 4 = 32

Simplify the left side and subtract 32 from both sides:

---> 5x^{2} - 4x - 28 = 0

Factor (or use the quadratic formula):

---> (5x - 14)(x + 2) = 0

Solve for x and put this value into the linear equation to find the corresponding y-value:

---> 5x - 14 = 0 ---> x = 14/5 = 2.8 and y = 0.4

or x + 2 = 0 ---> x = -2 and y = -2

The points of intersection are: (2.8, 0.4) and (-2, -2)

Use the midpoint formula to find the midpoint of these two points of intersection:

Their midpoint is: (0.2, -0.8)

geno3141
May 1, 2017