Maths coordinate geometry question!

One end of a diameter of a circle with equation x^2 + y ^2 - 4x +4y = 2 is (3, 1) Find the coordinates of the other end of the diameter? How would I do this?

The first thing you need to do is to find the equation of the circle. To do this you need to complete the square:

(x^2 - 4x + 4) + (y^2 + 4y + 4) = 2 + 4 + 4

(x - 2)^2 + (y + 2)^2 = 10

This means the center is located at (2,-2). Now find the equation of the line that passes through (2,-2) and (3,1):

m = (1 - -2)/(3-2) = 3/1 = 3

y = 3x + b -----> 1 = 3*3 + b -----> b = -8

y = 3x - 8

Now substitute the linear equation, y = 3x - 8, into the equation for the circle:

(x - 2)^2 + (3x - 8 + 2)^2 = 10 -----> (x - 2)^2 + (3x - 6)^2 = 10

Now solve for x:

x^2 - 4x + 4 + 9x^2 - 36x + 36 = 10

10x^2 - 40x + 30 = 0

Factor out 10 to get:

x^2 - 4x + 3 = 0

Now factor the equation to get the solutions for x:

(x - 3)(x - 1) = 0

This means that x = 3 or x = 1. You were given the solution when x = 3, the point (3,1). Now use the linear equation, y = 3x - 8, and x = 1 to determine the value of y:

y = 3*1 - 5 -----> y = -5

The other end of the diameter has coordinates of (1,-5).

Corey's answer is fine, but there's a very easy way to find the "other end of the diameter" once you know the circle's center: this "other end" and the point (3,1) are symmetrical in relation to the circle's center.