Question

Find the points of intersections of the circles:x^2 + y^2 - 2x -2y -2 = 0 and x^2 + y^2 + 2x + 2y -2 = 0. Draw the circles.

Answer 1

intersections:
x^2 + y^2 - 2x - 2y -2 = 0
x^2 + y^2 + 2x + 2y -2 = 0
---------------------------subtract
-4x -4y = 0
x = - y so along that line of slope -1
now where does x = -y on either circle?

2 y^2 + 2y -2y = 2
y^2 = 1
y = +/- 1
x = -/+ 1 in other words at (-1,1) and (1,-1)

to draw them:
x^2 - 2 x = -y^2+2y +2

x^2 - 2 x + 1 = -y^2+ 2 y + 3

(x-1)^2 = -y^2+2y+3

y^2 - 2 y = 3 - (x-1)^2

y^2 - 2 y + 1 = 4 -(x-1)^2
so when all is said and done
(x-1)^2 + (y-1)^2 = 2^2
circle of radius 2 with center at (1,1)
==================================
x^2 + 2x = -y^2 -2y + 2
x^2 + 2 x + 1 = -y^2 - 2 y + 3
-(x+1)^2 +3 = y^2 + 2y
-(x+1)^2 +4 = (y+1)^2

again radius = 2 but center at (-1,-1)