Question 390367
<pre><font size = 3 color = "indigo"><b>
Hi,
Showing graphically:  
x^2+y^2+2x+2y=0   |completing squares
x^2+2x+ y^2+2y =0 
(x+1)^2 -1 + (y+1)^2 -1 = 0
(x+1)^2 + (y+1)^2 = 2

x^2+y^2+4x+6y+12=0  |completing squares
x^2+4x + y^2+6y +12=0
(x+2)^2 -4 + (y+3)^2 -9 + 12 = 0
(x+2)^2 + (y+3)^2 -9 = 1
Algebraically:
 x^2+y^2+2x+2y = x^2+y^2+4x+6y+12
          -12 = 2x + 4y 
          -x/2 - 3 = y     
x^2 + (-x/2 -3)^2 + 2x + 2(-x/2-3) = 0
x^2 + x^2/4 +3x +9 +2x -x - 6 = 0
 4x^2 + x^2 + 12x + 36 + 8x - 4x - 24 = 0
  5x^2 +16x + 12 = 0  
  {{{x = (-16 +- sqrt( 16))/(10) }}}  x = -2  x = -1.2
-x/2 - 3 = y   | x = -2, y = -2    |x = -1.2, y = -2.4
Solution:  Pt(-2,-2) and Pt(-1.2,-2.4)
{{{drawing(300,300, -4, 2, -4, 2, grid(1),
circle(-1, -1,0.1),
circle(-1, -1,1.414),
circle(-2, -3,0.1),
circle(-2, -3,1),
circle(-2, -2,0.1),
circle(-1.2, -2.4,0.1),

graph( 300, 300, -4, 2, -4,2))}}}