# SOLUTION: How do I solve the following system of conic sections??? x^2+y^2+2x+2y=0 and x^2+y^2+4x+6y+12=0.

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 Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Quadratic-relations-and-conic-sections Question 390367: How do I solve the following system of conic sections??? x^2+y^2+2x+2y=0 and x^2+y^2+4x+6y+12=0.Answer by ewatrrr(10682)   (Show Source): You can put this solution on YOUR website!``` 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 = -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) ```