# Algebra

X^2+y^2+6y+3=0
x^2+y^2-9x-6=0

List the solutions:

1. You have two circles:

x^2 + (y+3)^2 = 6
(x - 9/2)^2 + y^2 = 105/4

Naturally, they intersect where

x^2+y^2+6y+3 = x^2+y^2-9x-6
6y = -9x-9
y = -(3x+3)/2

Subbing that into one of the equations,
x^2+(3x+3)^2/4-9x-6=0
13x^2 - 18x - 15 = 0
x = 1/13 (9±2√69)
y = -3/13 (11±√69)

posted by Steve

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