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)
Hmm. I had expected a bit simpler answer.
X^2+y^2+6y+3=0
x^2+y^2-9x-6=0
List the solutions:
Please and thank you!
1 answer
1 answer