Question

Solving a system.

x^2 + y^2 + 2x + 2y = 0
x^2 +y^2 + 4x + 6y + 12 = 0
hmm

Answer 1

I really like the "hmmm". LOL

Answer 2

Going about this algebraically will give me white hair and stress >.>
and I must show all my work too .

Answer 3

I have never solved this type of simultaneous equation.

Answer 4

They look like equations of two circles.

Answer 5

Yeah, they are circles.

Answer 6

I know the two solutions but getting there is just :(

Answer 7

Oh? what are x and y?

Answer 8

One solution is (-2,-2)

Answer 9

x^2 + y^2 + 2x + 2y = 0
y^2 = -x^2 - 2x - 2y

x^2 + y^2 + 4x + 6y + 12 = 0
x^2 - x^2 - 2x - 2y + 4x + 6y + 12 =0

2x + 4y + 12 = 0
x + 2y = 6
was gonna do something like this but its still two variables

Answer 10

x + 2y = -6*

Answer 11

x = -2y - 6
(-2y-6)^2 + y^2 + 4(-2y-6) + 6y + 12 = 0
?..

Answer 12

I'm not sure, never solved circle equations this way!

Answer 13

buddy why dont you substitute x^2+y^2 from 1st to 2nd
then the same for 2nd to 1st
u'll get 2 equations of x and y
then its just simple

Answer 14

Oh, why didnt i think of that LOL

Answer 15

x^2 + y^2 = -2x -2y
-2x -2y + 4x + 6y +12 = 0
2x + 4y + 12 = 0
-4x - 6y - 12 + 2x +2y = 0

so

2x + 4y = -12
-2x - 4y = 12
wait no that doesn't work, then its infinite solutions

Answer 16

either that or i cant do algebra...
prob the second one

Answer 17

thats interesting .... :P
then take eqn of y as x and put in 1s or 2nd one .......solve it as a quad and u'll get two values of x