SOLUTION: x^2+y^2+6x+4y+12=0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: x^2+y^2+6x+4y+12=0      Log On


   



Question 575999: x^2+y^2+6x+4y+12=0
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2+6x+4y+12=0
----
Complete the square on the x-terms and on the y-terms.
----
x^2+6x+9 + y^2+4y+4 + 12 -13 = 0
(x+3)^2 + (y+2)^2 = 1
---
Circle with center at (-3,-2); radius = 1
============================================
Cheers,
Stan H.
=============