The center of the circle x^2+y^2+2x-4y-4=0
Answers
Answered by
3
Answer:
x^2+y^2+2x-4y-4=0
x^2+2x+1+y^2-4y+4-1-4-4=0
(x+1)^2+(y-2)^2=(3)^2
center of the circle is(-1,2)
Similar questions