Question

what is the standard form of the equation of the circle X^2-4x+y^2+6y+4=0?

Answer 1

need to complete the square

Answer 2

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

Answer 3

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

Answer 4

add on (b/2)^2 to both sides, that is how i got the +4 and the +9

they are half of 4 or 2^2 and half of 6 squared or 9

Answer 5

what do you do after or are we done? I'm confused.

Answer 6

that is the standard form, all done

Answer 7

just complete the square to put into the form

(x-h)^2 + (y-k)^2 = r^2

Answer 8

ohh Ok Thanks for all your help!!!

Answer 9

yw

Answer 10

It is a circle centered at (h,k) = (2,-3) with a radius r=3