Question

What is the equation of the circle in general form? x2+y2−6x+10y+25=0 x2+y2+6x−10y+25=0 x2+y2−6x+10y+22=0 x2+y2+6x−10y+22=0?

Answer 1

x2+y2-6x+10y+25=0

Explanation:

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Answer 2

By completing squares for all the equations we have:
x2 + y2-6x + 10y + 25 = 0
x2 + y2-6x + 10y = -25
x2 + y2-6x + 10y + (- 6/2) ^ 2 + (10/2) ^ 2 = -25 + (- 6/2) ^ 2 + (10/2) ^ 2
x2 + y2-6x + 10y + 9 + 25 = -25 + 9 + 25
x2-6x + 9 + y2 + 10y + 25 = 9
(x-3) ^ 2 + (y + 5) ^ 2 = 9

x2 + y2 + 6x-10y + 25 = 0
x2 + y2 + 6x-10y = -25
x2 + y2 + 6x-10y + (6/2) ^ 2 + (- 10/2) ^ 2 = -25 + (6/2) ^ 2 + (- 10/2) ^ 2
x2 + y2 + 6x-10y + 9 + 25 = -25 + 9 + 25
x2 + 6x + 9 + y2-10y + 25 = 9
(x + 3) ^ 2 + (y-5) ^ 2 = 9

x2 + y2-6x + 10y + 22 = 0
x2 + y2-6x + 10y = -22
x2 + y2-6x + 10y + (- 6/2) ^ 2 + (10/2) ^ 2 = -22 + (- 6/2) ^ 2 + (10/2) ^ 2
x2 + y2-6x + 10y + 9 + 25 = -22 + 9 + 25
x2-6x + 9 + y2 + 10y + 25 = 12
(x-3) ^ 2 + (y + 5) ^ 2 = 12

x2 + y2 + 6x-10y + 22 = 0
x2 + y2 + 6x-10y = -22
x2 + y2 + 6x-10y + (6/2) ^ 2 + (- 10/2) ^ 2 = -22 + (6/2) ^ 2 + (- 10/2) ^ 2
x2 + y2 + 6x-10y + 9 + 25 = -22 + 9 + 25
x2 + 6x + 9 + y2-10y + 25 = 12
(x + 3) ^ 2 + (y-5) ^ 2 = 12