Radius of a circle?

Question

What is the radius of the circle given the equation:

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

how do you get this?

Answer 1

The equation of a circle with center (h,k) and radius r is:

(x - h)² + (y - k)² = r²

To get our equation in that form we need to complete two squares.

x² + y² + 4x - 6y + 12 = 0

x² + 4x + y² - 6y = -12

(x² + 4x + 4) + (y² - 6y + 9) = -12 + 4 + 9

(x + 2)² + (y - 3)² = 1

So the center is (-2, 3) and the radius is 1.

Answer 2

x^2 + y^2 + 4x - 6y + 12 = 0
x^2 + 4x + y^2 - 6y + 12 = 0
(x^2 + 4x) + (y^2 - 6y) + 12 = 0
(x^2 + 4x + 4 - 4) + (y^2 - 6y + 9 - 9) + 12 = 0
((x + 2)^2 - 4) + ((y - 3)^2 - 9) + 12 = 0
(x + 2)^2 - 4 + (y - 3)^2 - 9 + 12 = 0
(x + 2)^2 + (y - 3)^2 - 1 = 0
(x + 2)^2 + (y - 3)^2 = 1

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

(h,k) is the center and r is the radius

r^2 = 1
r = 1

Radius is 1

Answer 3

x^2 + 4x + y^2 + 6y = -12
(x^2 + 4x +4) + (y^2 + 6y + 9) = -12 + 4 + 9
(x+2)^2 + (y+3)^2 = 1

radius is 1..

Answer 4

The trick is to put the equation into the form

(x-a)^2 + (y-b)^2 = r^2

The radius is then r.

Answer 5

Impossible to get the answer (The question is incomplete)