equation of a circle?

Question

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

find the centre and radius of the circle

Answer 1

You need to get it into this form: (x-h)^2 + (y-k)^2=r^2. When you do this, P(h,k) will be your center, and r will be your radius. This requires completing two squares. Start by rearranging like this:

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

Then add the constants needed to complete the square which will be (b/2a)^2 where a is the leading coefficient and b is the second coefficient:

x^2-4x+4 + y^2-6y+9 = 12+4+9

Then you can make it into the squares:

(x-2)^2 + (y-3)^2 = 5^2

center = P(2,3)
radius = 5

There you go.

Answer 2

completing squares

x^2 - 4x + y^2 - 6y = 12
x^2 - 4x + 4 + y^2 - 6y + 9 = 12 + 4 + 9
(x-2)^2 + (y-3)^2 = 25
center @ (2,3) radius = 5

Answer 3

the answere is 7