Find the coordinates of the centre and the radius of the circle
x^2 + y^2 - 4x + 6y - 12 = 0
I need help with the second part. Anyway coordinates for centre of circle is (2,-3). However, I don't really quite get the question for 2nd part. Does it mean find the coordinate of the radius of the circle or just the radius?
Cuz the answer is y = 3/4 x - 9/2
Which leads to me think that this could be a wrong answer because how can there be a fixed equation for the radius of a circle? Since a circle goes in a round, for example at x - coordinate 5, there could be 2 answers for the y -coordinates, for the upper and lower part of the circle.
Do clear my doubts if possible. Thank you very much.
Hi,
x^2 + y^2 - 4x + 6y - 12 = 0
=> (x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0, by completing squares
=> (x - 2)^2 + (y + 3)^2 = 5^2.
So centre is at (2, -3) and radius is of length 5 units.
Thanks!
Cheers,
Wen Shih