How do you find the center and radius of the circle x^2 + y^2 - 2x + 6y + 3 = 0?

May 23, 2016

$\left(1 , - 3\right) , r = \sqrt{7}$

Explanation:

The general equation of a circle is

${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

with centre = (-g ,-f) and r $= \sqrt{{g}^{2} + {f}^{2} - c}$

the equation ${x}^{2} + {y}^{2} - 2 x + 6 y + 3 = 0 \text{ is in this form}$

comparing like terms : 2g = -2 → g = -1 , 2f = 6 → f = 3 , c = 3

thus : centre = (1 ,-3) and r$= \sqrt{{\left(- 1\right)}^{2} + {3}^{2} - 3} = \sqrt{7}$