# How do you find the center and radius of the circle given by the equation x^2+y^2-8 x- 6 y +21=0?

Mar 2, 2016

centre = (4 , 3 ) , r = 2

#### Explanation:

The general equation of a circle is :

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

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

compare this with ${x}^{2} + {y}^{2} - 8 x - 6 y + 21 = 0$

hence: 2g = -8 → g=-4 , 2f =-6 → f=-3 and c=21

centre=(4,3) and r = $\sqrt{{\left(- 4\right)}^{2} + {\left(- 3\right)}^{2} - 21} = \sqrt{4} = 2$