Question

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`?

Answer 1

centre = (4 , 3 ) , r = 2

Explanation:

The general equation of a circle is :

`x^2 + y^2 + 2gx + 2fy + c = 0`

centre = ( -g , -f ) and r =`sqrt(g^2+f^2 -c)`

compare this with `x^2 + y^2 - 8x -6y + 21 = 0`

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

centre=(4,3) and r = `sqrt((-4)^2+(-3)^2-21) = sqrt4 = 2`