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#?
1 Answer
Mar 2, 2016
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#
