How do you graph #x² + y² - 2x - 8y + 16 =0#?
1 Answer
Feb 17, 2016
circle: centre (1 , 4 ) , r = 1
Explanation:
The general equation of a circle is
#x^2 + y^2 + 2gx + 2fy + c = 0#
#x^2 + y^2 - 2x - 8y + 16 = 0 " is in this form " # and by comparison of the 2 equations
2g = -2 → g = -1 , 2f = - 8 → f = - 4 and c = 16
centre = ( - g , -f ) = ( 1 , 4 )
and
# r = sqrt(g^2 + f^2 - c ) = sqrt((-1)^2 +(-4)^2 - 16) = 1 #
