How do you find the radius of the circle: #x^2 + y^2 + 12x - 2y + 21 = 0#?

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How do you find the radius of the circle: #x^2 + y^2 + 12x - 2y + 21 = 0#?

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Answer 1 · 2016-03-09T13:26:54

#r=4#

Explanation:

#x^2+y^2+D x+E y+F=0#
#C(a,b)" coordinates for center of circle"#
#a=-D/2" "b=-E/2" "F=21#
#x^2+y^2+12x-2y+21=0#
#D=12" "E=-2#
#a=-12/2=-6" "b=2/2=1#
#r=1/2 sqrt(D^2+E^2-4F)#
#r=1/2 sqrt(12^2+(-2)^2-4*21)#
#r=1/2sqrt(144+4-84)#
#r=1/2sqrt64" "r=1/2*8#
#r=4#

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How do you find the radius of the circle: #x^2 + y^2 + 12x - 2y + 21 = 0#?

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