SOLUTION: Find the center, radius, and intercepts for the circle
3x^2 + 3y^2 + 8y - 4 = 0
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Question 53516: Find the center, radius, and intercepts for the circle
3x^2 + 3y^2 + 8y - 4 = 0
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Find the center, radius, and intercepts for the circle
3x^2 + 3y^2 + 8y - 4 = 0
x^2+y^2+8y/3 -4/3=0
(x-0)^2 + [(y)^2+2(y)(4/3)+(4/3)^2-(4/3)^2]-4/3 = 0
(x-0)^2 + {y+(4/3)}^2=4/3 +16/9 = 28/9
comparing with std.eqn.
(x-h)^2 + (y-k)^2 = r^2.....we have
h=0....k=-4/3.....centre =(0,-4/3).......r= sqrt(28)/3
x intercepts are when y=0....
3x^2=4
x= +2/sqrt(3) ...and....x = -2/sqrt(3)
y intercepts are when x=0...
3y^2+8y-4=0
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