Question 593945: 5x^2+5y^2-20=0 in standard form
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 5x^2+5y^2-20=0 in standard form
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5x^2+5y^2-20=0
divide by 5
5x^2+5y^2-20=0
x^2+y^2-4=0
x^2+y^2=4
This is an equation of a circle with center at (0,0) and radius=2, of the standard form:
(x-h)^2+(y-k)^2=r^2, (h,k)=(x,y) coordinates of center and r=radius
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