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Question 537542: what is the center and length of a circle with the equation 81x^2 + 81y^2 = 1?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! what is the center and length of a circle with the equation 81x^2 + 81y^2 = 1?
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Standard form of an equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k) being the (x,y) coordinates of the center, r=radius.
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81x^2 + 81y^2 = 1
Divide by 81
x^2+y^2=1/81
center: (0,0)
radius: √(1/81)=1/9 (I assumed this what you meant by length of a circle)
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