Question 592698: Find the center and radius of x^2+y^2+x+y-1/2=0
Answer by mamiya(56)   (Show Source):
You can put this solution on YOUR website! Find the center and radius of x^2+y^2+x+y-1/2=0
x^2+y^2+x+y-1/2=0 --> X^2 + X + Y^2 - 1/2 =0
--> X^2 + X + 1/4 -1/4 + Y^2 - 0 -1/2 = 0
--> (X + 1/2 )^2 + ( Y + 0)^2 - 1/4 - 1/2 = 0
--> ( X + 1/2)^2 + (Y+0)^2 = 3/4
--> [X - (-1/2)]^2 + ( Y - 0)^2 = (SQRT(3/4))^2
SO, BY LOOKING AT THIS PARAMETRIC EQUATION A CIRCLE, YOU CAN NOW ANSWER THE QUESTION:
* THE CENTER IS A POINT OF THE FOLLOWING COORDINATES ( -1/2, 0)
* THE RADIUS IS THE SQUARE ROOT OF 3/4, ( [sqrt(3)]/2 )
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