Question 771465: what is the radius of the circle x^2+y^2-2x-4y-31=0.
Answer by Cromlix(4381)   (Show Source):
You can put this solution on YOUR website! The general equation of a circle
= x^2 + y^2 + 2gx + 2fy + c = 0
Centre = (-g,-f)
Therefore the equation of the circle:
x^2 + y^2 - 2x - 4y - 31 = 0
has (1,2) as the coordinates of its centre.
The radius is found from the equation:
square root of(g^2 + f^2 - c)
square root of(1^2 + 2^2 -(-31))
square root of(1 + 4 + 31)
square root of (36)
= 6 units.
Hope this helps.
:-)
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