Question 771465
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.
:-)