Question 1026719
 
Question:
The equation x^2+y^2-2x-4y-4=0 represents a circle. What are the coordinates for the center of the circle? What is the length of the radius of the circle? 
 
Solution:
The general equation of a circle centred at (a,b) with radius r is written as:
{{{(x-a)^2+(y-b)^2=r^2}}}........ (1)
By completing squares, we have
{{{x^2+y^2-2x-4y-4}}}
={{{x^2-2x+1-1 + y^2-4y+4-4 -4}}}
={{{(x-1)^2 -1 + (y-2)^2 -4 -4 }}}
={{{(x-1)^2+(y-2)^2-9}}}
=0
Hence the equation of the circle can be rewritten as
(x-1)^2+(y-2)^2=3^3
By comparison with the general equation (1) above, we conclude that the centre of the circle is (a,b)=(1,2) and the radius is 3.