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Use completing the square to get them both in the form of :\[
(x-h)^2+(y-k)^2=r^2
\]

For example: \[
x^2-4x = (x-2)^2-4
\]

i still cant work it wio

General equation:
x^2 + y^2 +2gx + 2hy + c=0
Centre ( -g,-f)

x^2+y^2-4x-2y-11=0
Centre (2,1)

x^2+y^2+20x-12y+72=0
Centre (-10,6)

Hope this will help u..

\[r=\sqrt{g^2+f^2-c}\]

You really learn that in class?

Compare it with the General equation..

not on this

wio is that what yahoo did?cos i am lost on this method

I dunno what Yahoo! did...

wio i dont know much on completing the square

Look it up...

yap,and for x^2+y^2+20x-12y+72=0?

Try it out...

i lost in (x-2)^2 + (y-1)^2 = 16

i know you add +4+1 on both side

Since we added 4 and 1 on LHS we have to do in RHS tooooo

i know you can find x to subtitute to get y and have the points,but i dont know this method