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show that the two circlex^2+y^2-4x-2y-11=0 and x^2+y^2+20x-12y+72=0 do not intersect.Hint;find the distance between their centers

Mathematics
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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

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Other answers:

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..
Still gotta find radius though. @Yahoo!
\[r=\sqrt{g^2+f^2-c}\]
You really learn that in class?
@yahoo how do u get the center x^2+y^2+20x-12y+72=0 Centre (-10,6)?
Compare it with the General equation..
x^2 + y^2 + 2gx + 2fy + c=0 C ( -g,-f) so Here x^2+y^2+20x-12y+72=0 2g = 20 g = 10 2f = -12 f = -6 Since the Centre is -g,-f C ( -10 , - (-6) ) C ( -10 , 6)
@lissafi Do you know how to complete the square?
not on this
Just ignore the \(y\) terms and try to complete the square for (x\), then ignore the \(x\) terms and complete the square for \(y\)
wio is that what yahoo did?cos i am lost on this method
I dunno what Yahoo! did...
\[ x^2+y^2-4x-2y-11=0 \]Okay so we just ignore the \(y\) terms... \[ x^2-4x-11=0 \]How do you complete the square here?
wio i dont know much on completing the square
Look it up...
x^2+y^2-4x-2y-11=0 x^2-4x + y^2 -2y = 11 x^2 - 4x + 4 + y^2 - 2y + 1 = 11 + 4 + 1 (x-2)^2 + (y-1)^2 = 16 (x-2)^2 + (y-1)^2 = 4^2 now it is a circle with centre (2,1) and radius 4
@lissafi u got it..now
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
in \[x^2+y^2+20x-12y+72=0\] what term must be added to \[x^2+20x\] to get a perfect square trinomial that can be written in the form \[(x+\_\_)^2\]

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