lissafi
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
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wio
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Use completing the square to get them both in the form of :\[
(x-h)^2+(y-k)^2=r^2
\]
wio
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For example: \[
x^2-4x = (x-2)^2-4
\]
lissafi
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i still cant work it wio
Yahoo!
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General equation:
x^2 + y^2 +2gx + 2hy + c=0
Centre ( -g,-f)
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x^2+y^2-4x-2y-11=0
Centre (2,1)
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x^2+y^2+20x-12y+72=0
Centre (-10,6)
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Hope this will help u..
wio
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Still gotta find radius though. @Yahoo!
Yahoo!
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\[r=\sqrt{g^2+f^2-c}\]
wio
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You really learn that in class?
lissafi
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@yahoo how do u get the center x^2+y^2+20x-12y+72=0
Centre (-10,6)?
Yahoo!
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Compare it with the General equation..
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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)
wio
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@lissafi Do you know how to complete the square?
lissafi
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not on this
wio
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Just ignore the \(y\) terms and try to complete the square for (x\), then ignore the \(x\) terms and complete the square for \(y\)
lissafi
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wio is that what yahoo did?cos i am lost on this method
wio
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I dunno what Yahoo! did...
wio
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\[
x^2+y^2-4x-2y-11=0
\]Okay so we just ignore the \(y\) terms... \[
x^2-4x-11=0
\]How do you complete the square here?
lissafi
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wio i dont know much on completing the square
wio
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Look it up...
Yahoo!
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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
Yahoo!
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@lissafi u got it..now
lissafi
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yap,and for x^2+y^2+20x-12y+72=0?
Yahoo!
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Try it out...
lissafi
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i lost in (x-2)^2 + (y-1)^2 = 16
lissafi
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i know you add +4+1 on both side
Yahoo!
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Since we added 4 and 1 on LHS we have to do in RHS tooooo
lissafi
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i know you can find x to subtitute to get y and have the points,but i dont know this method
sirm3d
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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 \[(x+\_\_)^2\]