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show that the two circlex^2+y^24x2y11=0 and x^2+y^2+20x12y+72=0 do not intersect.Hint;find the distance between their centers
 one year ago
 one year ago
show that the two circlex^2+y^24x2y11=0 and x^2+y^2+20x12y+72=0 do not intersect.Hint;find the distance between their centers
 one year ago
 one year ago

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wioBest ResponseYou've already chosen the best response.0
Use completing the square to get them both in the form of :\[ (xh)^2+(yk)^2=r^2 \]
 one year ago

wioBest ResponseYou've already chosen the best response.0
For example: \[ x^24x = (x2)^24 \]
 one year ago

lissafiBest ResponseYou've already chosen the best response.0
i still cant work it wio
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.2
General equation: x^2 + y^2 +2gx + 2hy + c=0 Centre ( g,f)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.2
x^2+y^24x2y11=0 Centre (2,1)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.2
x^2+y^2+20x12y+72=0 Centre (10,6)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.2
Hope this will help u..
 one year ago

wioBest ResponseYou've already chosen the best response.0
Still gotta find radius though. @Yahoo!
 one year ago

wioBest ResponseYou've already chosen the best response.0
You really learn that in class?
 one year ago

lissafiBest ResponseYou've already chosen the best response.0
@yahoo how do u get the center x^2+y^2+20x12y+72=0 Centre (10,6)?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.2
Compare it with the General equation..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.2
x^2 + y^2 + 2gx + 2fy + c=0 C ( g,f) so Here x^2+y^2+20x12y+72=0 2g = 20 g = 10 2f = 12 f = 6 Since the Centre is g,f C ( 10 ,  (6) ) C ( 10 , 6)
 one year ago

wioBest ResponseYou've already chosen the best response.0
@lissafi Do you know how to complete the square?
 one year ago

wioBest ResponseYou've already chosen the best response.0
Just ignore the \(y\) terms and try to complete the square for (x\), then ignore the \(x\) terms and complete the square for \(y\)
 one year ago

lissafiBest ResponseYou've already chosen the best response.0
wio is that what yahoo did?cos i am lost on this method
 one year ago

wioBest ResponseYou've already chosen the best response.0
I dunno what Yahoo! did...
 one year ago

wioBest ResponseYou've already chosen the best response.0
\[ x^2+y^24x2y11=0 \]Okay so we just ignore the \(y\) terms... \[ x^24x11=0 \]How do you complete the square here?
 one year ago

lissafiBest ResponseYou've already chosen the best response.0
wio i dont know much on completing the square
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.2
x^2+y^24x2y11=0 x^24x + y^2 2y = 11 x^2  4x + 4 + y^2  2y + 1 = 11 + 4 + 1 (x2)^2 + (y1)^2 = 16 (x2)^2 + (y1)^2 = 4^2 now it is a circle with centre (2,1) and radius 4
 one year ago

lissafiBest ResponseYou've already chosen the best response.0
yap,and for x^2+y^2+20x12y+72=0?
 one year ago

lissafiBest ResponseYou've already chosen the best response.0
i lost in (x2)^2 + (y1)^2 = 16
 one year ago

lissafiBest ResponseYou've already chosen the best response.0
i know you add +4+1 on both side
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.2
Since we added 4 and 1 on LHS we have to do in RHS tooooo
 one year ago

lissafiBest ResponseYou've already chosen the best response.0
i know you can find x to subtitute to get y and have the points,but i dont know this method
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
in \[x^2+y^2+20x12y+72=0\] what term must be added to \[x^2+20x\] to get a perfect square trinomial that can be written in the form \[(x+\_\_)^2\]
 one year ago
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