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 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
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

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

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

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

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

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

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

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

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

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

lissafi
 one year ago
Best 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)?

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

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.2x^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)

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

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

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

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

wio
 one year ago
Best 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?

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

Yahoo!
 one year ago
Best ResponseYou've already chosen the best response.2x^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

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

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

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

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

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

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.0in \[x^2+y^2+20x12y+72=0\] what term must be added to \[x^2+20x\] to get a perfect square trinomial that can be written in the form \[(x+\_\_)^2\]
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