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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i still cant work it wio

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x^2+y^24x2y11=0 Centre (2,1)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x^2+y^2+20x12y+72=0 Centre (10,6)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hope this will help u..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Still gotta find radius though. @Yahoo!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[r=\sqrt{g^2+f^2c}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You really learn that in class?

anonymous
 3 years 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)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Compare it with the General equation..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x^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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@lissafi Do you know how to complete the square?

anonymous
 3 years 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 \(y\)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I dunno what Yahoo! did...

anonymous
 3 years 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wio i dont know much on completing the square

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x^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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@lissafi u got it..now

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know you add +4+1 on both side

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since we added 4 and 1 on LHS we have to do in RHS tooooo

anonymous
 3 years 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

anonymous
 3 years 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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