Question

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

Answer 1

Use completing the square to get them both in the form of :\[
(x-h)^2+(y-k)^2=r^2
\]

Answer 2

For example: \[
x^2-4x = (x-2)^2-4
\]

Answer 3

i still cant work it wio

Answer 4

General equation:

x^2 + y^2 +2gx + 2hy + c=0

Centre ( -g,-f)

Answer 5

x^2+y^2-4x-2y-11=0

Centre (2,1)

Answer 6

x^2+y^2+20x-12y+72=0

Centre (-10,6)

Answer 7

Hope this will help u..

Answer 8

Still gotta find radius though. @Yahoo!

Answer 9

\[r=\sqrt{g^2+f^2-c}\]

Answer 10

You really learn that in class?

Answer 11

@yahoo how do u get the center x^2+y^2+20x-12y+72=0

Centre (-10,6)?

Answer 12

Compare it with the General equation..

Answer 13

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)

Answer 14

@lissafi Do you know how to complete the square?

Answer 15

not on this

Answer 16

Just ignore the \(y\) terms and try to complete the square for (x\), then ignore the \(x\) terms and complete the square for \(y\)

Answer 17

wio is that what yahoo did?cos i am lost on this method

Answer 18

I dunno what Yahoo! did...

Answer 19

\[
x^2+y^2-4x-2y-11=0
\]Okay so we just ignore the \(y\) terms... \[
x^2-4x-11=0
\]How do you complete the square here?

Answer 20

wio i dont know much on completing the square

Answer 21

Look it up...

Answer 22

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

Answer 23

@lissafi u got it..now

Answer 24

yap,and for x^2+y^2+20x-12y+72=0?

Answer 25

Try it out...

Answer 26

i lost in (x-2)^2 + (y-1)^2 = 16

Answer 27

i know you add +4+1 on both side

Answer 28

Since we added 4 and 1 on LHS we have to do in RHS tooooo

Answer 29

i know you can find x to subtitute to get y and have the points,but i dont know this method

Answer 30

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 \[(x+\_\_)^2\]