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 Post subject: 2 tangent CirclesPosted: Fri, 23 May 2003 10:32:18 UTC
The circle C_{2} with the center in A_{2,1} is tangent to the circle
C_{1}: x^2+y^2+4x-8y+11=0

The eq. of C_[1} is x^2+y^2-4x-2y+5-r^2=0

Now i have to solve the following system:

x^2+y^2+4x-8y+11=0
x^2+y^2-4x-2y+5-r^2=0

I know that this system has only 1 solution and i still dont know how to solve it...

PS:The board doesnt let me tex this,where do i have written the syntax wrong?

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 Post subject: Posted: Fri, 23 May 2003 11:26:26 UTC
 S.O.S. Oldtimer

Joined: Mon, 19 May 2003 07:15:29 UTC
Posts: 204
Location: Los Angeles
You have:
(1):
and
(2):

Subtract (2) from (1):

which we rewrite as:

Dividing by 6:
(3):
You can go back to (1) and replace the occurances with the RHS of (3)
This will yield a quadratic in , for which you can find two solutions.
Then you can substitite these values of back into (1) to find the values for .

Hope this helps.

_________________
My thoughts on Pi:
How I need a drink, alcoholic of course, drunk for happy feelings ...

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 Post subject: Posted: Fri, 23 May 2003 12:35:33 UTC
Yes,but how do i get rid of the r?Or how do i determine it?Its the radius of the first circle...

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 Post subject: Posted: Fri, 23 May 2003 13:20:04 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 4 May 2003 16:04:19 UTC
Posts: 2906
circle 1: (2,-4) with radius 3
circle 2: (-2,-1) with radius r.

distance between the centers should be 3+r, assuming the circles are externally tangent. internally tangent is another case

_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:

where

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 Post subject: Posted: Fri, 23 May 2003 14:28:44 UTC
,how could i miss that ,thanks bugz and andyistic!

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