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Topic: Circles in a corner (Read 517 times) |
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Noke Lieu
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Whilst wrestling with somethingelse, this presented itself to me.
Take two circles of different radii and back them into a corner (as per the diagram)
What's the question though? That's something that's eluding me. I guess I'll settle for
The ratio of AB:BC
(because I haven't made it look particularly pretty, and hope someone around here can  )
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Obob
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Re: Circles in a corner
« Reply #1 on: Jan 22nd, 2008, 12:46am » |
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It is not too difficult to find the coordinates of the points of intersection of the two circles, from which a formula for the ratio follows.
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towr
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Re: Circles in a corner
« Reply #2 on: Jan 22nd, 2008, 1:12am » |
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If the ratio is constant, it seems 3+2sqrt(2)
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| « Last Edit: Jan 22nd, 2008, 11:31am by towr » |
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Obob
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Re: Circles in a corner
« Reply #3 on: Jan 22nd, 2008, 10:05am » |
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The way I interpreted the problem, the radi of the two circles are arbitrary. The circles are not necessarily orthogonal. And thus the ratio is not constant.
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towr
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Re: Circles in a corner
« Reply #4 on: Jan 22nd, 2008, 10:23am » |
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Actually, I'm not assuming the ratio would be constant because supposedly the circles are orthogonal. If the ratio isn't constant regardless, I doubt the value I found is the ratio in the case of orthogonal circles (because that's not the case I used to calculate it).
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Obob
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Re: Circles in a corner
« Reply #5 on: Jan 22nd, 2008, 10:29am » |
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Can we get some clarification as to where the point B is? Is it on the y-axis or is it the point where the two circles meet?
My calculations were based on B being on the y-axis.
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Noke Lieu
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Re: Circles in a corner
« Reply #6 on: Jan 22nd, 2008, 3:22pm » |
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on Jan 22nd, 2008, 10:29am, Obob wrote:Can we get some clarification as to where the point B is? Is it on the y-axis or is it the point where the two circles meet?
My calculations were based on B being on the y-axis. |
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I intended B to be on the y axis.
But your question makes me think of the lengths of the arcs.
I am curious about how you guys went about determining where the circles intercept. I went about it in a seemingly obvious fashion, then I noticed something cuter.
(Doesn't quite work when there is no intercept though)
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Obob
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Re: Circles in a corner
« Reply #7 on: Jan 22nd, 2008, 4:04pm » |
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I just took the equations (x-R)^2+(y-R)^2 = R^2 and (x-r)^2+(y-r)^2=r^2. Subtract the second equation from the first to get a linear equation relating x and y; solve for one of them and put it into the first equation to get a quadratic in x, say. Then use the quadratic formula to solve for x, and solving for y in the second equation gives the four solutions (x,y).
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Noke Lieu
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Re: Circles in a corner
« Reply #8 on: Jan 22nd, 2008, 7:48pm » |
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Exactly what I did.
Then I noticed that the result of
(x-R)^2+(y-R)^2 = R^2 minus
(x-r)^2+(y-r)^2=r^2
could be re-arranged to y= -x + (R+r)/2
(so on the line that is perpendicular to the line joining centres; intercepting the x and y axes at the average of the two radii)
obvious when one thinks about it.
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| « Last Edit: Jan 22nd, 2008, 7:49pm by Noke Lieu » |
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