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Finding centre of (moving) circle

  • Thread starter sa1988
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  • #1
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Homework Statement



Find the co-ordinate of the centre of the following circle as a function of time:

x2+y2 = C + 2 t x

Homework Equations




The Attempt at a Solution



No idea..!

It's part of a fluid dynamics problem, which I don't need to explain here, other than to say I plotted it on Mathematica and can see that the circle moves to the right at a rate of 't'.

So the answer is : Centre of circle = (t, 0) as a function of time.

But I don't know how to get to that answer.

The only thing that I think might be helpful is to rearrange it into the form:

x(x-2t)+y2 = C

As this looks a little more like a circle and conforms a little bit to the (x-a)2+(y-b)2 = R2 idea.

Thanks in advance!
 
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Answers and Replies

  • #2
SteamKing
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Homework Statement



Find the co-ordinate of the centre of the following circle as a function of time:

x2+y2 = C + 2 t x

Homework Equations




The Attempt at a Solution



No idea..!

It's part of a fluid dynamics problem, which I don't need to explain here, other than to say I plotted it on Mathematica and can see that the circle moves to the right at a rate of 't'.

So the answer is : Centre of circle = (t, 0) as a function of time.

But I don't know how to get to that answer.

Thanks in advance!
Have you tried re-writing the equation of this circle in its general form?

http://www.mathwarehouse.com/geometry/circle/equation-of-a-circle.php
 
  • #3
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21
Have you tried re-writing the equation of this circle in its general form?

http://www.mathwarehouse.com/geometry/circle/equation-of-a-circle.php
Yep, sorry I just edited my post to add that.

I've rewritten it as

x(x-2t)+(y-0)2 = C

As this looks a little more like a circle and conforms a little bit to the (x-a)2+(y-b)2 = R2 idea.

But I'm not sure where to take it from there to be honest, or even if I've rewritten in properly...

It's that extra x in the 2tx term which is confusing me...
 
  • #4
SteamKing
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Yep, sorry I just edited my post to add that.

I've rewritten it as

x(x-2t)+(y-0)2 = C

As this looks a little more like a circle and conforms a little bit to the (x-a)2+(y-b)2 = R2 idea.

But I'm not sure where to take it from there to be honest, or even if I've rewritten in properly...
No, this is still not the correct form of the equation.

You want (x - h)2 + (y - k)2 = R2

You might have to complete the square to obtain the necessary form of the equation.

Didn't you study conic sections in algebra class?
 
  • #5
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No, this is still not the correct form of the equation.

You want (x - h)2 + (y - k)2 = R2

You might have to complete the square to obtain the necessary form of the equation.

Didn't you study conic sections in algebra class?
Oh bugger, it's fantastically obvious now.

I'm actually halfway through a theoretical physics degree (!!!) so I'm putting my hands way up and admitting full, shameful ignorance here, although in my defence I've had no need to complete any squares for years. It just faded from memory, I guess. Time for some serious high school revision...!

:oops:

Thanks for the guidance :oldsmile:
 
  • #6
SteamKing
Staff Emeritus
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Oh bugger, it's fantastically obvious now.

I'm actually halfway through a theoretical physics degree (!!!) so I'm putting my hands way up and admitting full, shameful ignorance here, although in my defence I've had no need to complete any squares for years. It just faded from memory, I guess. Time for some serious high school revision...!

:oops:

Thanks for the guidance :oldsmile:
You're welcome. Good luck on the physics degree.
 

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