Finding centre of (moving) circle

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SUMMARY

The discussion focuses on finding the center of a moving circle defined by the equation x² + y² = C + 2tx. The center of the circle is determined to be (t, 0) as a function of time. Participants suggest rewriting the equation in the standard form of a circle, (x - h)² + (y - k)² = R², and recommend completing the square to achieve this. The conversation highlights the importance of algebraic manipulation in solving fluid dynamics problems.

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  • Familiarity with algebraic manipulation techniques, specifically completing the square
  • Basic knowledge of fluid dynamics concepts
  • Experience with mathematical software, such as Mathematica
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  • Learn how to complete the square for quadratic equations
  • Study the properties of conic sections, focusing on circles
  • Explore fluid dynamics problems that involve moving geometrical shapes
  • Practice using Mathematica for plotting and analyzing mathematical functions
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Students in physics or mathematics, particularly those studying fluid dynamics or conic sections, as well as educators looking for examples of algebraic manipulation in applied contexts.

sa1988
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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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sa1988 said:

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
 
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SteamKing said:
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...
 
sa1988 said:
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?
 
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SteamKing said:
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:
 
sa1988 said:
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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