Finding centre of (moving) circle

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The discussion revolves around finding the center of a moving circle represented by the equation x² + y² = C + 2tx. The center of the circle is determined to be (t, 0) as a function of time, indicating that it moves to the right at a rate of 't'. The participants discuss the need to rewrite the equation in a standard form for circles, suggesting the completion of the square to achieve this. One participant acknowledges their struggle with the concept due to a lack of recent practice, despite being in a theoretical physics program. The conversation concludes with encouragement and well wishes for academic success.
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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