# Finding centre of (moving) circle

1. Feb 12, 2016

### sa1988

1. The problem statement, all variables and given/known data

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

x2+y2 = C + 2 t x

2. Relevant equations

3. 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.

Last edited: Feb 12, 2016
2. Feb 12, 2016

### SteamKing

Staff Emeritus
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. Feb 12, 2016

### sa1988

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. Feb 12, 2016

### SteamKing

Staff Emeritus
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. Feb 12, 2016

### sa1988

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...!

Thanks for the guidance

6. Feb 12, 2016

### SteamKing

Staff Emeritus
You're welcome. Good luck on the physics degree.