# Find the differential equation of all circles of radius a [closed]

Can someone please post a detailed step-by-step procedure. Given the circle with a radius a, what is the differential equation of the circle.

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## closed as unclear what you're asking by Rory Daulton, yoknapatawpha, Alex Provost, ComplexPhi, kamil09875Jan 23 at 21:55

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quoting you, "what is the differential equation of the circle"? – Ilya Dec 3 '11 at 18:40
Does this other account belong to you as well? If so, we can flag the moderators to merge these two accounts; it'll make it convenient for you to keep track of your questions. Thanks, – Srivatsan Dec 3 '11 at 23:45
Nope, that account does not belong to me Srivatsan. – Nikhil Mulley Dec 4 '11 at 5:12

From the implicit equation of the circle $(x-u)^2+(y-v)^2=a^2$, you get $$x'(x-u)+y'(y-v)=0$$ by implicit differentiation. Add the initial condition $$x(0)=u+a, \quad y(0)=v$$

You can write the differential equations as $$x'=-y+v, \quad y' = x-u$$ which is especially nice for circles centered at the origin.

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Thanks Norbert and lhf. Appreciate your answers. I am however surprised to see clean mathematical notations here. Do you use any software as such to paste these equations here? (Pardon me, I am using math.stackexchange for the first time. – Nikhil Mulley Dec 3 '11 at 19:14
@Nikhil, this site understands TeX. View the source of the web pages to see how it's done. See also math.stackexchange.com/editing-help. Welcome to MSE. – lhf Dec 3 '11 at 19:18
Thanks :-) BTW, I was also wondering if there is any kind of math application, which would give me sort of help to find the steps as in for example: Evaluate the integral x^2 (sqrt(a pow3 + x pow3)) dx. – Nikhil Mulley Dec 13 '11 at 12:46

Circle equation $$(x-C_1)^2+(y-C_2)^2=a^2\quad (0)$$ Differentiate twice by $x$ $$(x-C_1)+y'(y-C_2)=0\quad (1)$$ $$1+(y-C_2)y''+(y')^2=0\quad (2)$$ From $(2)$ we obtain $$C_2=y+\frac{(y')^2+1}{y''}$$ Then substitute in $(1)$ and $(0)$ $$(x-C_1)-y'\frac{(y')^2+1}{y''}=0\quad(3)$$ $$(x-C_1)^2+\left(\frac{(y')^2+1}{y''}\right)^2=a^2\quad(4)$$ From $(3)$ we obtain $$x-C_1=y'\frac{(y')^2+1}{y''}$$ Then substitute in $(4)$ $$\left(y'\frac{(y')^2+1}{y''}\right)^2+\left(\frac{(y')^2+1}{y''}\right)^2=a^2$$ After some simplifications we get $$((y')^2+1)^3=(ay'')^2$$

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We can express standard result/formula for curvature rectangular coordinates

$$\frac{y''}{(1+y'^2)^{3/2}} =\frac{1}{a}$$

as the required differential equation.

EDIT 1:

In polar co-ordinates the ODE for curvature is

$$\dfrac { r^2 + 2 r ^{'2 }- r r ^{"} }{ (r^2 +r'^2)^{3/2}} \tag{1}$$

The above can be derived from intrinsic/natural differential equation of a circle is

$$\frac{d \phi }{ds} = \frac{d (\theta + \psi ) }{ds} \tag{2}$$

$$=\dfrac{ \sin \psi}{r} + \frac{d}{ds} ( \tan ^{-1} \frac{r}{r'} ) \tag{3}$$

where $\phi$ is angle to x-axis, $\psi$ is between arc and radius vector, $$\tan \psi = \dfrac {r}{r^{'}} \tag{4}$$

Introducing above into (3) and differentiating,

$$\frac{1}{\sqrt{ r^2 +r'^2 }} + \frac{r'^2 - r r ^{''}}{ r^2 +r'^2 } \frac{1}{\sqrt{ r^2 +r'^2 }} \tag{5=1}$$

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