# Draw a circle with radius of x using pi = 22/7 [closed]

I'm using this free word command in Mathematica 8:

"draw a circle with radius of x using pi = 22/7"

but understandable, it doesn't use my version of pi for calculating perimeter and area. Is it somehow possible in Mathematica? Or does this involve some serious alteration of the drawing routines of the program?

I'm using some of the "primitive" approximations of the pi presented on: Wikipedia: Approximations_of_π

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Why would you insist on $\pi$ being equal to $22/7$ anyway? – J. M. Nov 12 '12 at 13:10
When I draw a circle, I take a compass, use a specific radius and draw. Where do you need pi in that?? – halirutan Nov 12 '12 at 13:26
As @halirutan notes, one does not at all need an approximation of $\pi$ to draw a circle, whether on paper or on the screen. – J. M. Nov 12 '12 at 14:00
The only way I can make sense of the question is if the circle was drawn in a non-euclidean space, e.g. on a sphere. The set of points on a sphere with a certain distance from a certain point is a circle, and it's length need not be x*Pi*2. Is that what you mean? Otherwise, how should the shape of the circle change as you change Pi? Should it grow/shrink? Would it stay closed or become a segment? Would it become an ellipse or some other shape? – nikie Nov 12 '12 at 14:07
Since $\pi\neq 22/7$, the question is non-sensical. The closest we can come to making sense of the question is to ask: "What does a circle look like if it's circumference $C$ and radius $r$ satisfy $C=44\pi/7$. The answer is that it must live in hyperbolic space. We could generate such a thing in the Poincare disk, or even on a surface of constant negative curvature. – Mark McClure Nov 12 '12 at 14:30

## closed as not a real question by Artes, F'x, Sjoerd C. de Vries, Mr.Wizard♦Nov 12 '12 at 17:42

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

If for a given norm, we define a circle of radius $r$ centered around $q$ as all points at a distance $r$ from $q$: $\{x \in R^2 ; \|x-q\|=r \}$

Then we can define $\pi$ as the circumference of this curve over $2r$.

Let's use the p-norms: $\|x\|_p = (|x_1|^p + |x_2|^p)^{\frac{1}{p}}$

(* Unit circle parametrization in p-norm *)
unitCircle[t_, p_] := {Cos[t], Sin[t]}/Norm[{Cos[t], Sin[t]}, p];

(* Derivative of parametrization *)
tangent[t_, p_] :=
Evaluate[FullSimplify[D[unitCircle[t, p], t], p > 1 && t > 0]];

(* Calculate circumference *)
circumf[p_] := NIntegrate[Norm[tangent[t, p], p], {t, 0, 2 Pi}];

pi[p_] := circumf[p]/2;


Using FindRoot we can see which norm would give $\pi=22/7$

p = p /. FindRoot[Re[pi[p]] == 22/7, {p, 2., 1.5, 2.5}]
Abs[pi[p]-22/7]


2.07016

4.44089*10^-16

And finally we can see how different this circle is from our normal circle:

ParametricPlot[
{unitCircle[t, p], unitCircle[t, 2]},
{t, 0, 2 Pi},
PlotStyle -> {Automatic, Directive[Dashed, Thin]}]


Note that this is entirely dependent on the types of norms I decided to use, you can get pretty much any shape you please

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 Good answer indeed. – PlatoManiac Nov 12 '12 at 16:02 Fantastic! Glad to see, someone didn't have any difficulty to reply with a detailed answer to my question. – PHPGAE Nov 12 '12 at 21:36

I'd leave Pi protected personally and assign a different variable to be used in its place, so rather than

c=2πr


c would be equal to x2r, with x as another variable, so you can then make x equal to Pi, 22/7, 4, 3.14, 3.2, 3.1, my shoe size... whatever.

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Welcome David on Mathematica.SE. I edited your answer a bit. I suggest the following: 1) Read the FAQs! 2) When you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign – halirutan Nov 12 '12 at 13:59
"...a different variable to be used in its place..." - \[DoubledPi]` would be useful... – J. M. Nov 12 '12 at 14:24