1
$\begingroup$

I am trying to find an equation for a curve along circle with radius $5$ and centre at $(-2,3)$ and between points $(3,3)$ and $(-2,-2)$ on circle

Now I know that equation for this circle is $$(x+2)^2 + (y-3)^2 = 25$$

enter image description here

Here it is how it looks on graph

How can I find equation for a curve between points $A$ and $B$ going counter clockwise and then going $B$ to $A$ in line

$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

I think the better Idea would be to first apply basic scaling transformation on your circle to make its origin at (0,0) so that calculations become easier and so the points that you mentioned would become points as (5,0) and (0,-5). i.e. ( +2 in x value and -3 in y value ) for shifting origin and points. Finally, there will be two curves C1 and C2. C1 for that curve that goes forward and C2 for that line that comes back. Now for curve part we might want to use poolar coordinates. as :

(x,y) = (5cos(t),5sin(t)); for center at origin.

now reversing transformation for curve part yields:

Curve C1: (x,y) = (5cos(t) -2 ,5sin(t) + 3)

where t goes from 3π/2 to 2π. This give anticlockwise curve.

And for reverse line part. is simple line equation as :

Curve C2: y-3 = x-3 ; as x,y goes from 3 to -2.

I Tried to put it in desmos. Desmos Link: https://www.desmos.com/calculator/gjrtzitvmo

Picture From Desmos

Only Curve Image for clarity

Disclaimer: This answer is not by an expert. Seniors are requested to rectify if any mistake is done here. Thanks.

$\endgroup$
5
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Sep 24 at 2:30
  • $\begingroup$ @AbdurRehman,I think What OP intends to ask is The exact equation, Which will give only the "curve" that is required and not the extra part even if you put the formula directly on desmos $\endgroup$
    – Dheeraj Gujrathi
    Sep 24 at 5:34
  • $\begingroup$ @Dheeraj I tried to think harder on a single formula that could govern whole of the curve. But, according to my limited knowledge, It has to be parametrized b/c it is not like any regular geometric shape that might have closed form solutions, but again as to my knowledge. If you have some better closed-form solution then please do enlighten us. $\endgroup$
    – AbdurRehman
    Sep 24 at 7:15
  • $\begingroup$ @AbdurRehman,Your parametric equation works well ,I was just Asking you to confirm If you had equation without parametric form as well, I have added my attempt as an answer below $\endgroup$
    – Dheeraj Gujrathi
    Sep 24 at 8:45
  • $\begingroup$ You answer also gives an exact formula for that but There is a confusion that I faced is how we define the path, direction and the limits of those directions. Your formula is just as amazing but I think it would make the calculus work very difficult. $\endgroup$
    – AbdurRehman
    Sep 25 at 10:36
1
$\begingroup$

This way Of combining functions and restricting domain and ranges might work

$$\left(x-y\right)\left(\left(x+2\right)^{2}+\left(y-3\right)^{2}-25\ \right)\left(e^{\left(\operatorname{sgn}\left(\sqrt{\left(3-x\right)}\right)+\operatorname{sgn}\left(\sqrt{\left(3-y\right)}\right)+\operatorname{sgn}\left(\sqrt{\left(x+2\right)}\right)+\operatorname{sgn}\left(\sqrt{\left(y+2\right)}\right)\right)}\right)=0$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .