Finding an equation for a curve along circle between two points

Question

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 $$(x+2)^2 + (y-3)^2 = 25$$

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

Answer 1

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.

Answer 2

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

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