@Kitonum and MB86.

Using Kitonum's code I tried this:
 

restart;

sys_ode := diff(x(t), t) = -x(t)^2/(4*Pi*y(t)*(x(t)^2 + y(t)^2)), diff(y(t), t) = y(t)^2/(4*Pi*x(t)*(x(t)^2 + y(t)^2));
ics := x(0) = 1, y(0) = 1:

ode1,ode2:=sys_ode;
ode2/ode1;
ode:=diff(y(x),x)=-(y(x)/x)^3;
dsolve(ode);
sol:=dsolve({ode,y(1)=1});
plot(rhs(sol),x=1/sqrt(2)..5,y=0..3,color=red); p1:=%: #Saving the plot

Sol := dsolve({sys_ode, ics}, {x(t),y(t)}, numeric);

plots:-odeplot(Sol,[x(t),y(t)], t=-45..45, color=blue,view=0..3);  p2:=%: # Saving the plot
plots:-display(p1,p2); # blue
plots:-display(p2,p1); #red except at the end
plots:-display(Array([p1,p2]));

If you use this events version you make sure that x(t) stops at 5.
 

Sol := dsolve({sys_ode, ics}, {x(t),y(t)}, numeric,events=[[x(t)=5,halt]]);

@Carl Love Thanks Carl.
Two comments:
1.
I was just now looking at the help page for type/satisfies. It says:

"Predicates are expected to return normally, without raising exceptions."
My predicates don't meet that condition.

2.

The example of the use of overload(callseq_only) in the help page for overload is AppendUnique, which has nothing to do with jumping from procedure 1 to procedure 2 as the statement in help page talks about.

@Carl Love Thanks for commenting on this.

Here is the example:
overload_callseq_only_2.mw

@Ronan I have submitted an SCR about option overload(callseq_only).
I made an admittedly contrived example, where an exception (i.e. error) is raised in the parameter checking phase.
The execution didn't proceed from p1 to p2.

@Ronan I'm beginning to see that your original foo1 is behaving correctly, though not as you intended:
 

restart;
foo1:=overload([                        
                  proc(P1::list,P2::list,P3::list,$)
                     option overload;
                     print("3 lists");
                  end proc,

                  proc(P1::list,P2::list,a::algebraic:=4,$)
                     option overload;
                     print("2 lists");
                  end proc
                       ]):
foo1([1,2],[3,4]); #Output correct because P1 and P2 are lists
foo1([1,2],[3,4],4); 
foo1([1,2],[3,4],[4,7]);

Now, however, I don't see any need for using option overload(callseq_only).
I have been looking for examples in the documentation, but have not found any.

@Ronan Yes your example foo3 is puzzling.

Contrast the example with this:

p1:=proc(P1::list,P2::list,P3::list,$)
          P3
end proc;
#######     

p1([1,2],[3,4]); # Error message

But if the body of the procedure p1 doesn't make use of P3 there will be no error.

foo3 didn't produce an error message, it just jumped to the second procedure.

@Prakash J As I said above, dsolve/numeric/bvp uses a finite difference method.
So what more do you want?
To follow a little of what's happening you can try setting

infolevel[`numeric/bvp`]:=5:

just before executing dsolve/numeric/bvp

The initial mesh is by default 8. To see what happens when you set that value yourself, try
res:=dsolve(eval(sys,{Parameters1}),numeric,initmesh=16);

@dharr Maybe you are right:
Anyway I had fun making the animations over two ranges for inf.
There are some ranges where dsolve/numeric/bvp doesn't converge. That is why there are two animations.

#############

@Thomas Richard I noticed that simplify remembers:
 

restart;
f := (BesselI(0, alpha)*alpha - 2*BesselI(1, alpha))/(BesselK(0, alpha)*BesselI(1, alpha)*BesselI(0, alpha)*alpha^2 + BesselK(1, alpha)*BesselI(0, alpha)^2*alpha^2 - 2*BesselI(1, alpha));
simplify(f);
simplify(f, 'wronskian'); # Not 1
restart;
f := (BesselI(0, alpha)*alpha - 2*BesselI(1, alpha))/(BesselK(0, alpha)*BesselI(1, alpha)*BesselI(0, alpha)*alpha^2 + BesselK(1, alpha)*BesselI(0, alpha)^2*alpha^2 - 2*BesselI(1, alpha));
#simplify(f);
simplify(f, 'wronskian'); # 1

I tried forget(simplify);  just after simplify(f);  but it doesn't seem to work (at least not consistently).
Note: I was using Maple 2023.2.

@acer Thank you very much acer.
Setting _EnvUsePlotThing:=false:  produces a nice plot in Maple 2023.2.
I will put it into my maple.ini file.

Maple 2022.2 also sets _EnvUsePlotThing:=false in `plots/animate`, but it appears not necessary in this case with plot(x*exp(-x),x=0..20) with Digits set very high.
The speed is affected a lot, though, as you also point out.
I tried this in Maple 2022.2:

restart;
Digits:=100;
CodeTools:-Usage([seq(plot(x^n*exp(-x),x=0..20),n=1..20)]):
restart;
Digits:=100;
_EnvUsePlotThing:=false;
CodeTools:-Usage([seq(plot(x^n*exp(-x),x=0..20),n=1..20)]): # Roughly 10 times faster.

@fnavarro It is a really bad idea to use plot as a toy model.
As dharr is pointing out there is nothing wrong with evaluation, but plot is rather special for the reasons I mentioned.
Take a look at this code.

restart;
Digits:=40:
E:=x*exp(-x);
V:=Vector([seq](evalf(eval(E,x=i)),i=0..20));
max(V); # 0.3678794411714423215955237701614608674458
min(V); # 0.

@sand15 There is some confusion: You write to begin with:
"But it seems that : test and `test` do not mean exactly the same thing here:",

but you use unevaluation quotes in select(type, {anames()}, suffixed('test'));

When test:=0 (or assigned to anything else)  'test'; first evaluates to test, then to 0.
 

restart;
test:=47;
'test'; # test
%; # 47
## In contrast:
`test`; # 47

@Carl Love Yes, although ? keywords brings up a help page with the title
Maple Keywords (Reserved Words).

Furthermore the statement really refers to symbols:

type(sin,symbol); #true
type(Pi,symbol);  #true
evalb(`sin`=sin); #true
evalb(`a[1]`=a[1]);  #false
type(`a[1]`,symbol); #true
type(a[1],symbol);   # false
type(a[1],name);     #true

@Carl Love It could be pointed out that in these examples the same holds:

`sin`(Pi/3); # 1/2*sqrt(3)
`exp`(ln(x)); # x
sin(`Pi`/3); # 1/2*sqrt(3)
Pi:=6; # Error

The attempt to assign to Pi results in the error:
Error, attempting to assign to `Pi` which is protected.  Try declaring `local Pi`; see ?protect for details.

Similar errors follow attempts to assign to sin or exp.

In the help page ? names we find this statement:

Any valid Maple name formed without using left single quotes is precisely the same as the name formed by surrounding the name with left single quotes.  Therefore, x and `x` both refer to the same name x. However, a keyword cannot be used as a name unless it is enclosed in left single quotes.

@Axel Vogt You didn't imply that your analysis shows an error and there isn't any.
Since B > 0, A+B*s will be positive if just s is large enough. It is correct that A < 0 under the assumptions.
The requirement is s > -A/B and that is exactly the same as returned by solve in my res[6].
In detail:
 

restart;
Q:=(a - c)^2*y*(3*b*y - 1)/(2*(2*b*y - 1)^2) + 
  (-a*(a + c)/(2*b) + (a - c)^2/(8*b) + s*(a - c)^2/(8*b^2));

theAssumptions:=(2*c < a, a/c < 2*b*y, 0 < a, 0 < b, 0 < c, 0 < s, 0 < y); 

A:=coeff(Q,s,0);

B:=coeff(Q,s,1); # Obviously positive

is(A < 0) assuming theAssumptions; # true

#The requirement for s being positive is:
sol:=s > -A/B; # -A/B is positive

#The result that I found using solve with the assumptions were (in your notation):
res:=solve(Q>0,useassumptions) assuming theAssumptions;

res6:=simplify(res[6]);
simplify(lhs(sol)-lhs(res6)); # 0