@masoud moeini 

restart;

alias(z(t) = exp(t)+a+sqrt(exp(t)-a+b)):

A(t):=ln(z(t)):

diff(A(t), t);

@rlopez  Thank you for an elegant solution to the problem! I am amazed at your deep knowledge of the package.

@c4meleon 

Your loop does not make a single step, because  xnew-xold<0 . In addition, at each step, do nothing.

Corrected code:

restart:

 xold:=1.5:

 f:=x->tan(x)-x-1:

 divis:= evalf(subs(x=xold,f(x))/subs(x=xold,diff(f(x),x))):

 xnew:=xold-divis:

 dp:=7:

 while abs(xnew-xold)>5.0*10.0^(-(dp+1)) do

  divis:= evalf(subs(x=xnew,f(x))/subs(x=xnew,diff(f(x),x))):

  xold:=xnew: xnew:=xnew-divis:

od:

 xnew;

Your task is similar to the nonlinear programming problem  http://en.wikipedia.org/wiki/Nonlinear_programming . In Maple  these problems solves  Optimization  package. Give specific wording of your problem. 

@Markiyan Hirnyk  Read carefully my original question. Error does not appear immediately, and if second time to run the code with changing the parameters.

@Axel Vogt  I wrote that the error only occurs with repeated use of the procedure with modified  arguments. But if you run the procedure again with the same arguments (after error), the error disappears. Errors are the same in Maple 12 (classic) and M16 (standard).

@Markiyan Hirnyk  Your code does not work for an equilateral triangle, because in this case, O2 = O3. In general, the code is very cumbersome , because to find the point   O1, O2, O3  and  etc. you actually use the same calculations that can be replaced by a single procedure.

@Markiyan Hirnyk  In your code you have problems with  xC := 1/2:  yC := (1/2)*sqrt(3):

@Markiyan Hirnyk   It is not interesting. To prove just replace  a= ln(4)+sin(1)

@Markiyan Hirnyk  We must use the exact math! Replace  1.1  with  11/10 .

There is a more interesting case, when the code is not working. Try to find it!

@Markiyan Hirnyk  Thank you for the detailed checking my code! Of course, if we consider only the proof of the original assertion, the proof itself conducted to the general case (for 6 points). If in the particular case  a few points are merged into one, then by continuity it follows that these points also lie on a circle.

Here is the code that works for all cases and even for degenerate triangle and  returns also the center of the circle and its radius.

restart;

Cc:=proc(A,B,C)

local x1, y1, x2, y2, x3, y3, x, y;

x1,y1:=op(A);  x2,y2:=op(B);  x3,y3:=op(C);

if x2*y3-x2*y1-x1*y3-x3*y2+x3*y1+x1*y2=0 then error "No triangle. Three points are collinear" fi;

solve({(x2-x1)*(x-(x1+x2)/2)+(y2-y1)*(y-(y1+y2)/2)=0, (x2-x3)*(x-(x2+x3)/2)+(y2-y3)*(y-(y2+y3)/2)=0},{x,y});

assign(%);

[simplify([x,y]), simplify(sqrt((x-x1)^2+(y-y1)^2))];

end proc:

The proof:

A, B, C:=[x1,y1], [x2,y2], [x3,y3]:

A1, B1, C1, M:=(B+C)/2, (A+C)/2, (A+B)/2, (A+B+C)/3:

P1:=Cc(A,M,B1)[1]: P2:=Cc(B1,M,C)[1]: P3:=Cc(C,M,A1)[1]:

P4:=Cc(A1,M,B)[1]: P5:=Cc(B,M,C1)[1]: P6:=Cc(C1,M,A)[1]:

P:={seq(P||i, i=1..6)}:

Cc1:=Cc(P[1],P[2],P[3]);

if is(eval((x-Cc1[1,1])^2+(y-Cc1[1,2])^2, {x=P4[1],y=P4[2]})=Cc1[2]^2) and

is(eval((x-Cc1[1,1])^2+(y-Cc1[1,2])^2, {x=P5[1],y=P5[2]})=Cc1[2]^2) and

is(eval((x-Cc1[1,1])^2+(y-Cc1[1,2])^2, {x=P6[1],y=P6[2]})=Cc1[2]^2) then true else false fi;

Why do you ask the same question?

http://www.mapleprimes.com/questions/201010-How-To-Computer-GSO-In-Maple#answer203907

@Carl Love   Of course, you're right! It is my carelessness.

@toandhsp   

eq1:=(x+y)*(x^2+y^2) = 5500;

eq2:=(x-y)*(x^2-y^2) = 352;

solve[RealDomain]({eq1,eq2});

@Markiyan Hirnyk  Maple does not compute this integral, since the exact form of the term  O(x - Pi)  as a function of  x  is unknown. But it is clear that this function is limited, and integral of limited function over a finite interval is limited. Therefore, it is obvious even without Maple that the sum of infinitely large and  limited functions is infinitely large function.