sorry question misplaced; I did not confuse f1 and f2I which must be replaced by f
the code works for   f := (x, y) -> 3*x^2 - 3*y*x + 6*y^2 - 6*x + 7*y - 9 and don't work for    
                                f := (x, y) -> 9*x^2 - 24*y*x + 16*y^2 + 10*x - 70*y + 175

Based on your advice, this is what I get :
 

intersections(X^2 + Y^2 - 1, X - Y, X);
                          R￩sultant :

                               2    
                            2 Y  - 1

X=-.7071067812,   Y=-.7071067812   --->  0.\\nX=-.7071067812,   Y=.7071067812   --->  1.414213562\\nX=.7071067812,   Y=-.7071067812   --->  1.414213562\\nX=.7071067812,   Y=.7071067812   --->  0.\\nNombre de solutions :  2\n
 {[-0.7071067812, -0.7071067812], [0.7071067812, 0.7071067812]}
This is not satisfactory.
I would like a single X= and a single Y= on a line without n; Thank you

Sorry idon't understand your answer.
 

intersections(X^2 + Y^2 - 1, X - Y, X);              

X=-.7071067812,   Y=-.7071067812   --->  0.\n X=-.7071067812,   Y=.7071067812   --->  1.414213562\n X=.7071067812,   Y=-.7071067812   --->  1.414213562\n X=.7071067812,   Y=.7071067812   --->  0.\n Nombre de solutions :  2\n
 {[-0.7071067812, -0.7071067812], [0.7071067812, 0.7071067812]}
How to arrange X and Y on a line without \n; Thank you.

s := curve([A, M, N, P, A], color = red, filled = [color = "pink", transparency = 0.7])

Sorry, I am finding the coordinates O of the circle (C,D,K,H)

How to obtain the coordinates of the O centre of the circle (A,B,C,D) from the M matrix ? Thank you.

with the numerical values x := 0.645980840726419; y := 0.207986256183860; z := 0.146032903089721;
the solution given is :
{0.146032903089721 = 0.1460329031, a = -1.354182738 c, 

  b = -1.149311492 c, c = c}, {0.146032903089721 = 0.1460329031, 

  a = -1.354182738 c, b = 1.149311492 c, c = c}, 

  {0.146032903089721 = 0.1460329031, a = 1.354182738 c, 

  b = -1.149311492 c, c = c}, {0.146032903089721 = 0.1460329031, 

  a = 1.354182738 c, b = 1.149311492 c, c = c}

What are the values of a, b, and c ?

How to avoid a poster asking me : enter name of horizoltal axis then veticial axis ? Thank you

How to find the barycentric coordinates of the orthocenter of the ABC triangle (in the ABC repere) using cartesian coordinates ? Thank you..

cf mathcurve la Hire
droite de la Hire
La Hire Line
I would like to build this figure.

restart;
Proc := proc(t) local a1, t4, l3, R, r, dr, eq, sol; _EnvHorizontalName := 'x'; _EnvVerticalName := 'y'; t4 := thickness = 4; l3 := linestyle = dot;
R := 9; r := 1/2*R;
geometry:-point(OO, 0, 0);
geometry:-circle(Cir, [OO, R]);
geometry:-point(K, R*cos(t), R*sin(t));
geometry:-point(Omega, r*cos(t), r*sin(t));
geometry:-circle(cir, [Omega, r]);
eq := geometry:-Equation(cir);
geometry:-line(XXp, y = 0);
geometry:-line(YYp, x = 0);
geometry:-line(L1, y = x);
geometry:-line(L2, y = -x);
geometry:-projection(M1, K, XXp);
geometry:-coordinates(M1);
geometry:-point(K2, geometry:-coordinates(M1)[1] - 2*R, 0);
geometry:-coordinates(K2); geometry:-segment(sT, K2, M1);
geometry:-point(N1, 0, R*sin(t)); subs(y = x, eq);
sol := solve(%, x); geometry:-point(N2, sol[2], sol[2]);
subs(y = -x, eq);
sol := solve(%, x);
geometry:-point(N3, sol[2], -sol[2]);
plots:-display([plottools[arc]([op(geometry:-coordinates(Omega))], r, t .. t + 1/2*Pi, color = red, t4),
plottools[arc]([op(geometry:-coordinates(Omega))], r, t + Pi .. t + 3/2*Pi, color = coral, t4),
plottools[arc]([op(geometry:-coordinates(Omega))], r, t - 1/2*Pi .. t, color = cyan, t4),
plottools[arc]([op(geometry:-coordinates(Omega))], r, t + 1/2*Pi .. t + Pi, color = green, t4)],
geometry:-draw([Cir(color = blue, t4), cir(color = grey, t4), sT(color = black, t4), XXp(color = black, l3), YYp(color = black, l3), L1(color = black, l3), L2(color = black, l3), N1(color = blue, symbol = solidcircle, symbolsize = 15), N2(color = blue, symbol = solidcircle, symbolsize = 15), N3(color = blue, symbol = solidcircle, symbolsize = 15), M1(color = blue, symbol = solidcircle, symbolsize = 15)]), axes = none, view = [-30 .. 10, -10 .. 10], size = [800, 800]); end proc;
plots:-animate(Proc, [t], t = 0 .. 2*Pi, frames = 30);

Despite your animations I can not answer these questions :
1.-Why 2 triangles CMP and ABH are similar ?
2.-At what curve E is the line MP tangent ?
Can you help me ?

It is a corollary of Poncelet’s theorem : The line that joins a focal spot F to a point P common in M and M' to an ellipse is bisector inside the angle MFM"

Do you want to tell me what is the interest of the calculation of alpha1 compared to alpha ? alpha can be drawn;

Problem with the 3 last lines of the code..
(sin(Setof[2]))^2
Error, invalid subscript selector
simpascn:=arcsin(sqrt(factor(  ))); # the t's eliminate
Error, Invalid label reference
 alpha1:=(subs(Setof[2]=simpascn,alpha))*180/Pi
Error, invalid subscript selector
plot(alpha1,t=-10..10,view=[-10..10,0..180])
Warning, expecting only range variable t in expression alpha1 to be plotted but found name alpha1