The circle is $x^2+y^2=1$. Euler showed it is also $e^{i\theta}$.
Euler Line: $O,G,H$ collinear, $OG:GH=1:2$. Nine-Point: 3 midpoints + 3 feet + 3 Euler points = one circle, center $N$ midpoint of $OH$, radius $R/2$. Basel: $\sin x/x=\prod(1-x^2/n^2\pi^2)$ gives $\pi$ from integers.
theta=linspace(0,2*pi,1000); plot(cos(theta),sin(theta),'LineWidth',2); hold on; plot(real(exp(1i*theta)), imag(exp(1i*theta)), 'r--'); axis equal; grid on;
t=linspace(0,2*pi,300); for k=1:length(t) zk=exp(1i*t(k)); clf; th=linspace(0,2*pi,200); plot(cos(th), sin(th),'k-'); hold on; axis equal; axis([-1.3 1.3 -1.3 1.3]); grid on; plot([0 real(zk)],[0 imag(zk)],'b-o','LineWidth',2); drawnow; pause(0.02); end
res=exp(1i*pi)+1; abs(res)
fprintf('e^{i pi}=%.15f\n', real(exp(1i*pi)));N=10000; n=1:N; s=sum(1./n.^2); sqrt(6*s) semilogx(n, sqrt(6*cumsum(1./n.^2))); yline(pi,'r--');
function euler_nine_point(A,B,C)
D=2*(A(1)*(B(2)-C(2))+B(1)*(C(2)-A(2))+C(1)*(A(2)-B(2)));
O=[(A(1)^2+A(2)^2)*(B(2)-C(2))+(B(1)^2+B(2)^2)*(C(2)-A(2))+(C(1)^2+C(2)^2)*(A(2)-B(2));
(A(1)^2+A(2)^2)*(C(1)-B(1))+(B(1)^2+B(2)^2)*(A(1)-C(1))+(C(1)^2+C(2)^2)*(B(1)-A(1))]'/D;
H=A+B+C-2*O; Nc=(O+H)/2; R=norm(A-O); th=linspace(0,2*pi,300);
plot([A(1) B(1) C(1) A(1)],[A(2) B(2) C(2) A(2)],'k-','LineWidth',2); hold on;
plot(O(1)+R*cos(th), O(2)+R*sin(th),'b--'); plot(Nc(1)+R/2*cos(th), Nc(2)+R/2*sin(th),'r-','LineWidth',2);
axis equal; grid on;
enda=norm(B-C); b=norm(A-C); c=norm(A-B); s=(a+b+c)/2; area=sqrt(s*(s-a)*(s-b)*(s-c)); R=a*b*c/(4*area); r=area/s; % O,I as above, then norm(O-I)^2 vs R*(R-2*r)
N=7; t=linspace(0,2*pi,1000); x=zeros(size(t)); y=x; for k=1:2:2*N-1 coeff=4/(pi*k); x+=coeff*cos(k*t); y+=coeff*sin(k*t); end plot(x,y); axis equal;