Matlab | visualization of mathematical properties related to three interesting circles

Quasi circle of ellipse

proposition 1: A point outside the ellipse makes two tangents to the ellipse , If two tangents are perpendicular , Then the locus of the point is a circle .

Simply explain why ：
For ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, Assume its fault $(m,n)$ The tangent equation of the point is $y-n=k(x-m)$, Bring in the elliptic equation ：

$$\begin{array}{rl}& b^2{x}^2+a^2(k^2(x-m{)}^2+2nk(x-m)+n^2)-x^2{b}^2=0\\ & (a^2{k}^2+b^2)x^2+2k{a}^2(n-mk)x+a^2(km-n{)}^2-a^2{b}^2=0\end{array}$$

Because it is a tangent, so $\Delta =0$, namely $(m^2-a^2)k^2-2mnk+n^2-b^2$, This is one about $k$ Constraints of , The slopes of the two tangents are obtained from the Weida theorem $k_1,k_2$ Meet the conditions ：
$$k_1{k}_2=\frac{n^2-b^2}{m^2-a^2}$$
Because the two tangents are perpendicular , therefore $k_1{k}_2=-1$, So there is :
$$m^2+n^2=a^2+b^2$$
It's a radius of $\sqrt{a^2+b^2}$ The circle of .

Visualization ：

MATLAB Code :

function directorC
% @author : slandarer
%  official account   : slandarer essays 
%  You know     : hikari
a=3;b=2;syms k
% axes Basic property settings 
ax=gca;
hold on;grid on
ax.XLim=[-6,6];
ax.YLim=[-4,4];
ax.DataAspectRatio=[1,1,1];
ax.XAxisLocation='origin';
ax.YAxisLocation='origin';
ax.LineWidth=1.5;
ax.XMinorTick='on';
ax.YMinorTick='on';
ax.GridLineStyle='-.';
ax.GridAlpha=.09;
%  Draw circles and ellipses 
t=linspace(0,2*pi,200);
plot(a.*cos(t),b.*sin(t),'LineWidth',2.5,'Color',[0,118,168]./255)
plot(sqrt(a^2+b^2).*cos(t),sqrt(a^2+b^2).*sin(t),'LineWidth',2.5,'Color',[169,64,71]./255)
%  Draw tangents and intersections 
thetaList=linspace(0,2*pi,200);
kline1=plot(0,0,'-.','LineWidth',1.5,'Color',[1,1,1].*.5);
kline2=plot(0,0,'-.','LineWidth',1.5,'Color',[1,1,1].*.5);
kpnt=scatter(a,0,'filled','CData',[1,1,1].*.3);
X=-6:.1:6;
%  Circular drawing 
for i=1:length(thetaList)
    theta=thetaList(i);
    m=sqrt(a^2+b^2).*cos(theta);
    n=sqrt(a^2+b^2).*sin(theta);
    kk=double(solve((m^2-a^2)*k^2-2*m*n*k+n^2-b^2==0)); 
    Y1=kk(1).*(X-m)+n;Y2=kk(2).*(X-m)+n;
    kline1.XData=X;kline1.YData=Y1;
    kline2.XData=X;kline2.YData=Y2;
    kpnt.XData=m;
    kpnt.YData=n;
    drawnow;
    pause(.01)
end
end

The illusion of rotation

proposition 2: Suppose there is a radius of $\frac{r}{2}$ The small circle of has a radius of $r$ Roll in the big circle , The locus of points on a small circle is a straight line .

The proof is simple , Suppose there is a little contact between the small circle and the big circle , At this time, the line between the center of the small circle and the origin is x The included angle in the positive direction of the axis is ${\theta }_0$, Now suppose that the rolling of the small circle causes the change of the included angle of the center of the small circle $\theta$, Because the radius of the small circle is half of that of the big circle , So when you roll the same way , The included angle of the points on the small circle changes in the opposite direction $2\theta$ Therefore, the parameter equation of the position change of the point on the small circle can be obtained ：

$$\begin{array}{rl}& \{\begin{array}{c}x=\frac{r}{2}cos({\theta }_0+\theta )+\frac{r}{2}cos({\theta }_0+\theta -2\theta )\\ y=\frac{r}{2}sin({\theta }_0+\theta )+\frac{r}{2}sin({\theta }_0+\theta -2\theta )\end{array}\\ \Rightarrow & \{\begin{array}{c}x=(\frac{r}{2}cos(\theta ))cos({\theta }_0)\\ y=(\frac{r}{2}cos(\theta ))sin({\theta }_0)\end{array}\end{array}$$

This is a cross origin , The angle is ${\theta }_0$ Straight line , The proof is complete , All points on the small circle go straight , For visualization to look good , Here, draw several equal points to make some visualization ：

Visualization ：

MATLAB Code :

function PointOnLineInCircle(N)
% @author : slandarer
%  official account   : slandarer essays 
%  You know     : hikari
if nargin<1||N<2
    N=4;
end
% 0.2157    0.3765    0.5725
% 0.3098    0.5059    0.7412

% figure And axes Basic property settings 
fig=gcf;
fig.Name='PoLiC';
fig.Position=[50 100 600 600];
fig.NumberTitle='off';
fig.MenuBar='none';
%
ax=axes(fig);
ax.Position=[0 0 1 1];
ax.XLim=[-2.05 2.05];
ax.YLim=[-2.05 2.05];
ax.XColor='none';
ax.YColor='none';
hold(ax,'on');
% grid on

thetaB=0;
thetaS=0;


%  Draw a straight line 
pntsTheta=0:2*pi/N:2*pi;
for i=1:(length(pntsTheta)-1)
    plot([cos(pntsTheta(i)),cos(pntsTheta(i)+pi)].*2,...
         [sin(pntsTheta(i)),sin(pntsTheta(i)+pi)].*2,...
         'Color',[0.8 0.8 0.8],'LineWidth',1.5)
end
if mod(N,2)==0
    for i=1:(length(pntsTheta)-1)
        plot([cos(pntsTheta(i)+2*pi/N/2),cos(pntsTheta(i)+2*pi/N/2+pi)].*2,...
             [sin(pntsTheta(i)+2*pi/N/2),sin(pntsTheta(i)+2*pi/N/2+pi)].*2,...
             'Color',[0.8 0.8 0.8],'LineWidth',1.5)
    end
end
%  Draw a big circle 
tempT=0:0.1:2*pi+0.1;
CPnts=[cos(tempT);sin(tempT)];
plot(CPnts(1,:).*2,CPnts(2,:).*2,'Color',[0.2098 0.4059 0.6412],'LineWidth',2);
%  Draw a small circle 
sCircleHdl=plot(cos(thetaB)+CPnts(1,:),sin(thetaB)+CPnts(2,:),...
    'Color',[0.3098 0.5059 0.7412],'LineWidth',2);
%  Draw a positive inscribed polygon 
pntsHdl=plot(cos(pntsTheta+thetaS)+cos(thetaB),...
    sin(pntsTheta+thetaS)+sin(thetaB),...
    'Color',[0.2157 0.3765 0.5725],'LineWidth',2,'Marker','o',...
    'MarkerFaceColor',[0 0 0],'MarkerEdgeColor',[0 0 0],'MarkerSize',6);

patchHdl=fill(cos(pntsTheta(1:end-1)+thetaS)+cos(thetaB),...
              sin(pntsTheta(1:end-1)+thetaS)+sin(thetaB),...
              [0.8500 0.3250 0.0980],'EdgeColor','none','FaceAlpha',0.1);
%  Circular drawing 
while 1
    thetaB=thetaB-2*pi/200;
    thetaS=thetaS+2*pi/200;
    sCircleHdl.XData=cos(thetaB)+CPnts(1,:);
    sCircleHdl.YData=sin(thetaB)+CPnts(2,:);
    pntsHdl.XData=cos(pntsTheta+thetaS)+cos(thetaB);
    pntsHdl.YData=sin(pntsTheta+thetaS)+sin(thetaB);
    patchHdl.XData=cos(pntsTheta(1:end-1)+thetaS)+cos(thetaB);
    patchHdl.YData=sin(pntsTheta(1:end-1)+thetaS)+sin(thetaB);
    pause(0.05);
end
end 

Lissajous figure

There's nothing to say about this , It's good to draw a good picture hiahiahia:

Visualization ：

MATLAB Code :

function lissajous
% @author : slandarer
%  official account   : slandarer essays 
%  You know     : hikari

% figure And axes Basic property settings 
fig=gcf;
fig.Position([2,4])=...
[fig.Position(2)-fig.Position(3)+fig.Position(4),fig.Position(3)];
%
ax=gca;hold on
ax.Color=[102,117,149]./255;
ax.XColor='none';ax.XLim=[.4,8.6];ax.YTick=[];
ax.YColor='none';ax.YLim=[.4,8.6];ax.XTick=[];
ax.DataAspectRatio=[1,1,1];
ax.InnerPosition=[0,0,1,1];

%  Draw circles 
t=linspace(0,2*pi,200)';
X=cos(t).*0.42;Y=sin(t).*0.42;
plot(X+ones(1,7),Y+(1:7),'Color',[1,1,1,.7],'LineWidth',2);
plot(X+(2:8),Y+ones(1,7).*8,'Color',[1,1,1,.7],'LineWidth',2);
%  Draw points and base lines 
[Xpnt,Ypnt]=meshgrid(2:8,1:7);
pntHdl=plot(Xpnt+0.42,Ypnt,'Color',[1,1,1],'LineStyle','none',...
    'Marker','o','MarkerFaceColor',[1,1,1]);
XlineCol=2:8;YlineCol=8.*ones(1,7);
XlineRow=ones(1,7);YlineRow=1:7;
lineColHdl=plot([XlineCol;Xpnt(1,:)]+0.42,[YlineCol;Ypnt(1,:)],...
    'Color',[1,1,1,.5],'LineStyle','-.','Marker','o','MarkerFaceColor',[1,1,1]);
lineRowHdl=plot([XlineRow;Xpnt(:,end)']+0.42,[YlineRow;Ypnt(:,end)'],...
    'Color',[1,1,1,.5],'LineStyle','-.','Marker','o','MarkerFaceColor',[1,1,1]);
%  Draw polyline handle 
for m=1:7
    for n=1:7
        lineMatHdl(m,n)=plot(Xpnt(n,m),Ypnt(n,m),...
            'Color',[1,1,1,.6],'LineWidth',1.2);
    end
end
ratio=1:7;
thetaList=linspace(0,2*pi,300);
%  Circular drawing 
for i=1:length(thetaList)
    theta=thetaList(i).*ratio;
    tX=cos(theta).*0.42;
    tY=sin(theta).*0.42;
    for j=1:7
        pntHdl(j).XData=Xpnt(:,j)+tX(j);
        pntHdl(j).YData=Ypnt(:,j)+tY(end:-1:1)';
        lineColHdl(j).XData=[XlineCol(j);Xpnt(1,j)]+tX(j);
        lineColHdl(j).YData=[YlineCol(j);Ypnt(1,j)]+[tY(j);tY(7)];
        lineRowHdl(j).XData=[XlineRow(j);Xpnt(j,end)']+[tX(8-j);tX(7)];
        lineRowHdl(j).YData=[YlineRow(j);Ypnt(j,end)']+tY(8-j);
    end
    %
    ttList=thetaList(1:i);
    for m=1:7
        for n=1:7
            lineMatHdl(m,n).XData=cos(ttList.*m).*0.42+Xpnt(n,m);
            lineMatHdl(m,n).YData=sin(ttList.*(8-n)).*0.42+Ypnt(n,m);
        end
    end
    drawnow
    pause(.01)
end
end