How do I receive the touching points of a circle with given radius inside a curve?

Hello,

I am trying to receive the two touching points of a circle inside a body that is not symmetrically shaped, to then receive the center point of this circle. For example an object with four corners, in which there is a circle of a smaller radius. The object is defined through a lot of x- and y coordinates that are loaded into matlab. I know the radius of that circle.

I know the circle equations and have tried to equal these and then solve the quadratic equation that gives me the following:

for i=1:1:length(x1)
    for j=1:1:length(x2)
        if x1(i)~=x2(j) && y1(i)~=y2(j) 
            a=1+((2*y2(j)-2*y1(i))^2)/(2*x2(j)-2*x1(i))^2;
            b=(x1(i)*2*(2*y2(j)-2*y1(i))/(2*x2(j)-2*x1(i))+((-2*x2(j)^2*(2*y2(j)-y1(i))-2*y2(j)^2*(2*y2(j)-2*y1(i))+2*x1(i)^2*(2*y2(j)-2*y1(i))+2*y1(i)^2*(2*y2(j)-2*y1(i)))/(2*x2(j)-           2*x1(i))^2)-2*y1(i));
            c=x1(i)^2-((2*x1(i)*(x2(j)^2+y2(j)^2-x1(i)^2-y1(i)^2))/(2*x2(j)-2*x1(i)))+((x2(j)^4+2*x2(j)^2*y2(j)^2-2*x2(j)^2*x1(i)^2-2*x2(j)^2*y1(i)^2+y2(j)^4-2*y2(j)^2*x1(i)^2-2*y2(j)^2*y1(i)^2+x1(i)^4+2*x1(i)^2*y1(i)^2+y1(i)^4)/(2*x2(j)-x1(i))^2)+y1(i)^2-rsoll^2;
            My1=(-b+sqrt(b^2-4*a*c))/2*a;
            My2=(-b-sqrt(b^2-4*a*c))/2*a;
                if abs(My1)<abs(My2)
                    My=My1;
                else My=My2;
                end
            Mx=(x2(j)^2+y2(j)^2-x1(i)^2-y1(i)^2-My*(2*y2(j)-2*y1(j)))/(2*x2(j)-2*x1(i));

But I don't know how to continue. If I try to run the function with all data it never stops running! Furthermore I defined this

for o=1:1:length(x1)
                tt=sqrt((x1(o)-Mx).^2+(y1(o)-My).^2);
                disp('tt:')
                disp(tt)
                m=sqrt(Mx^2+My^2);
                disp('m:')
                disp(m)
                if min(tt)==rsoll
                    cx=Mx;
                    cy=My;
                    break
                   
                    
                else continue
                    m00=m0;
                    m0=m;
                    
                end
            end

But then the programm never stops running.

I also thought of using fminsearch but with my function:

z0=[0,0];
fun=@(z)(abs(((xx-z(1)).^2+(yy-z(2)).^2).^0.5-r^2));
z=fminsearch(fun,z0,optimset('TolX',1e-10,'TolFun',1e-10));
cx=z(1);
cy=z(2);

I only receive an error

Subscripted assignment dimension mismatch.

Error in fminsearch (line 191)
fv(:,1) = funfcn(x,varargin{:});

Does anybody have any idea on how I could solve this problem?

[Athul Prakash]

Hey Nina,

I see you're looping through every possible pair of (x,y) points and solving the Circle Equation:

However, in your input 'fun' to fminsearch, you seem to be using an extra square root operation (.^0.5) which doesn't produce the correct minimizing equation.

Beyond that, solving this equation with 2 (x,y) points would give you a circle which passes through both the points, not necessarily touching (tangent to) the body at those points.

I also understand that your 2-D body is described using a set of points. Are you certain that the required circle would touch the body at these exact points? For a given shape, it could be possible that the 2 tangent points you are looking for are not a part of the sample of points that you have.

[Image Analyst]

Please include a screenshot of your data. I want to know what it looks like.

Answer 1

[KSSV]

Use this:

https://in.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections?focused=5165138&tab=function

Also you need not to use loops as Athul Prakash commented.

[Athul Prakash]

Since the OP has only the radius of the circle, not points on the circle, how would you apply InterX to solve his problem?

Also, we still need to eliminate the intersection points which are not tangential since the circle should be touching the given shape and InterX naturally finds points where they pass through as well.

[KSSV]

I feel OP shold have center of the circle too.