How can I draw a tangent line from a given point to a circle in Matlab

Hi I am trying to draw a tangent line from a given external point to a circle of known radius and origin as the center

Please help

[Shashank Prasanna]

If you solve for the point on that circle using standard methods, you can then use line or plot to draw the line.

If you are struggling with your MATLAB code, please provide what you have tried already and we can help you out.

[dpb]

The obvious is that the radius and the tangent line must be normal to each other so the three points have to form a right triangle...

[Suhas]

I am facing problem with identifying the point of tangency actually. This is what i did so far. Consider a point(p,q)external to circle. eqn: y-q=m(x-p)

This line is normal to line connecting center and tangent point

consider point(a,b) is the tangent point.slope of line connecting center(0,0) and tangent point is (b/a)

Hence slope of line connecting the external point and tangent point is (-a/b) -----(1)

Also slope of line connecting external point(p,q) and tangency point (a,b) can be written as (b-q)/(a-p) ----(2)

solving (1) and (2) we get ap+bq=r^2 , where r is the radius of circle

Now I need to find values for (a,b)

Answer 1

[Roger Stafford]

Let P1 be a 1 x 2 row vector of the external point's cartesian coordinates and P2 a similar vector of the circle's center coordinates. Let r be the circle's radius. Then Q1 and Q2 below will be the two possible points of tangency of a line from P1. Q1 is to the right and Q2 to the left as viewed from P1 looking toward P2.

 P = P1-P2;
 d2 = dot(P,P);
 Q0 = P2+r^2/d2*P;
 T = r/d2*sqrt(d2-r^2)*P*[0,1;-1,0];
 Q1 = Q0+T;
 Q2 = Q0-T;

[dpb]

Was hoping the OP could be goaded to that point on own, perhaps...

[Suhas]

Thanks a lot Roger

[Antonio Cantiani]

I would like to understand the mathematics behind this. Can someone explain it, please?

[Roger Stafford]

@Antonio: This is a brief outline of the derivation of the above expressions. If the points Q1 and Q2 are connected with a straight line, this line will cross the line between P1 and P2 at a right angle by symmetry, and the point Q0 is defined as the point where these lines cross. The four points P1, P2, Q1,and Q0 when connected form three similar right triangles and the expression for Q0 as P2+r^2/d2*P can be deduced from some of the like ratios involved in these triangles. The vector P*[0,1;-1,0] points at right angles to the vector P, and the expression for T which points from Q0 to Q1 can also be derived from these similar triangle ratios, along with the Pythagorean theorem. By symmetry -T points in the opposite direction from Q0 to Q2.