Finding equations of tangent lines to a circle

Given the position vectors of a point A and the center of a circle B of radius r,

pA = [pAx; pAy];
pB = [pBx; pBy];
r  = 2;

Using this data alone, is it possible to find the position vectors of tangent points H and K drawn from point A to the circle B? I would then use this information to form equations of tangent lines AH and AK

Answer 1

[John D'Errico]

WTP? When all else fails, apply Pythagoras. I'll be lazy, and just throw it into solve.

P1 = [22; 20];
P2 = [31; 26];
r = 2.5; 

syms xh yh
Q = [xh;yh];

% Q lies on the circle.
E1 = r^2 - [1 1]*(P2 - Q).^2 == 0;

% Pythagoras applies
E2 = [1 1]*(P1 - P2).^2 == r^2 + [1 1]*(Q - P1).^2

res = solve(E1,E2);
res.xh
ans =
 (5*443^(1/2))/78 + 1587/52
 1587/52 - (5*443^(1/2))/78

vpa(res.xh)
ans =
 31.868433665374947976737252727146
 29.170027873086590484801208811316

res.yh
ans =
 2003/78 - (5*443^(1/2))/52
 (5*443^(1/2))/52 + 2003/78

vpa(res.yh)
ans =
 23.655682835270911368227454242615
  27.70329152370344760613152011636

Two solutions, because there are two tangents. Surely you can do the rest.

Be careful when you plot things, because if the x and y axes do not use the same scale, the lines won't appear to be tangent. Of course then the circle won't appear to be circular either.

t = linspace(0,2*pi,100);
plot([P1(1),P2(1)],[P1(2),P2(2)],'-ro')
hold on
axis equal
grid on
plot(P2(1) + r*cos(t),P2(2) + r*sin(t),'b-')
plot(double(res.xh(1)),double(res.yh(1)),'go')
plot(double(res.xh(2)),double(res.yh(2)),'go')

Looks right to me. So WTP?

Just for kicks, there is another approach we can try.

Given the two points P1 and P2, and a radius r, solve for the point Q=(x,y), such that the vectors formed from (Q-P1) and (Q-P2) are orthogonal, as well, we need the distance, norm(Q-P2) to be r.

That is, we have one equation

(Q-P1).'*(Q-P2) == 0

as well as

norm(Q-P2) == r

The latter can be written as

(Q-P2).'*(Q-P2) == r^2

Again, solve could be used here to solve the problem, and we should recover the same two solutions.

Hmm. How about a simple alternative solution that has no need for the symbolic toolbox? Again, we rely on the presence of a right triangle. But this time, we will use trig.

P1 = [22; 20];
P2 = [31; 26];
r = 2.5;

Assume the angle between the line connecting P1 and P2 and the tangent line is theta. Assume the tangent line touches the circle at point Q. Then {P1,Q,P2} forms a right triangle, with hypotenuse as the edge {P1,P2}. We get theta from the sine relation:

sin(theta) = r/norm(P2-P1)

Therefore

theta = asin(r/norm(P2-P1))

By default here, theta will be positive, but -theta will also suffice to get the other tangent line. So what is the angle of the tangent line? The two tangent lines fall at angles:

psi = atan2(P2(2)-P1(2),P2(1)-P1(1)) + [-1 1]*theta
psi =
      0.35477      0.82124

How far along those tangent lines do they touch the circle? So what is the length of that segment?

L = cos(theta)*norm(P2-P1)
L =
       10.524

This makes perfect sense, since remember the distance between points P1 and P2 was

norm(P1-P2)
ans =
       10.817

Q = [P1,P1] + L*[cos(psi);sin(psi)]
Q =
       31.868        29.17
       23.656       27.703

Q has two columns, corresponding to the two tangent points. Happily, they are the same that we found using the symbolic toolbox. Note that this last expression can be made slightly simpler if we are using the latest (R2017) MATLAB release. I could then have written it as

Q = P1 + L*[cos(psi);sin(psi)]
Q =
       31.868        29.17
       23.656       27.703

[Lee Quan]

If I'm not mistaken, vpa(res.xh) gives the x-coordinates of the tangent points and vpa(res.yh) gives the y-coordinates of the tangent points, yes? Is this how I should get the points Q1=[Q1x; Q1y] and Q2=[Q2x; Q2y]? Also, is there a way to round the numbers upto like 5 places?

x_co = vpa(res.xh);
y_co = vpa(res.yh);
Q1 = [x_co(1); y_co(1)]
Q2 = [x_co(2); y_co(2)]

[John D'Errico]

Yes. That gives you the two tangent points.

Of course, it is also possible to do the computations without using solve. Lots of ways that will work here.

[Lee Quan]

 x_co = vpa(res.xh);
 y_co = vpa(res.yh);
 roundn = @(x,n) round(x*10^n)./10^n;
 Q1 = [x_co(1); y_co(1)];
 Q1 = roundn(Q1,5)
 Q2 = [x_co(2); y_co(2)];
 Q2 = roundn(Q2,5)

I tried this and now it shows in fractions. Ugh!

[Image Analyst]

This is old, and I guess you figured it out since you accepted the answer. If it's not solved, then unaccept it and let us know, otherwise I'm assuming it's solved now and am not going to answer any questions anymore (because it's solved, so there's no need anymore).

[John D'Errico]

Fractions are there because you have somehow gotten them into symbolic form. Convert to doubles, using double. Or convert to floating point symbolic, using vpa.

If you actually read my answer, then you should have seen that I used either vpa or double, as was appropriate at the time.

[John D'Errico]

Image Analyst: I'm confused. Why are you suggesting it be unaccepted? This response solves the problem. In fact, I've now solved the basic problem using three distinct methods, all equally valid.

[Image Analyst]

Well I saw he asked a question of me regarding some error, then three more questions of you, so it seemed like it wasn't working yet.

[John D'Errico]

He had gotten a code from Roger. I assume Stafford. But finding a problem in uncommented code that solves a problem is often way harder than solving the problem 3 different ways. So I just took the approach of solving the underlying problem, rather than fixing the code error.

[Image Analyst]

That's fine. No worries. You provided an excellent answer (that I voted for).

[Lee Quan]

Wonderfully explained! Thanks again, John!

Sorry for the trouble, Image Analyst.

Answer 2

[Image Analyst]

Yes. Find point H, knowing that angle AHB is a right angle. Then knowing points H and A (or K and A) you can use the regular point-slope formula for the equation of a line.

[John D'Errico]

Basic geometry. The trick is in recognizing where you can draw a right triangle. Once you have that, the problem becomes trivial. Pythagoras saves the day again.

[Lee Quan]

Based on Roger's code,

 P1 = [22; 20];
 P2 = [31; 26];
 r = 2.5; 

 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];     % points from Q0 to Q1
 Q1 = Q0+T
 Q2 = Q0-T

Could you please tell me how to correct the error in the expression for T?