Calculating coordinates for Inner / transverse common tangent points

Hello all,

I've been trying to calculate this for sometime and have not had much luck. I am trying to calculate the coordinates for the four points in which two inner common tangent lines intersect with two circles. It is assumed that the two circles at coordinates C1 = [x1 y1] and C2 = [x2 y2] circles are of equal radii.

I have found something similar

http://www.mathworks.com/matlabcentral/answers/162912-tangent-to-two-circles

However, this calculates the coordinates for the outer common tangent points. I have found plenty of resources regarding the lengths of the tangent lines but I am trying to determine the exact coordinates along the circle in which they intersect/touch.

Similar to this image:

Any assistance would be greatly appreciated,

Thank you!

Answer 1

[John D'Errico]

Sooooo many ways to do this. Here are two.

#1. Just basic algebra. You have two circles.

(x-x1)^2 + (y-y1)^2 = r^2
(x-x2)^2 + (y-y2)^2 = r^2

You are looking for the pair of lines that cross, so are tangent to opposite sides of the circle.

Where is the intersection of those lines? An easy conclusion, since the circles have the same radii, is that the intersection point MUST lie at the mid-point between the two circle centers. Thus:

x0 = (x1 + x2)/2
y0 = (y1 + y2)/2

The lines MUST EACH pass through that point, (x0,y0). Those lines can then be written as

y - y0 = s*(x-x0)
y - y0 = t*(x-x0)

where s and t are the respective slopes of the pair of lines.

Now, just find the intersection of one of those lines with the circles. Do so by substituting for y in the circle equation:

subs((x-x1)^2 + (y-y1)^2 - r^2,y,y0+s*(x-x0))
ans =
(x - x1)^2 + (y1/2 - y2/2 + s*(x1/2 - x + x2/2))^2 - r^2

Solve for x, such that we have an intersection. Of course, we get two solutions.

solve((x - x1)^2 + (y1/2 - y2/2 + s*(x1/2 - x + x2/2))^2 - r^2,x)
ans =
  (2*x1 + s*y1 - s*y2 + s^2*x1 + s^2*x2 + (4*r^2*s^2 + 4*r^2 - s^2*x1^2 + 2*s^2*x1*x2 - s^2*x2^2 + 2*s*x1*y1 - 2*s*x1*y2 - 2*s*x2*y1 + 2*s*x2*y2 - y1^2 + 2*y1*y2 - y2^2)^(½))/(2*(s^2 + 1))

    (2*x1 + s*y1 - s*y2 + s^2*x1 + s^2*x2 - (4*r^2*s^2 + 4*r^2 - s^2*x1^2 + 2*s^2*x1*x2 - s^2*x2^2 + 2*s*x1*y1 - 2*s*x1*y2 - 2*s*x2*y1 + 2*s*x2*y2 - y1^2 + 2*y1*y2 - y2^2)^(½))/(2*(s^2 + 1))

But that was a quadratic polyomial, and we know that we wish a single solution for the intersection. So solve for the slope that yields a discriminant of zero in that solution.

s = solve(4*r^2*s^2 + 4*r^2 - s^2*x1^2 + 2*s^2*x1*x2 - s^2*x2^2 + 2*s*x1*y1 - 2*s*x1*y2 - 2*s*x2*y1 + 2*s*x2*y2 - y1^2 + 2*y1*y2 - y2^2==0,s)

s =
(x1*y2 - x1*y1 + x2*y1 - x2*y2 + 2*r*(- 4*r^2 + x1^2 - 2*x1*x2 + x2^2 + y1^2 - 2*y1*y2 + y2^2)^(1/2))/(4*r^2 - x1^2 + 2*x1*x2 - x2^2)

 -(x1*y1 - x1*y2 - x2*y1 + x2*y2 + 2*r*(- 4*r^2 + x1^2 - 2*x1*x2 + x2^2 + y1^2 - 2*y1*y2 + y2^2)^(1/2))/(4*r^2 - x1^2 + 2*x1*x2 - x2^2)

So we have found the two possible slopes. We could go ahead and compute the intersection points, etc.

=======================================

#2. Basic geometry.

Given circles, centered at [x1,y1] and [x2,y2], each of radius r. Find the tangent points.

Answer: USE RIGHT TRIANGLES!

As we stated before, the intersection point of the lines happens at

[x0,y0] = [(x1 + x2)/2, (y1 + y2)/2]

The distance between the circle centers is

D = norm([x1,y1] - [x2,y2])

So the distance from the center of either circle to the mid-point is D/2.

The radius of each circle is r. And the line tangent to a circle MUST make a right triangle where it touches that circle as a tangent. The right triangle has hypotenuse of length D/2, one side of length r, and therefore the third side, by virtue of the pythagorean theorem has length...

L = sqrt(D^2/4 - r^2)

The angles of that triangle are now trivially found from a little trigonometry. How did it go? SOHCAHTOA. Easy, peasy.

Use method 2. It is a lot less messy.

[Jonathan Rojas]

John,

Thank you, I was making it more difficult than it had to be so I appreciate the clarification

- Jon

[John D'Errico]

Interestingly, I thought the analytical approach would be a lot cleaner. Then I thought of the geometrical approach, which is quite simple. I'm sure there are other solutions too though.

[Mohamad Harb]

Hello dear, could anyone help me to get a solution for a problem? Having two touching circles, find the third one so it's touching both

Answer 2

[Brian]

I have made a pragmatic solution to do what is asked for based on these sources:

https://en.wikipedia.org/wiki/Homothetic_center

https://en.wikipedia.org/wiki/Tangent_lines_to_circles

https://se.mathworks.com/matlabcentral/answers/86365-how-can-i-draw-a-tangent-line-from-a-given-point-to-a-circle-in-matlab

I tried finding a nicer solotion for the tangential points of the inner tangents, but I could not find it.

clc;
clear all;
close all;

% circle 1
x1 = 0; 
y1 = 0.4; 
r1 = 0.4;

% circle 2
x2 = 1;
y2 = 0;
r2 = 0.1;

% plotting the two circles
figure
circle_(x1,y1,r1,'k')
hold on 
circle_(x2,y2,r2,'k')
axis equal

% The outer tangent lines https://en.wikipedia.org/wiki/Tangent_lines_to_circles
Deltax = x2-x1;
Deltay = y2-y1;
Deltar = r2-r1;
d = sqrt(Deltax^2+Deltay^2);
X = Deltax/d;
Y = Deltay/d;
R = Deltar/d;
k = 1;
a = R*X-k*Y*sqrt(1-R^2);
b = R*Y+k*X*sqrt(1-R^2);
c = r1-(a*x1 + b*y1);
x = -0.2:0.01:1.2;
y = (a*x + c)/(-b);
plot(x,y)
k = -1;
a = R*X-k*Y*sqrt(1-R^2);
b = R*Y+k*X*sqrt(1-R^2);
c = r1-(a*x1 + b*y1);
x = -0.2:0.01:1.2;
y = (a*x + c)/(-b);
plot(x,y)

%internal homothetic center https://en.wikipedia.org/wiki/Homothetic_center
hcx = r2/(r1+r2)*x1 + r1/(r1+r2)*x2;
hcy = r2/(r1+r2)*y1 + r1/(r1+r2)*y2;
plot(hcx,hcy,'*')

%tangent points on each cicle from a straight line going through the
%homothetic center.
%https://se.mathworks.com/matlabcentral/answers/86365-how-can-i-draw-a-tangent-line-from-a-given-point-to-a-circle-in-matlab 
P = [hcx hcy]-[x1 y1];
d2 = dot(P,P);
Q0 = [x1 y1]+r1^2/d2*P;
T = r1/d2*sqrt(d2-r1^2)*P*[0,1;-1,0];
Q1_1 = Q0+T;
Q1_2 = Q0-T;
plot(Q1_1(1), Q1_1(2),'*')
plot(Q1_2(1), Q1_2(2),'*')

P = [hcx hcy]-[x2 y2];
d2 = dot(P,P);
Q0 = [x2 y2]+r2^2/d2*P;
T = r2/d2*sqrt(d2-r2^2)*P*[0,1;-1,0];
Q2_1 = Q0+T;
Q2_2 = Q0-T;
plot(Q2_1(1), Q2_1(2),'*')
plot(Q2_2(1), Q2_2(2),'*')

plot([Q1_1(1),Q2_1(1)], [Q1_1(2),Q2_1(2)],'-*')
plot([Q1_2(1),Q2_2(1)], [Q1_2(2),Q2_2(2)],'-*')

function circle_(x,y,r,opt)
th = 0:pi/50:2*pi;
xunit = r * cos(th) + x;
yunit = r * sin(th) + y;
plot(xunit, yunit, opt);
end