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I have two circles with the same radius and I want to calculate the points of tangency.

For example, in the picture below, I want to calculate $(x_3, y_3)$ and $(x_4,y_4)$. I have the radius and the distance between the two circles as shown below:

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A tangent and a radius to a circle at any point are perpendicular. You might find that fact helpful. – ncmathsadist Jul 14 '12 at 18:41
up vote 1 down vote accepted

The gradient of the tangent to any circle is given by $$-\frac{(x'-a)}{(y'-b)}$$ Where (x',y') is the point of tangency and (a,b) is the center of the circle. Now the gradient of the line joining the centers of the two circles is same as the gradient of the tangent. Hence in this case this essentially translates to the following equation $$ -\frac{(x_3-x_1)}{(y_3-y_1)}=\frac{(y_2-y_1)}{(x_2-x_1)}$$ The other equation is $$ (x_3-x_1)^2+(y_3-y_1)^2= R^2$$ Solving the above two equations you will get two points for $(x_3,y_3)$. This shows the existence of two parallel tangents. Similarly you can solve for $(x_4,y_4)$

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?thanks,it seems great,i didnt understand about (a,b) this the center point of which circle? – Arash Jul 15 '12 at 13:59
This is the center of the circle to which you are drawing the tangent. In your case you can take it as $(x_1,y_1) or (x_2,y_2)$. But then the equation of the circle should be consistent. I have edited my comments and removed the ambiguity. – Barun Dasgupta Jul 15 '12 at 14:47
thanks,in the first equation i have x1,x2,y1,y2.i can use just this equation to calculate x3,y3 and x4,y4 .so why should i use second equation?thanks so much – Arash Jul 15 '12 at 16:13
If you do get the correct answer, it is alright. But just double check before you conclude. – Barun Dasgupta Jul 15 '12 at 16:29
thanks so much for your help – Arash Jul 15 '12 at 16:39

\begin{eqnarray} y_3 = y_1 + R \cos \theta \\ x_3= x_1 + R \sin \theta \\ y_4= y_2 +R \cos \theta \\ x_4 = x_2 +R \sin \theta \\ \end{eqnarray}

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thanks but i dont know the value of theta – Arash Jul 14 '12 at 19:04
If you have only the radius and the distance between the circles. I do not think is possible to find points. – user29999 Jul 14 '12 at 19:13
If this is a problem in coordinate geometry, you may know the coordinaes of the centres. And knowing the radius and the distance between centres, you know the trig functions of the little angle at $(x_2,y_2)$. – André Nicolas Jul 14 '12 at 19:18
can i calculate hypotenuse of below triangle with Pythagorean formula and then calculate lengh of tangent line(again using Pythagorean) and then calculate (x3,y3) and (x4,y4) ? – Arash Jul 14 '12 at 19:21
@AndréNicolas:i did not understand what you said – Arash Jul 14 '12 at 19:23

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