# Circle to Circle Tangent Equation

1. Jan 8, 2010

### billinr

I am trying to write an Excel template which will chart two circles and their external tangent lines - similar to a belt and two pulley system.

I have the formulas to calculate and chart the circles. I am looking for a formula to calculate the tangent points between the circles, given the following criteria:

Circle 1 will always have a center at 0,0
Circle 2 will always be larger than circle 1
Center points of both circles will always be on the X axis
Diameters and center distances are known, and will change with the system design

Is there such a formula published?

Thank you

2. Jan 8, 2010

### billinr

I should have included an example....

Circle 1 has a diameter of 21
Circle 2 has a diameter of 106.7
Center distance is 152.2

Thanks

3. Jan 8, 2010

### zgozvrm

In regard to the "upper belt":

Define the center of circle 1 as point A and the center of circle 2 as point B.

The line segment tangent to the top of both circles will intersect circle 1 at point C and circle 2 at point D. Radius AC of circle 1 will be parallel to radius BD of circle 2 (since both are perpendicular to the "belt" CD).

You can then construct a line segment parallel to line segment CD (the upper belt) from the center of circle 1 (point A) to radius BD of circle 2. Call this intersection point E. This line segment AE will have the same length as segment CD. It will also divide radius BD of circle 2 into 2 segments; DE with a length equal to radius AC of circle 1 (21) and BE with a length equal to the difference between the two radii

$$\left(\frac{106.7}{2} - \frac{21}{2}\right) = 42.85$$.

Now you have right triangle ABE with right angle AEB and hypoteneuse AB = 152.2. The angle of radius BD of circle 2 (angle ABE) can be found by

$$\cos^{-1}\left(\frac{42.85}{152.2}\right)$$

Or, more generally

$$\cos^{-1} \left(\frac{R2 - R1}{D} \right)$$

where R1 is the radius of circle 1, R2 is the radius of circle 2, and D is the distance between the centers of the circles.

The tangent point of circle 1 is then at [itex](x_1, y_1)[/tex]
The tangent point of circle 2 is at [itex](x_2, y_2)[/tex]

$$X_1 = \frac{R1^2 - (R1 \times R2)}{D}$$

$$Y_1 = \sqrt{R1^2 - {X_1}^2}$$

$$X_2 =D - \frac{R2^2 - (R1 \times R2)}{D}$$

$$Y_2 = \sqrt{R2^2 - (D - X_2)^2}$$

Use similar logic to find the "lower belt"

Last edited: Jan 9, 2010
4. Jan 9, 2010

### zgozvrm

I made a couple of corrections to my last post.

1) I originally mistook the diameter values given as radius values

2) There were typos in my formulas

5. Jan 11, 2010

### billinr

Thank you very much for the help.