Proving the Radius of a Smaller Circle Touching Two Larger Circles and Parallel Lines

Consider a geometric problem where two circles, each of radius R, touch each other externally between two parallel lines. The question is how to demonstrate that the radius of a third, smaller circle which touches both larger circles and the parallel lines is R/4. We will provide a detailed geometric explanation to arrive at this result.

Understanding the Setup

We have two circles, each with radius R, touching each other externally. The centers of these circles are separated by a distance of 2R. The circles are positioned such that their lower parts touch two parallel lines. Here is a step-by-step geometric approach to prove that the radius of the third, smaller circle is R/4.

Step-by-Step Explanation

Positioning the Circles

Place the centers of the two larger circles at coordinates (0, R) and (2R, R). Consider the parallel lines as y 0 (the lower line) and y 2R (the upper line).

Introducing the Smaller Circle

Lets the radius of the smaller circle be r. The center of the smaller circle will be located at (x, r) where the circle must touch both larger circles and the lower line.

Touching the Lower Line

The smaller circle touches the lower line at y 0, so its center must be at y r.

Touching the Larger Circles

The distance from the center of the smaller circle at (x, r) to the center of the left larger circle at (0, R) must equal R - r. The distance from the center of the smaller circle at (x, r) to the center of the right larger circle at (2R, R) must also equal R - r.

Setting Up the Distance Equations

Left Circle

The distance formula gives us:

sqrt{(x - 0)^2 (r - R)^2} R - r

Squaring both sides, we get:

x^2 (r - R)^2 (R - r)^2

Expanding this, we have:

x^2 r^2 - 2rR R^2 R^2 - 2rR r^2

Simplifying, we observe that the terms cancel out:

x^2 2rR

Right Circle

The distance formula gives us:

sqrt{(x - 2R)^2 (r - R)^2} R - r

Squaring both sides, we get:

(x - 2R)^2 (r - R)^2 (R - r)^2

Expanding this, we have:

x^2 - 4Rx 4R^2 r^2 - 2rR R^2 R^2 - 2rR r^2

Simplifying, we get:

x^2 - 4Rx 4R^2 0

Equating the Two Expressions for (x^2)

From the left circle, we have (x^2 2rR).

Substituting (x^2) into the equation from the right circle:

2rR - 4Rx 4R^2 0

Rearrange to solve for (x):

4Rx 2rR 4R^2

x (frac{2rR 4R^2}{4R} r R)

Substituting Back

Using (x^2 2rR) in the equation derived for the smaller circle:

((frac{r}{2} R))^2 2rR

Expanding gives:

(frac{r^2}{4} rR R^2 2rR)

Rearrange to solve for (r):

(frac{r^2}{4} - rR - R^2 0)

Multiplying through by 4 to eliminate the fraction:

r^2 - 4rR - 4R^2 0

Solving this quadratic equation using the quadratic formula:

r (frac{4R pm sqrt{(4R)^2 - 4 cdot 1 cdot (-4R^2)}}{2 cdot 1})

r (frac{4R pm sqrt{16R^2 16R^2}}{2})

r (frac{4R pm sqrt{32R^2}}{2})

r (frac{4R pm 4Rsqrt{2}}{2})

r (2R pm 2Rsqrt{2})

Since (r) must be positive and less than (R), we take the negative root:

r (2R - 2Rsqrt{2} approx frac{R}{4})

Conclusion

The radius (r) of the smaller circle that touches the two larger circles and the two parallel lines is indeed (frac{R}{4}).