Determine the radius of the smallest circle that can be positioned tangent to two larger circles with given radii while sharing a common tangent line | Step-by-Step Solution

1. What We're Solving

We need to find the radius of a small circle that touches two larger circles (both with radius 4) and a middle circle (radius 2), where all these circles are externally tangent to each other and they all share a common tangent line. Think of it like arranging bubbles that just touch each other while resting on a flat surface!

2. The Approach

This is a beautiful problem that combines coordinate geometry with the properties of tangent circles. Our strategy will be to:

• Set up a coordinate system using the common tangent line as our reference
• Use the fact that externally tangent circles have centers separated by the sum of their radii
• Apply the constraint that all circles must touch the same tangent line

The key insight is that if circles are tangent to the same line, their centers are positioned at distances equal to their respective radii from that line.

3. Step-by-Step Solution

Step 1: Set up coordinates Place the common tangent line along the x-axis (y = 0). Since each circle is tangent to this line, the center of each circle will be at a height equal to its radius above the x-axis.

Step 2: Position the known circles

• Let the two large circles (radius 4) have centers at (-d, 4) and (d, 4) for some distance d
• Let the middle circle (radius 2) have center at (0, 2)
• Let our unknown small circle (radius r) have center at (x, r)

Step 3: Use the external tangency conditions For externally tangent circles, the distance between centers equals the sum of their radii.

Between the middle circle and each large circle:

• Distance from (0, 2) to (±d, 4) = 2 + 4 = 6
• Using distance formula: √(d² + 4) = 6
• Solving: d² + 4 = 36, so d² = 32, therefore d = 4√2

Step 4: Apply tangency for the small circle The small circle must be tangent to both large circles and the middle circle.

Distance from small circle center (x, r) to middle circle center (0, 2): √(x² + (r-2)²) = r + 2

Distance from small circle center (x, r) to right large circle center (4√2, 4): √((x-4√2)² + (r-4)²) = r + 4

Step 5: Solve the system From the first equation: x² + (r-2)² = (r+2)² Expanding: x² + r² - 4r + 4 = r² + 4r + 4 Simplifying: x² = 8r, so x = 2√(2r)

Substituting into the second equation and solving: √((2√(2r) - 4√2)² + (r-4)²) = r + 4

After algebraic manipulation (squaring both sides and simplifying): This leads to: r² - 16r + 16 = 0

Using the quadratic formula: r = (16 ± √(256-64))/2 = (16 ± 8√3)/2 = 8 ± 4√3

4. The Answer

Since we want the smallest circle, we take the smaller root: r = 8 - 4√3 ≈ 1.072

5. Memory Tip

Remember that in tangent circle problems, the magic happens when you use "center distance = sum of radii" for external tangency. Also, when circles share a tangent line, think of them as "sitting" on that line with their centers at heights equal to their radii - this gives you a great coordinate system to work with!

The beautiful symmetry in this problem (with the middle circle perfectly centered) is what makes the algebra work out so nicely. Problems like this show how geometry and algebra work together harmoniously!