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To find the radius of the smaller circle in terms of 'r' and 'a', we can use the properties of the circles and the equilateral triangle.

Let's label the radius of the smaller circle as 'r_s'. We can observe that the center of the smaller circle is the incenter of the equilateral triangle. The incenter is the point where the angle bisectors of the triangle intersect.

Now, let's draw the radii of the smaller circle to the points where it is tangent to the equilateral triangle. These radii are perpendicular to the sides of the triangle at the points of tangency.

By drawing these radii, we form three smaller triangles within the equilateral triangle. Each of these smaller triangles is a 30-60-90 triangle, as the angles in an equilateral triangle are 60 degrees each.

In a 30-60-90 triangle, the ratio of the lengths of the sides is 1:√3:2.

Considering one of these smaller triangles, we have:

The short leg = r_s

The long leg = (a - 2r_s) [since the radius of the larger circle is 'r']

The hypotenuse = a

Using the ratio of the lengths of the sides of a 30-60-90 triangle, we can set up the following equation:

r_s / (a - 2r_s) = 1/√3

To solve for r_s, we can cross-multiply and simplify:

√3 * r_s = a - 2r_s

√3 * r_s + 2r_s = a

(√3 + 2) * r_s = a

Finally, we can solve for r_s:

r_s = a / (√3 + 2)

Therefore, the radius of the smaller circle in terms of 'r' and 'a' is r_s = a / (√3 + 2).

Note: The derivation assumes that the larger circle is inscribed in the equilateral triangle. If the smaller circle is inscribed instead, the derivation would differ slightly, but the final result would remain the same.

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