Two circles touch each other at a single point X. Two common tangents to the circles meet at an external point P, and neither tangent passes through X. These tangents touch the larger circle at points B and C. If the radius of the larger circle is 15 centimetres and CP = 20 centimetres, what is the radius, in centimetres, of the smaller circle?

Introduction / Context:
This is a more advanced circle geometry problem involving two tangent circles and their common tangents. The two circles touch each other at a single point, and two external common tangents meet at a point P. One of the tangent segments from P to the larger circle is known, and we must deduce the radius of the smaller circle. The problem uses tangent length properties and the concept of homothety between circles.

Given Data / Assumptions:

• Two circles touch externally at point X.
• The larger circle has radius R = 15 cm and centre O1.
• The smaller circle has radius r cm and centre O2 (r is unknown).
• P is the intersection point of two common external tangents.
• The tangents from P touch the larger circle at points B and C, with CP = 20 cm.
• Tangent lengths from the same external point to the same circle are equal, so PB = PC = 20 cm.
• All geometry is in a Euclidean plane.

Concept / Approach:
First, we use the right triangle formed by the radius to the point of tangency and the tangent segment to find the distance from P to the centre of the larger circle. Then, we use homothety of the two circles. For two externally tangent circles, the intersection point of their external common tangents, the two centres and the homothety centre are collinear, and the distances from this external homothety centre to the two centres are in the ratio of their radii. Together with the known distance between the centres (R + r), this allows us to solve for r.

Step-by-Step Solution:
Step 1: In triangle O1CP, O1C is a radius of length 15 cm and CP is a tangent of length 20 cm. Since radius is perpendicular to tangent at the point of contact, triangle O1CP is right angled at C. Step 2: Use Pythagoras theorem to find O1P: O1P^2 = O1C^2 + CP^2 = 15^2 + 20^2 = 225 + 400 = 625, so O1P = 25 cm. Step 3: Let O2 be the centre of the smaller circle with radius r. Since the circles touch externally, O1O2 = R + r = 15 + r. Step 4: The point P is the external homothety centre of the two circles, so O1, O2 and P are collinear and the distances satisfy O1P / O2P = R / r = 15 / r. Step 5: Place P outside the segment between O1 and O2 so that O1O2 = O1P − O2P. Thus 15 + r = O1P − O2P = 25 − O2P, giving O2P = 10 − r. Step 6: Use the ratio O1P / O2P = 15 / r: 25 / (10 − r) = 15 / r. Step 7: Cross multiply: 25r = 15(10 − r) = 150 − 15r. Step 8: Add 15r to both sides: 40r = 150, so r = 150 / 40 = 15 / 4 = 3.75 cm. Step 9: Check consistency: O2P = 10 − 3.75 = 6.25 cm and O1O2 = 15 + 3.75 = 18.75 cm; also O1P − O2P = 25 − 6.25 = 18.75 cm, which matches.

Verification / Alternative check:
Once r is known, we can confirm the homothety ratio. The ratio R : r = 15 : 3.75 = 4 : 1. The distances O1P : O2P are 25 : 6.25, which also simplifies to 4 : 1, confirming the homothety condition. Additionally, the centre distance 18.75 equals R + r, which is 15 + 3.75, so the circles indeed touch externally with radii 15 cm and 3.75 cm. These checks confirm that no algebraic mistake was made.

Why Other Options Are Wrong:
The values 3.5, 4.25 and 4.45 give ratios of radii that do not match the ratio of distances from P to the centres. If one inserts these in the equation 25 / (10 − r) = 15 / r, the two sides do not agree. The value 5.0 would also contradict the centre distance requirement and the tangent geometry. Only 3.75 cm satisfies both the distance equation and the homothety ratio exactly.

Common Pitfalls:
Many learners incorrectly assume that the tangent lengths from P to both circles are equal, which is not true here; they are only equal for tangents to the same circle. Another pitfall is neglecting the concept of homothety and trying to work purely with right triangles, which becomes very messy. It is also easy to mistake the order of points on the line O1O2P and write O1O2 = O1P + O2P, which leads to negative or impossible radii. Drawing a clear diagram and using the simple ratio O1P / O2P = R / r is the most efficient route.

Final Answer:
The radius of the smaller circle is 3.75 cm.