A and B are centres of two circles with radii 2 cm and 1 cm respectively, where AB = 5 cm. C is the centre of another circle of radius r cm, which touches each of the above two circles externally. If ∠ACB = 90°, then find the value of r.

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Given:
- Circle A with center A and radius 2 cm
- Circle B with center B and radius 1 cm
- Distance AB = 5 cm
- Circle C with center C and radius r cm touches both circles externally
- ∠ACB = 90°

Since circle C touches both circles externally, the distance between centers A and C is (2 + r) and between B and C is (1 + r).

Let AC = 2 + r and BC = 1 + r.

Given ∠ACB = 90°, triangle ACB is right angled at C.

By Pythagoras theorem:
AB² = AC² + BC²

Substitute values:
5² = (2 + r)² + (1 + r)²
25 = (2 + r)² + (1 + r)²

Expand:
25 = (r² + 4r + 4) + (r² + 2r + 1)
25 = 2r² + 6r + 5

Bring all terms to one side:
2r² + 6r + 5 - 25 = 0
2r² + 6r - 20 = 0

Divide entire equation by 2:
r² + 3r - 10 = 0

Solve quadratic equation:
r = [-3 ± sqrt(3² - 4*1*(-10))]/(2*1)
= [-3 ± sqrt(9 + 40)]/2
= [-3 ± sqrt(49)]/2
= [-3 ± 7]/2

Two roots:
r = (-3 + 7)/2 = 4/2 = 2
r = (-3 - 7)/2 = -10/2 = -5 (not possible as radius can't be negative)

Therefore, r = 2 cm.

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