Three circles with centres A, B, C respectively, touch each other externally. If AB = 5cm, BC = 7cm and CA = 6cm, then the radius of the circle with centre A is:

Let the radii be a, b, c for circles A, B, C. When two circles touch externally, the distance between their centres equals the sum of their radii. So a + b = 5, b + c = 7, and c + a = 6.

Adding all three gives 2(a + b + c) = 18, so a + b + c = 9. Subtracting b + c = 7 from this gives a = 2 cm. Option B is correct.

• A) 1.5 cm fails, because it would make b + c = 7.5, not 7.
• C) 3 cm fails, because it would make b + c = 6, which contradicts BC = 7.
• D) 4 cm fails, because it would make b + c = 5, which contradicts BC = 7.

The core concept is the tangency condition: external contact means centre distance = r1 + r2, while internal contact means centre distance = |r1 − r2|. Setting up such constraints as simultaneous equations is a routine tool in coordinate and mensuration problems. The result is unchanged since the exam year.

Trap: aspirants often use the centre distance directly as a radius, or use the difference of radii as for internal tangency. Read the word "externally" carefully.