Two circles C₁ and C₂ touch each other externally. The radius of C₁ = 16 cm and the radius of C₂ = 8 cm. Find the length (in cm) of their common tangent.

Correct — B. The direct common tangent of two circles measures √(d² − (r₁ − r₂)²), where d is the distance between the centres.

Touching externally means d = r₁ + r₂ = 16 + 8 = 24
r₁ − r₂ = 16 − 8 = 8
Length = √(24² − 8²) = √(576 − 64) = √512
512 = 256 × 2, so √512 = 16√2 → option (b)

The same result in one line: for externally touching circles the formula reduces to 2√(r₁r₂) = 2√128 = 16√2.

Why the others are wrong

• (a)
  8√2 is √128, which is √(r₁r₂) on its own. The touching-circles formula is 2√(r₁r₂), so this is exactly half the tangent.
• (c)
  8√3 is √192, and neither 576 − 64 = 512 nor r₁r₂ = 128 comes to 192. A √3 cannot appear here, because 512 and 128 are both powers of two.
• (d)
  16√3 is √768. The 16 in front is right but the surd is not: the tangent squared is 512, not 768.

Circles that touch externally have their centres r₁ + r₂ apart, so the word externally hands you d and it never has to be computed.

Substituting d = r₁ + r₂ into √(d² − (r₁ − r₂)²) and expanding, (r₁ + r₂)² − (r₁ − r₂)² = 4r₁r₂, so the length is 2√(r₁r₂). That shortcut is worth memorising for this exact configuration.

The transverse common tangent uses (r₁ + r₂) where the direct one uses (r₁ − r₂). For touching circles that gives √(576 − 576) = 0, which is why the direct tangent is the one with a length to find.

Key facts

• Circles touching externally have centres r₁ + r₂ apart.
• Direct common tangent length = √(d² − (r₁ − r₂)²).
• For externally touching circles that reduces to 2√(r₁r₂), here 2√128 = 16√2.

Common traps

• Using the transverse formula √(d² − (r₁ + r₂)²), which is zero for touching circles.
• Reporting √(r₁r₂) and forgetting the factor of 2 in front.
• Leaving the answer as √512 when the options are written in surd form.

No directly related past PYQ was found.