If the length of a common external tangent to two circles is 9 and that of a common internal tangent is 5, then the product of the radii of the two circles is:

Correct — B. Write both tangent formulas and subtract — the unknown centre distance cancels itself.

External: L₁² = d² − (r₁ − r₂)², so d² − (r₁ − r₂)² = 9² = 81
Internal: L₂² = d² − (r₁ + r₂)², so d² − (r₁ + r₂)² = 5² = 25

Subtract the second from the first:
81 − 25 = (r₁ + r₂)² − (r₁ − r₂)²
56 = 4r₁r₂

r₁r₂ = 56⁄4 = 14 → option (b).

Why the others are wrong

• (a)
  12 — Needs 4r₁r₂ = 48, i.e. L₁² − L₂² = 48, which would pair an external tangent of 9 with an internal one of √33 ≈ 5.7. The stem's 5 fixes the difference at 56.
• (c)
  10 — Needs 4r₁r₂ = 40, which would follow from an internal tangent of √41 ≈ 6.4. With 9 and 5 the difference of squares is 56, and 56⁄4 is 14.
• (d)
  8 — The squares are subtracted, not the lengths. 9 − 5 = 4 is not the quantity in the identity; 81 − 25 = 56 is, and dividing that by 4 gives 14.

For two circles whose centres are d apart, with radii r₁ and r₂:

Direct (external) tangent: L₁ = √(d² − (r₁ − r₂)²)
Transverse (internal) tangent: L₂ = √(d² − (r₁ + r₂)²)

Square both and subtract, and d disappears:
L₁² − L₂² = (r₁ + r₂)² − (r₁ − r₂)² = 4r₁r₂

So the two tangent lengths alone fix the product of the radii.

Key facts

• L₁² − L₂² = 4r₁r₂, with L₁ the external and L₂ the internal common tangent.
• Here 81 − 25 = 56, so r₁r₂ = 14.
• The identity behind the shortcut is (r₁ + r₂)² − (r₁ − r₂)² = 4r₁r₂.
• The internal tangent is always the shorter of the two, so L₁ > L₂ whenever both exist.

Common traps

• Subtracting the lengths, 9 − 5 = 4, instead of their squares.
• Dividing 56 by 2 after misremembering the identity as 2r₁r₂.
• Trying to solve for r₁ and r₂ separately, which this data cannot do.

No directly related past PYQ was found.