The distance between the centers of two circles is d. The lengths of their direct and transverse common tangents are L and M, respectively. If L² + M² = 200 and the sum of the squares of their radii is 100, what is the value of d?

Correct — B. Write both tangent formulas and add them, so the individual radii drop out.

Direct: L² = d² − (r₁ − r₂)²
Transverse: M² = d² − (r₁ + r₂)²
Add: L² + M² = 2d² − [(r₁ − r₂)² + (r₁ + r₂)²]

Expand the bracket: (r₁ − r₂)² + (r₁ + r₂)² = 2(r₁² + r₂²)
Substitute: 200 = 2d² − 2 × 100, so 2d² = 400
Solve: d² = 200, d = 10√2 → option (b)

Why the others are wrong

• (a)
  10 — 10 is what L² + M² = 2d² gives if the radius terms are dropped. With them back in, d = 10 makes L² + M² = 200 − 200 = 0, not 200.
• (c)
  5√2 — 5√2 makes d² = 50, so L² + M² = 100 − 200 = −100. A sum of two squares cannot be negative.
• (d)
  20 — 20 is √400, the root of 2d² before halving. d = 20 makes L² + M² = 800 − 200 = 600, not 200.

The two tangent formulas differ in the sign between the radii. Adding them cancels the cross term 2r₁r₂, because (r₁ − r₂)² + (r₁ + r₂)² = 2r₁² + 2r₂².

That leaves L² + M² = 2d² − 2(r₁² + r₂²), which ties d to the sum of squares of the radii without knowing either radius.

Subtracting instead gives L² − M² = 4r₁r₂, which ties the tangents to the product of the radii.

Key facts

• L² + M² = 2d² − 2(r₁² + r₂²).
• L² − M² = 4r₁r₂.
• (a − b)² + (a + b)² = 2(a² + b²).

Common traps

• Dropping the radius terms and writing L² + M² = 2d², which leads to d = 10.
• Taking the square root of 2d² = 400 before dividing by 2, which leads to 20.

No directly related past PYQ was found.