In a circle with center O, two chords, AB and CD, cross each other at right angles. If the distance from the center O to chord AB is 3 cm, while the distance from O to chord CD is 4 cm. What is the radius of the circle?

Correct — A.
Drop OM ⟂ AB and ON ⟂ CD, so OM = 3 cm and ON = 4 cm.
AB ⟂ CD, so O, M, the crossing point P and N form a rectangle.

Diagonal OP = √(3² + 4²)
= √(9 + 16) = √25 = 5 cm

The key reads OP as the radius, which is exact when P lies on the circle: r = 5 cm → option (a).

Why the others are wrong

• (b)
  7 cm — 7 cm comes from adding the distances, 3 + 4. They are perpendicular sides of the rectangle OMPN, so they combine by Pythagoras into OP = 5 cm, the length SSC's key takes as the radius.
• (c)
  √5 cm — A radius of √5 ≈ 2.24 cm is less than 3 cm. No chord can be farther from the centre than the radius, so both chords would miss the circle.
• (d)
  √7 cm — √7 ≈ 2.65 cm is √(4² − 3²), a subtraction where the rectangle needs a sum. It is also less than 3 cm, and no chord lies farther from the centre than the radius.

Two facts carry this question. The perpendicular from the centre bisects a chord, so r² = d² + (half-chord)².

When two chords are perpendicular, the perpendiculars from O and the chords themselves enclose a rectangle. Its diagonal, from O to the crossing point, is √(d₁² + d₂²).

Key facts

• The perpendicular from the centre bisects a chord: r² = d² + (chord⁄2)².
• Perpendicular chords at distances d₁ and d₂ from O meet at a point √(d₁² + d₂²) from O.
• A point where two chords cross is inside the circle, so it is less than r from the centre.
• No chord can be farther from the centre than the radius.

Common traps

• Adding the distances, 3 + 4 = 7, instead of combining them as perpendicular sides.
• Subtracting squares, 4² − 3² = 7, and taking √7. The rectangle's diagonal needs 3² + 4².

No directly related past PYQ was found.