In a circle with radius 5 cm, a diameter AB and a chord PQ (which is not a diameter) intersect at point X such that PQ is perpendicular to AB at X. If AX : XB = 3 : 2 along the diameter AB, what is the length (in cm) of the chord PQ?

Introduction / Context:
This problem deals with the geometry of a circle involving a diameter, a chord perpendicular to that diameter, and a given ratio on the diameter. It tests understanding of how to use basic coordinate geometry or Pythagoras theorem inside a circle. By using the given ratio on the diameter and the fact that the chord is perpendicular to the diameter, we can find the exact position of the intersection point and then compute the chord length using the circle equation or distance formula.

Given Data / Assumptions:

• Circle radius r = 5 cm.
• AB is a diameter of the circle.
• PQ is a chord that is not a diameter.
• PQ is perpendicular to AB at their intersection point X.
• AX : XB = 3 : 2 along the diameter AB.
• We need the length of chord PQ.

Concept / Approach:
A convenient way to solve this is to use a coordinate system. We can place the centre of the circle at the origin and take the diameter AB along the horizontal axis. Once we know the coordinates of point X on AB using the given ratio, we exploit the fact that PQ is perpendicular to AB, so it must be a vertical line through X. The endpoints of the chord PQ are then the intersection of this vertical line with the circle. Using the circle equation x^2 + y^2 = r^2, we can find the y-coordinates of the chord endpoints and thus the length of PQ.

Step-by-Step Solution:
Let the centre of the circle O be at (0, 0) with radius 5 cm.Take diameter AB along the x-axis, so A = (−5, 0) and B = (5, 0).Given AX : XB = 3 : 2, and AB = 10 cm, so AX = 6 cm and XB = 4 cm.Starting from A at x = −5, point X is 6 cm to the right, so X = (−5 + 6, 0) = (1, 0).Chord PQ is perpendicular to AB at X, so PQ lies on the vertical line x = 1.Points P and Q are where this line meets the circle x^2 + y^2 = 25.Substitute x = 1: 1^2 + y^2 = 25 so y^2 = 24.Thus y = ±√24 = ±2√6.So P = (1, 2√6) and Q = (1, −2√6).Length of chord PQ = difference in y-coordinates = 2√6 − (−2√6) = 4√6 cm.

Verification / Alternative check:
We can verify using the chord length formula. The distance from the centre to the chord is the x-coordinate of X, which is 1 cm. For a circle of radius r, a chord at distance d from the centre has length L = 2 * √(r^2 − d^2). Here, r = 5 and d = 1, so L = 2 * √(25 − 1) = 2 * √24 = 4√6 cm. This matches our coordinate geometry result and confirms the correctness of the calculation.

Why Other Options Are Wrong:
The options 2√13 cm, 5√3 cm, and 6√5 cm all correspond to chord lengths that would require different distances from the centre or different radii. None satisfy the exact relation with a radius of 5 cm and a chord at distance 1 cm from the centre. The option 8 cm is a simple rational guess and does not match the exact geometric requirements. Only 4√6 cm is consistent with the circle equation and perpendicular chord construction.

Common Pitfalls:
Some learners may misplace the centre or choose the wrong orientation for the diameter, leading to incorrect coordinates for X. Another common error is not using the ratio AX : XB correctly along the full length of 10 cm. Mistakes in solving 1^2 + y^2 = 25 or incorrectly calculating the chord length from the two y-values also occur. Being systematic about coordinate placement and carefully applying the chord length formula helps avoid these issues.

Final Answer:
The length of the chord PQ is 4√6 cm.