P is a point outside a circle with centre O and it is 14 cm away from the centre A secant PAB drawn ...........

Quantitative Aptitude (CPO) Geometry

P is a point outside a circle with centre O, and it is 14 cm away from the centre. A secant PAB drawn from P intersects the circle at

the points A and B such that PA = 10 cm and PB = 16 cm. The diameter of the Circle is:

This questions was previously asked in

SSC CPO 9th Dec 2019 Shift-2

10 cm

13 cm

12 cm

11 cm

- Given: P is outside the circle with center O, and OP = 14 cm.

- PAB is a secant line with PA = 10 cm and PB = 16 cm.

- According to the Power of a Point theorem: \(OP^2 = PA \times PB + OA^2\).

- Calculate: \(OP^2 = PA \times PB\): \(14^2 = 10 \times 16 + OA^2\).

- Solve: \(196 = 160 + OA^2\), so \(OA^2 = 36\).

- Find OA: \(OA = 6\).

- The diameter of the circle is twice the radius (OA), hence 12 cm.

- Correct Answer: Option 3 - 12 cm.

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