SSC CGL 2024 Tier-I · 26 September 2024, Shift 3 · Quantitative Aptitude · Question 25 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

A secant PAB is drawn from an external point P to the circle with the centre at O, intersecting it at A and B. If OP = 17 cm, PA = 12 cm and PB = 22.5 cm, then the radius of the circle is:

Answer

Why

Correct — A. Use the power of an external point: for any secant through P, PA × PB = OP² − r².

PA × PB = 12 × 22.5 = 270
OP² = 17² = 289

Rearranged, r² = OP² − PA × PB
= 289 − 270 = 19

r = √19 cm, which is the value printed in option (a).

Why the others are wrong

(b)
Option (b) shows √17 cm. It comes from treating the 17 cm as r² itself, or from reading OP as the radius. OP is measured from the centre out to P, which lies outside the circle.
(c)
Option (c) shows √21 cm, which would need OP² − PA × PB to come to 21, that is a product of 268. The product is 12 × 22.5 = 270 exactly, so 289 − 270 leaves 19.
(d)
Option (d) shows √23 cm, which would need a product of 266. Multiplying 12 by 22.5 gives 270, and no reading of the given lengths produces 266.

Concept

P lies outside the circle and a line through P cuts it at A and then at B. The power of the point P is the product PA × PB, and it is the same for every line drawn through P.

That power also equals OP² − r², which is what links the secant lengths to the radius.

Here PA = 12 and PB = 22.5 give a power of 270, and OP² = 289, so r² = 289 − 270 = 19.

The tangent length from P is the square root of the same power, √270 — the tangent-secant rule and this one are the same statement written two ways.

PB is measured from P all the way to the far intersection, so PB = PA + AB and the chord AB is 22.5 − 12 = 10.5 cm, not 22.5.

Reading PB as the chord is the slip these numbers are built to catch: it gives a product of 12 × 10.5 = 126 and a radius of √163, which no option offers.

Key facts

For an external point P, PA × PB = OP² − r² for every secant drawn through P.
Here 12 × 22.5 = 270 and 17² = 289, so r² = 19 and the radius is √19 cm.
PB is the whole segment from P to the far intersection, so the chord AB equals PB − PA.

Study next

The tangent-secant relation PT² = PA × PB and its link to the power of a pointTwo secants from one external point, where PA × PB = PC × PDFinding the distance from the centre to a chord using the perpendicular from the centre

Common traps

Putting AB = 10.5 into the product in place of PB = 22.5.
Taking OP = 17 as the radius, since that length is also measured from the centre.
Adding OP² and the product instead of subtracting, which gives √559.

SSC works the power of a point from several directions — the radius asked from a secant, a missing secant length, or a tangent length.

Comparable items are set at 19 Sep 2024, 09:00, Quant Q.15, where the radius is asked from a secant, and at 23 Sep 2024, 12:30, Quant Q.2, where two secants are drawn from one point.

Related PYQs

No directly related past PYQ was found.