SSC CGL 2024 Tier-I · 23 September 2024, Shift 3 · Quantitative Aptitude · Question 23 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.

The length of the tangent drawn from a point 10 cm away from the centre of a circle with radius 4 cm, is:

Answer

Why

Correct — A. A tangent meets the radius at the point of contact at 90°. So the centre O, the point of contact T and the external point P form a right triangle, right-angled at T, with OP as the hypotenuse.

OP = 10 cm, OT = radius = 4 cm
PT² = OP² − OT²
= 100 − 16 = 84

PT = √84 = √(4 × 21) = 2√21 cm ≈ 9.17 cm

The four options are printed as images, and the one reading 2√21 is option (a).

Why the others are wrong

(b)
Option (b) reads 2√14, which is √56. The right triangle needs 10² − 4² = 84, not 56 — 2√14 would follow from a radius of √44 ≈ 6.6 cm.
(c)
Option (c) reads 3√21, which is √189 ≈ 13.7 cm. That is longer than the 10 cm hypotenuse, and a leg cannot exceed the hypotenuse.
(d)
Option (d) reads 3√14, which is √126 ≈ 11.2 cm. Again longer than the 10 cm distance from P to the centre, so it fails the same size check.

Concept

Every tangent question begins from one fact: the radius drawn to the point of contact is perpendicular to the tangent. That single right angle converts a circle problem into a Pythagoras problem.

With d the distance from the external point to the centre and r the radius, the tangent length is √(d² − r²). The distance d is always the hypotenuse, so the tangent is always the shorter of the two.

Here √(100 − 16) = √84, and √84 simplifies to 2√21 because 84 = 4 × 21. The options here are printed in simplified surd form, so an unsimplified √84 will appear to match nothing.

Because 2√21 ≈ 9.17 cm is only a little under 10 cm, the size check alone will not separate it from option (b) at 7.48 cm. It does rule out both options above 10 cm without any arithmetic.

Key facts

A tangent is perpendicular to the radius at the point of contact.
Tangent length from an external point = √(d² − r²), where d is the distance to the centre.
Here √(10² − 4²) = √84 = 2√21 cm, about 9.17 cm.
The two tangents drawn from the same external point to a circle are equal in length.

Study next

Equal tangents from an external point and the proofs that use themChord length from the perpendicular distance to the centre, the same right triangle reversedPower of a point: PT² = PA × PB for any secant PAB through the same external point

Common traps

Adding rather than subtracting: √(100 + 16) treats the 10 cm as a leg instead of the hypotenuse.
Leaving the answer as √84 and finding no option that matches, because the options are simplified surds.
Taking the 10 cm as the distance to the circle rather than to the centre, which would make the radius part of a different triangle.

The same right triangle inside a circle is asked in both directions, and the given quantity tells you which side is the hypotenuse. Quant Q.3 of this paper runs it the other way, giving a 14 cm chord 24 cm from the centre and asking for the radius.

Related PYQs

No directly related past PYQ was found.

Topic: Circle and its chords, tangents, angles subtended by chords of a circle, common tangents to two or more circles — all 67 SSC CGL questions on it, every solved shift.