SSC CGL 2024 Tier-I · 12 September 2024, Shift 2 · Quantitative Aptitude · Question 18 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 6 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.

From a point T, the length of the tangent to a circle is 32 cm and the distance of T from the centre is 40 cm. The diameter (in cm) of the circle is:

(a)48
(b)16
(c)32
(d)36

Answer

Why

Correct — A. A tangent touches a circle at right angles to the radius, so the centre O, the point of contact P and the external point T form a right triangle right-angled at P.

OT² = OP² + PT²
40² = r² + 32²
1600 = r² + 1024
r² = 576

r = 24 cm
Diameter = 2 × 24 = 48 cm → option (a).

Why the others are wrong

(b)
16 — 16 comes from subtracting the lengths, 40 − 32 = 8, and doubling. Pythagoras works on the squares of the sides, not on the sides, so this shrinks the radius from 24 cm to 8 cm.
(c)
32 — 32 is the tangent length itself, handed back as a diameter. The tangent is one leg of the right triangle; the circle's width is built from the other leg.
(d)
36 — 36 matches no step of the working. With OT = 40 and the tangent 32, the radius is pinned at 24 cm, so the diameter can only be 48 cm.

Concept

The tangent–radius theorem: a tangent is perpendicular to the radius drawn to the point of contact. That single right angle turns every external-point question into Pythagoras.

With O the centre, T the external point and P the contact point, triangle OPT is right-angled at P, so (distance from centre)² = (radius)² + (tangent length)².

Any two of those three give the third. SSC hands you two and asks for the third, sometimes dressed as a diameter so that the radius has to be doubled at the end.

The 40 cm is measured from the centre, not from the edge of the circle. Read it as the distance to the nearest point of the circle and the triangle changes, and so does the answer.

Key facts

A tangent is perpendicular to the radius at the point of contact, so the right angle sits on the circle, not at the centre.
For an external point T: (distance from centre)² = (radius)² + (tangent length)².
Here 40² − 32² = 1600 − 1024 = 576, so the radius is 24 cm.
The question asks for the diameter, which is 2 × 24 = 48 cm.

Study next

Length of the direct common tangent to two circles, √(d² − (r₁ − r₂)²)Length of the transverse common tangent, √(d² − (r₁ + r₂)²)Tangents from an external point are equal, and the angle that pair subtends at the centre

Common traps

Stopping at the radius, 24 cm, when the question asks for the diameter.
Subtracting 40 − 32 instead of 40² − 32².
Treating the 40 cm as the distance from the circle rather than from the centre.

SSC keeps the same right triangle and moves the unknown around. The tangent length is asked from radius and distance on 13 Sep 2024, 12:30, Quant Q.7 and on 23 Sep 2024, 16:00, Quant Q.23; the two-circle version, where the tangent runs between two circles, is asked on 10 Sep 2024, 16:00, Quant Q.22.

Related PYQs

No directly related past PYQ was found.