SSC CGL 2025 Tier-I · 17 September 2025, Shift 1 · Quantitative Aptitude · Question 23 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 13 September 2026.

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SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

From an outside point P, two tangents, PA and PB, are drawn to a circle. Given that the length of PA is 12 cm, what is the length of PB?

(a)6 cm
(b)10 cm
(c)12 cm
(d)24 cm

Answer

Why

Correct — C.
Join the centre O to A, B and P.
OA = OB (both radii), and OP is common.
∠OAP = ∠OBP = 90° (a radius meets a tangent at right angles).

So △OAP ≅ △OBP by RHS, which makes PA = PB.
PB = PA = 12 cm → option (c)

Why the others are wrong

(a)
6 cm — 6 cm is half of PA, but no step halves a tangent. PA and PB are matching sides of the congruent triangles OAP and OBP, so PB is 12 cm.
(b)
10 cm — 10 cm cannot come from the given data. With only PA known, the equal-tangents rule is the one link to PB, and it gives 12 cm.
(d)
24 cm — 24 cm is PA + PB, the two tangents together. The question asks for PB alone, which equals PA = 12 cm.

Concept

From a point outside a circle you can draw two tangents to it, and they are the same length.

The proof uses the right angle at the point of contact: a radius drawn to where a tangent touches is perpendicular to that tangent. So triangles OAP and OBP are right-angled, share the hypotenuse OP and have equal radii OA and OB.

They are congruent, so PA = PB. The same pair of triangles shows that OP bisects the angle APB.

The radius and the distance OP are not given, and they are not needed: the equal-tangents rule links PA to PB directly.

Key facts

Tangents drawn from an external point to a circle are equal in length.
A radius drawn to the point of contact is perpendicular to the tangent.
The line from the external point to the centre bisects the angle between the two tangents.
Tangent length = √(d² − r²), where d is the distance from the point to the centre and r is the radius.

Study next

Tangent length from the distance to the centre and the radius, √(d² − r²)The angle between two tangents and the angle AOB at the centre, which add to 180°Perimeter of the triangle cut off by a third tangent, found with equal tangents

Common traps

Adding the two tangents and choosing 24 cm when the question asks for one of them
Hunting for a radius or a distance to the centre, when the equal-tangents rule needs neither

At 23 Sep 2025, 16:00, Quant Q.25 a third tangent cuts PA and PB at R and S, and equal tangents from P, R and S make the perimeter of △PRS equal PA + PB = 20 cm.

19 Sep 2025, 09:00, Quant Q.23 adds the radius: tangents of 10 cm and a radius of 6 cm put P about 12 cm from the centre.

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 140 SSC CGL questions on it, every solved shift.