SSC CGL 2025 Tier-I · 19 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 11 September 2026.

Where this paper and its answers come from

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.

The length of each of two tangents drawn from an external point to a circle is 10 cm. What is the approximate distance from the external point to the center of the circle, if the radius is 6 cm?

(a)4 cm
(b)10 cm
(c)12 cm
(d)16 cm

Answer

Why

Correct — C.
Call the external point P, the contact point A and the centre O.
Tangent meets radius at 90°: ∠OAP = 90°

Pythagoras in △OAP: OP² = PA² + OA²
= 10² + 6² = 100 + 36 = 136
OP = √136 ≈ 11.66 cm

Nearest option: 12 cm → option (c)

Why the others are wrong

(a)
4 cm — 4 cm is 10 − 6, a subtraction with no geometric meaning here. OP is the hypotenuse of △OAP, so it must be longer than the 10 cm tangent.
(b)
10 cm — 10 cm is the tangent itself, which ends at the contact point A. OP runs on to the centre and, as the hypotenuse, exceeds 10 cm: √136 ≈ 11.66.
(d)
16 cm — 16 cm is 10 + 6, adding the legs instead of combining them by Pythagoras. √(10² + 6²) ≈ 11.66 cm, well short of 16.

Concept

A tangent is perpendicular to the radius at the point of contact. So the centre O, the contact point A and the external point P form a right triangle, with the right angle at A and OP as the hypotenuse.

Any two of radius, tangent length and distance give the third: OP² = PA² + r². The two tangents from P are equal in length, which is why the stem can say each is 10 cm.

√136 = 2√34 is not a whole number, which is why the stem asks for the approximate distance. 11.66 rounds to 12.

Key facts

The radius to the point of contact is perpendicular to the tangent.
Tangents drawn from an external point to a circle are equal in length.
(Distance from the centre)² = (tangent length)² + (radius)².

Study next

Angle between two tangents and the angle at the centre: they add to 180°Area of the kite formed by two tangents and two radiiDirect and transverse common tangents of two circles

Common traps

Subtracting 6 from 10, or adding them, instead of using Pythagoras
Taking the tangent length as the distance to the centre

18 Sep 2025, 12:30, Quant Q.23 is the same right triangle with radius 7 cm and tangent 24 cm: distance = √(49 + 576) = the keyed 25 cm.

17 Sep 2025, 16:00, Quant Q.23 runs it backwards: radius 8 cm and distance 17 cm give tangent √(289 − 64) = the keyed 15 cm.

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 84 SSC CGL questions on it, every solved shift.