SSC CGL 2025 Tier-I · 18 September 2025, Shift 2 · 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.

A circle has radius 7 cm. A tangent is drawn from an external point P. If the length of the tangent is 24 cm, what is the distance from P to the center of the circle?

(a)17 cm
(b)25 cm
(c)31 cm
(d)576 cm

Answer

Why

Correct — B. The radius drawn to the point of contact T is perpendicular to the tangent, so △OTP has its right angle at T and OP as the hypotenuse.

Pythagoras in △OTP: OP² = OT² + PT²
Substitute: OP² = 7² + 24² = 49 + 576
Add: OP² = 625
Take the root: OP = √625 = 25 cm → option (b)

Why the others are wrong

(a)
17 cm — 17 cm is 24 − 7, a plain subtraction of lengths. The hypotenuse must be longer than the 24 cm tangent, and √(7² + 24²) = 25 cm.
(c)
31 cm — 31 cm is 24 + 7. Adding the two sides ignores the right angle at T. The hypotenuse of a 7–24 right triangle is √625 = 25 cm.
(d)
576 cm — 576 is 24², the square of the tangent alone. OP² needs 7² added as well, giving 625, and then the square root: 25 cm.

Concept

A tangent is perpendicular to the radius at the point of contact. So the centre, the contact point and the external point form a right triangle, and Pythagoras links its three sides.

The distance from the centre is always the hypotenuse: OP² = r² + (tangent)². Any two of radius, tangent and distance give the third. Here 7, 24, 25 is a Pythagorean triple, so the root comes out whole.

Key facts

A tangent to a circle is perpendicular to the radius drawn to the point of contact.
(Distance from the centre)² = radius² + (tangent length)².
7, 24, 25 is a Pythagorean triple: 49 + 576 = 625.
8, 15, 17 is another: 64 + 225 = 289.

Study next

Tangent length from a given radius and distance: PT = √(OP² − r²)Two tangents from one external point and the angle between themDirect and transverse common tangents of two circles

Common traps

Treating the tangent as the hypotenuse and working out √(24² − 7²) ≈ 23 cm: OP is opposite the right angle at T, so OP is the hypotenuse.
Squaring and never taking the root, which leaves 576 or 625.

17 Sep 2025, 16:00, Quant Q.23 runs the triangle the other way: radius 8 cm and distance 17 cm give a tangent of √(289 − 64) = 15 cm.

12 Sep 2024, 12:30, Quant Q.18 asks for the circle instead: tangent 32 cm and distance 40 cm give a radius of 24 cm, so a diameter of 48 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.