SSC CGL 2025 Tier-I · 23 September 2025, Shift 3 · Quantitative Aptitude · Question 21 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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Two circles have radii of 8 cm and 3 cm. The distance between their centers is 15 cm. What is the length of a direct common tangent?

(a)10 cm
(b)12 cm
(c)10√2 cm
(d)12√2 cm

Answer

Why

Correct — C. Direct common tangent = √(d² − (r₁ − r₂)²).

Difference of radii: 8 − 3 = 5, squared = 25
Square the centre distance: 15² = 225
Subtract: 225 − 25 = 200
Take the root: √200 = √(100 × 2) = 10√2 cm → option (c)

Why the others are wrong

(a)
10 cm — 10 cm is √100, the square root of 200 with the √2 lost. √200 = √100 × √2 = 10√2 ≈ 14.14 cm.
(b)
12 cm — 12 cm needs 15² − 12² = 81 as (r₁ − r₂)², a radius difference of 9 cm. The radii here differ by 5 cm.
(d)
12√2 cm — 12√2 ≈ 16.97 cm is longer than the 15 cm between the centres. The direct tangent is √(d² − (r₁ − r₂)²), so it can never exceed d.

Concept

Both radii to the points of contact are perpendicular to the tangent. Through the smaller circle's centre, draw a line parallel to the tangent until it meets the larger radius.

That makes a right triangle: hypotenuse = the centre distance d, one leg = the tangent length, the other leg = r₁ − r₂. For a transverse tangent that leg becomes r₁ + r₂.

The transverse common tangent of these circles would be √(225 − 11²) = √104 ≈ 10.2 cm, which is not among the options. Using the sum of the radii by mistake therefore leads nowhere.

Key facts

Direct common tangent = √(d² − (r₁ − r₂)²).
Transverse common tangent = √(d² − (r₁ + r₂)²), which exists only when d > r₁ + r₂.
A radius drawn to the point of contact is perpendicular to the tangent.
√200 = 10√2 ≈ 14.14.

Study next

Transverse common tangent and when it existsNumber of common tangents for separate, touching and intersecting circlesDirect common tangent of circles touching externally: 2√(r₁r₂)

Common traps

Using r₁ + r₂ in place of r₁ − r₂, which is the transverse formula and gives √104.
Simplifying √200 to 10 by dropping the √2.

The direct common tangent is also asked 9 Sep 2024, 09:00, Quant Q.17 (radii 22 and 10 cm, centres 37 cm apart: √(1369 − 144) = 35 cm).

15 Sep 2025, 09:00, Quant Q.22 runs it backwards: tangent 15 cm, radii 12 and 4 cm, so d = √(225 + 64) = 17 cm.

24 Sep 2025, 12:30, Quant Q.21 asks the transverse tangent instead: radii 6 and 2 cm, 10 cm apart, √(100 − 64) = 6 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 140 SSC CGL questions on it, every solved shift.