SSC CGL 2024 Tier-I · 13 September 2024, Shift 1 · Quantitative Aptitude · Question 13 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 10 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.

A and B are centres of two circles with radii 2 cm and 1 cm respectively, where AB = 5 cm. C is the centre of another circle of radius r cm, which touches each of the above two circles externally. If ∠ ACB=90° , then the value of r is:

(a)2 cm
(b)4 cm
(c)3 cm
(d)5 cm

Answer

Why

Correct — A. External contact makes each centre-to-centre distance the sum of the radii.

AC = 2 + r and BC = 1 + r
∠ACB = 90°, so Pythagoras holds in triangle ACB: AC² + BC² = AB²
(2 + r)² + (1 + r)² = 5² = 25

r² + 4r + 4 + r² + 2r + 1 = 25
2r² + 6r − 20 = 0, i.e. r² + 3r − 10 = 0
(r + 5)(r − 2) = 0

A radius cannot be −5, so r = 2 cm → option (a).

Why the others are wrong

(b)
4 cm — 4 cm gives AC = 6 and BC = 5, so AC² + BC² = 61 against AB² = 25. The angle at C would be acute, not the right angle the stem fixes.
(c)
3 cm — 3 cm gives AC = 5 and BC = 4, and 25 + 16 = 41, again not 25. It is also not a root of r² + 3r − 10 = 0, whose roots are 2 and −5.
(d)
5 cm — 5 cm is AB copied straight out of the stem. It makes AC = 7 and BC = 6, and 49 + 36 = 85 is nowhere near AB² = 25.

Concept

Tangency is a statement about centre distance, and that is the only thing that turns this picture into algebra.

Two circles touching externally have their centres apart by the sum of the radii; touching internally, by the difference. Circle C touches both of the given circles externally, so AC = 2 + r and BC = 1 + r.

The right angle at C then puts those two lengths on the legs of a right triangle whose hypotenuse AB = 5 is given, and Pythagoras turns the geometry into a quadratic with one usable root.

No figure is printed with this question, so the triangle has to be drawn from the words. Nothing in the stem places C on either side of AB, and it does not matter — only the three lengths enter the working.

Key facts

Two circles touching externally have centre distance equal to the sum of their radii.
Two circles touching internally have centre distance equal to the difference of their radii.
Here (2 + r)² + (1 + r)² = 25 reduces to r² + 3r − 10 = 0, whose positive root is r = 2.

Study next

Direct and transverse common tangent lengths between two circlesInternal versus external contact and the centre-distance ruleTurning a right-angle condition into a quadratic in an unknown radius

Common traps

Using the difference of the radii, which is the internal-contact rule
Placing the right angle at A or B rather than at C where the stem puts it
Keeping the negative root and offering it as a second answer

The same centre-distance rule can be asked for a radius, as here, or for a tangent length.

Quant Q.9 of the 10 Sep 2024, 09:00 shift uses the same external-contact set-up for circles of radii 18 cm and 12 cm and asks for their direct common tangent.

Related PYQs

No directly related past PYQ was found.