Enforcement Officer / Accounts Officer (EO/AO) 2017 recruitment test. Transcribed from the Commission's own booklet (series A, GRS-A-TNS); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

The area of the smallest circle which contains a square of area 4 cm² inside is

(a)π cm²
(b)2π cm²
(c)3π cm²
(d)4π cm²

Answer

Why

Correct — B, (b) 2π cm². The smallest circle that contains a square is the circle through its four corners — its circumcircle — and the diameter of that circle is the diagonal of the square, not its side. That is the step which decides the item; the rest is two lines of arithmetic.

The square has area 4 cm², so its side is 2 cm. Its diagonal is side × √2 = 2√2 cm. That diagonal is the diameter of the smallest containing circle, so the radius is √2 cm, and the area of the circle is π r² = π × (√2)² = 2π cm². That is option (b).

Why the circumcircle and not something smaller? Any circle that contains the square must contain all four corners, and the two ends of a diagonal are 2√2 cm apart, so no circle of diameter less than 2√2 cm can hold both of them — a circle of diameter d contains no two points further apart than d. A circle of exactly that diameter, centred at the point where the diagonals cross, does hold all four corners, and since the square is the region enclosed by those corners it holds the whole square. So 2√2 cm is both a lower limit and an achievable one, which is what makes it the answer.

The companion figure is worth learning at the same time. The largest circle that fits inside the square is the incircle, which touches the four sides and has diameter equal to the side, giving a radius of 1 cm and an area of π cm² here. For any square the circumradius is √2 times the inradius, so the circumcircle always has exactly twice the area of the incircle — 2π against π in this case.

Why the others are wrong

(a)π cm² — An area of π cm² means a radius of 1 cm, which is half the side of the square. That is the incircle — the largest circle that fits inside the square, touching the middle of each of the four sides. It does not contain the square at all: the four corners lie √2 cm from the centre, which is outside a circle of radius 1, so each corner sticks out. The option is chosen by a candidate who treats the side of the square as the diameter of the containing circle, which is the standard confusion between the inscribed and the circumscribed circle. The two are easy to keep apart with one picture in the head: the circle inside touches the sides and its diameter is the side, while the circle outside passes through the corners and its diameter is the diagonal. This item asks for the second.
(c)3π cm² — An area of 3π cm² means a radius of √3 cm, which is about 1·73 cm. Such a circle does contain the square, since the corners are only √2 cm, about 1·41 cm, from the centre — so this option is not geometrically impossible in the way option (a) is. It is simply not the smallest, and the stem asks for the smallest. That is the discipline the question is really testing: a candidate must reject not only circles that fail to hold the square but also circles that hold it with room to spare. Nothing in the figure produces √3 as a natural length, which is a second reason to distrust the option. The lengths that arise from a square of side 2 are 2 for the side, 2√2 for the diagonal, 1 for the inradius and √2 for the circumradius, and any answer built from a different length has come from somewhere outside the problem.
(d)4π cm² — An area of 4π cm² means a radius of 2 cm, equal to the whole side of the square, and it comes from treating the side as the radius rather than as twice the inradius. Like option (c) this circle really does contain the square — comfortably, since a radius of 2 cm is well beyond the √2 cm needed — but it is far from the smallest one, having twice the area of the correct answer. It is also the option a candidate reaches by pure substitution, taking the number 4 from the stem and putting it into πr² without asking what the 4 measures; the 4 in the stem is an area in square centimetres, not a radius. Reading each number with its unit attached is the cheapest safeguard in mensuration, and here it disposes of this option immediately.

Concept

Two circles are naturally associated with a square. The incircle is the largest circle that fits inside it; its centre is the centre of the square, it touches the midpoint of each side, and its diameter equals the side, so its radius is half the side. The circumcircle is the smallest circle that contains the square; it has the same centre, passes through all four corners, and its diameter equals the diagonal, so its radius is half the diagonal. For a square of side s the diagonal is s√2, so the circumradius is s ÷ √2 and the inradius is s ÷ 2. The circumradius is therefore √2 times the inradius and the circumcircle has exactly twice the area of the incircle, whatever the size of the square. The general principle behind the phrase 'smallest containing circle' is worth stating separately, because it applies to any shape: a circle of diameter d cannot contain two points more than d apart, so the smallest circle containing a figure can never be smaller than the greatest distance between two points of that figure. For a square that greatest distance is the diagonal, and a circle on the diagonal as diameter does contain the square, so the lower limit is achieved and the answer is exact.

Plane geometry in an EPFO paper is nearly always about the relationship between a figure and a circle drawn inside or around it, and the whole family is answered from two facts about the square: the side is the diameter of the inner circle and the diagonal is the diameter of the outer one. The examiner tests whether a candidate has those two the right way round, and offers the answer to the other one among the options — here π cm², the incircle, sits at the head of the list. A second habit is being tested as well. The stem gives the square by its area rather than by its side, so the first move must be to take a square root and recover the side. Candidates who substitute the given number straight into a formula produce 4π, which is exactly the fourth option. Converting every given quantity into the dimension the formula actually needs, before any calculation, is the general lesson.

Key facts

The smallest circle containing a square is its circumcircle, which passes through all four corners.
The diameter of that circumcircle is the diagonal of the square, which for side s is s√2.
A square of area 4 cm² has side 2 cm, diagonal 2√2 cm and therefore circumradius √2 cm.
The area of the circle is π × (√2)² = 2π cm².
The incircle, the largest circle fitting inside the square, has diameter equal to the side and area π cm² here.
For any square the circumradius is √2 times the inradius, so the circumcircle has exactly twice the area of the incircle.
A circle of diameter d cannot contain two points more than d apart, which is why the diagonal fixes the minimum size.
Circles of radius √3 cm or 2 cm would also contain this square, but the question asks for the smallest such circle.

The smallest containing circle is the one through the corners

Area 4 cm² → side 2 cm → diagonal 2√2 cm→The diameter of the smallest circle containing a square is the diagonal of the square, not its side. That is the step which decides the item; the rest is two lines of arithmetic.

Radius √2 cm, so area = π × (√2)² = 2π cm²→The diagonal is the diameter, so the radius is half of it.

Why the circumcircle, and nothing smaller→Any circle containing the square must contain all four corners, and the two ends of a diagonal are 2√2 cm apart — a circle of diameter d contains no two points further apart than d. A circle of exactly that diameter, centred where the diagonals cross, does hold all four corners, so 2√2 cm is both a lower limit and an achievable one.

(a) π cm² — the incircle→Radius 1 cm, half the side: the largest circle that fits inside the square, touching the middle of each of the four sides. It does not contain the square at all, since the corners lie √2 cm from the centre and stick out. The standard confusion between the inscribed and the circumscribed circle — one touches the sides, the other passes through the corners.

(c) 3π cm² and (d) 4π cm² — containing, but not smallest→Radii of √3 cm and 2 cm both hold the square comfortably, and the stem asks for the smallest. Nothing in the figure produces √3 as a natural length; and 4π comes of taking the number 4 from the stem and putting it into πr² without asking what the 4 measures — it is an area in square centimetres, not a radius.

The pair worth learning together→For any square the circumradius is √2 times the inradius, so the circumcircle always has exactly twice the area of the incircle — 2π against π here. The general principle applies to any shape: the smallest containing circle can never be smaller than the greatest distance between two points of the figure.

Study next

Incircle and circumcircle of a square, a rectangle and an equilateral triangle, with the radius in each caseThe diagonal of a square and of a rectangle, and the Pythagorean relation behind themAreas of shaded regions formed between a polygon and a circle drawn inside or around itThe circumcircle of a triangle and how its centre is located from the perpendicular bisectorsProblems asking for the largest figure of one kind that fits inside a figure of another

Common traps

Using the side of the square instead of the diagonal as the diameter of the containing circle.
Answering with the incircle, which touches the sides but leaves the four corners outside.
Accepting any circle that contains the square when the stem asks specifically for the smallest one.
Substituting the given area straight into the circle formula, without first taking the square root to get the side.
Losing the square of the radius: (√2)² is 2, and dropping the squaring turns the answer into π√2.

Geometry items in these papers cluster around a small number of standard configurations, and the square with a circle drawn inside or around it is the commonest of them. The examiner varies the item by changing which quantity is given and which is wanted: the side may be given and the outer circle's area wanted, the circle's area given and the square's side wanted, or the shaded region between the two figures asked for. Occasionally the same configuration appears in coordinate form, with three points given and the radius of the circle through them required. All of these reduce to the two relations between a square and its circles, together with the Pythagorean theorem that produces the diagonal. Committing those to memory as pictures rather than as formulas is what makes the family quick.

Related PYQs

EPFO_EOAO_2017_Q120Open & attempt →

The circumference of a circle is 2π cm. Then the area of a square inscribed in the circle is

(a) π/2 cm²
(b) 1 cm²
(c) 2π cm²
(d) 2 cm²

Answer(d) 2 cm²

The same square-and-circle relation reversed, three items later in this paper: a circle of known circumference with a square inscribed in it, and the square's area wanted.

EPFO_APFC_2023_Q73

What is the length of the radius of the circle that passes through the points (0, 0), (0, 3) and (2, 0)?

(a) 2√3
(b) 2√5
(c) √11/2
(d) √13/2

Answer(d) √13/2

The circumcircle in coordinate form — the radius of the circle passing through three given points.

Practice

practice — not a real PYQ
What is the area of the smallest circle that contains a square of side 4 cm?
- (a)4π cm²
- (b)8π cm²
- (c)16π cm²
- (d)32π cm²
Answer(b) 8π cm²
practice — not a real PYQ
What is the radius of the largest circle that can be drawn inside a square of side 10 cm?
- (a)5 cm
- (b)5√2 cm
- (c)10 cm
- (d)10√2 cm
Answer(a) 5 cm

Why