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Geometry - Circles - Previous Year CAT/MBA Questions

The best way to prepare for Geometry - Circles is by going through the previous year Geometry - Circles CAT questions. Here we bring you all previous year Geometry - Circles CAT questions along with detailed solutions.

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It would be best if you clear your concepts before you practice previous year Geometry - Circles CAT questions.

1. CAT 2024 QA Slot 2 | Geometry - Circles CAT Question

Three circles of equal radii touch (but not cross) each other externally. Two other circles, X and Y, are drawn such that both touch (but not cross) each of the three previous circles. If the radius of X is more than that of Y, the ratio of the radii of X and Y is

(a)
4 + 2√3 : 1
(b)
4 + √3 : 1
(c)
2 + √3 : 1
(d)
7 + 4√3 : 1

2. CAT 2024 QA Slot 3 | Geometry - Circles CAT Question

A circular plot of land is divided into two regions by a chord of length 10√3 meters such that the chord subtends an angle of 120° at the center. Then, the area, in square meters, of the smaller region is

(a)
$25 (\frac{4 π}{3} - \sqrt{3})$
(b)
$20 (\frac{4 π}{3} + \sqrt{3})$
(c)
$20 (\frac{4 π}{3} - \sqrt{3})$
(d)
$25 (\frac{4 π}{3} + \sqrt{3})$

3. CAT 2022 QA Slot 1 | Geometry - Circles CAT Question

All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is

(a)
$\frac{P^{2}}{2}$ - 2PR
(b)
$\frac{P^{2}}{8}$ - 2R²
(c)
$\frac{P^{2}}{8}$ - $\frac{R^{2}}{2}$
(d)
$\frac{P^{2}}{16}$ - R²

Answer: Option B

Text Explanation :

Let the longer side of the rectangle be a and shorter side by b.

In the figure given, (2R)² = a² + b²

⇒ P = 2(a + b)

⇒ $\frac{P}{2}$ = a + b

Squaring both sides.

⇒ $\frac{P^{2}}{4}$ = a² + b² + 2ab

⇒ $\frac{P^{2}}{4}$ - 4R² = 2ab

⇒ $\frac{P^{2}}{8}$ - 2R² = ab

Area of the rectangle = a × b = ab = $(\frac{P^{2}}{8} - {2R}^{2})$

Hence, option (b).

4. CAT 2022 QA Slot 3 | Geometry - Circles CAT Question

In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then, the area in sq. cm, of the overlapping region between the two circles is

(a)
16(π - 1)
(b)
32(π - 1)
(c)
16π
(d)
32π

5. CAT 2021 QA Slot 1 | Geometry - Circles CAT Question

A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC = 90°. If BC = 10 inches then the area of triangle in square inches is.

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6. CAT 2020 QA Slot 2 | Geometry - Circles CAT Question

Let C₁ and C₂ be concentric circles such that the diameter of C₁ is 2 cm longer than that of C₂. If a chord of C₁ has length 6 cm and is a tangent to C₂, then the diameter, in cm, of C₁ is

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7. CAT 2020 QA Slot 2 | Geometry - Circles CAT Question

Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 meters north of O and B is located 3 meters east of O. Then, the length of PQ, in meters, is nearest to

(a)
8.8
(b)
6.6
(c)
7.8
(d)
7.2

Answer: Option A

Text Explanation :

The following figure can be drawn from the information given in the question.

In right triangle AOB, AB = 5 (Pythagorean triplet)

Also, OM is the altitude = $\frac{3 \times 4}{5}$ = 2.4

In right triangle OMP,

OP² = OM² + MP²

⇒ 5² + (2.4)²+ MP²

⇒ MP = $\sqrt{19.24}$

⇒ PQ = 2MP = 2 × $\sqrt{19.24}$ ≈ 2 × 4.4 = 8.8.

Hence, option (a).

8. CAT 2020 QA Slot 3 | Geometry - Circles CAT Question

The vertices of a triangle are (0, 0), (4, 0) and (3, 9). The area of the circle passing through these three points is

(a)
123π/7
(b)
205π/9
(c)
14π/3
(d)
12π/5

Answer: Option B

Text Explanation :

The diagram can be drawn as follows:

Height of the given triangle = 9 and base = 4

∴ Area of the triangle = $\frac{1}{2} \times 9 \times 4$ = 18

We know area of a triangle = abc/4R

(where a, b, and c are the sides of the triangle and R is the circumradius)

a = AB = 4 units
b = BC = $\sqrt{9^{2} + 1^{2}} = \sqrt{82}$ units
c = AC = $\sqrt{3^{2} + 9^{2}} = \sqrt{90}$ units

The sides of the triangle are 4, √82 and √90

⇒ 18 = $\frac{4 \times \sqrt{82} \times \sqrt{90}}{4 R}$

⇒ R = $\frac{\sqrt{205}}{3}$

∴ Area of the circumcircle = πR² = $\frac{205 π}{9}$

Hence, option (b).

9. CAT 2019 QA Slot 1 | Geometry - Circles CAT Question

In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is

(a)
0.5
(b)
2.5
(c)
3.5
(d)
1.5

10. CAT 2019 QA Slot 1 | Geometry - Circles CAT Question

AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

(a)
7.8
(b)
8.5
(c)
9.3
(d)
9.1

11. CAT 2019 QA Slot 2 | Geometry - Circles CAT Question

Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is

(a)
π/3
(b)
1
(c)
1/√2
(d)
√2

12. CAT 2019 QA Slot 2 | Geometry - Circles CAT Question

Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

(a)
10
(b)
8√2
(c)
6√2
(d)
5

13. CAT 2018 QA Slot 1 | Geometry - Circles CAT Question

In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

(a)
√12
(b)
√13
(c)
√11
(d)
√14

Answer: Option B

Text Explanation :

OP⊥ PQ and OQ⊥ AB

In a circle perpendicular to the chord bisects the chord.
∴ CP = PD = 3 and AQ = QB = 2

Let OP = x cm

OC = OB

In ∆OPC and ∆OQB
∴ Radius = $\sqrt{P C^{2} + P O^{2}}$ = $\sqrt{Q B^{2} + Q O^{2}}$

∴ $\sqrt{3^{2} + x^{2}}$ = $\sqrt{2^{2} + {(x + 1)}^{2}}$

∴ x = 2

∴ Radius = $\sqrt{9^{2} + 2^{2}}$ = √13

Hence, option (b).

14. CAT 2018 QA Slot 1 | Geometry - Circles CAT Question

In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

(a)
(π/6)^1/2
(b)
(π/3√3)^1/2
(c)
(π/4√3)^1/2
(d)
(π/4)^1/2

Answer: Option B

Text Explanation :

Let OC = OD = x

△OCD is an equilateral triangle of side x. [All angles are 60 degree.]

A(△OCD) = $\frac{\sqrt{3}}{4} x^{2}$

A(sector O-AB) = $\frac{60}{360} \times π {(1)}^{2}$ = $\frac{π}{6}$

By the given condition,

$\frac{\sqrt{3}}{4} x^{2}$ = $\frac{1}{2} \times \frac{π}{6}$

∴ x² = $\frac{π}{3 \sqrt{3}}$

∴ x = ${(\frac{π}{3 \sqrt{3}})}^{1 / 2}$

Hence, option (b).

15. CAT 2018 QA Slot 2 | Geometry - Circles CAT Question

A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is

(a)
$5 \sqrt{3}$
(b)
$6 \sqrt{2}$
(c)
2π
(d)
8

Answer: Option A

Text Explanation :

The triangle formed by joining the endpoints of the chord of a circle with the center of the circle is an isosceles triangle because the two sides of the triangle (radii of the circle) are congruent. If the angle subtended by the chord at the center of the circle is 60 degrees, the triangle is an equilateral triangle. Therefore the side of the triangle and the radius of the circle is equal to the side of the triangle = 5 cm.

Now we have the following

OP is perpendicular to the chord AB. Therefore P is the midpoint of chord AB. Further, OA and OP are the radii of the circle. Therefore triangles OAP and OBP are congruent. Therefore triangle OAP and OBP are 30-60-90 triangles.

We have, $\cos 30 = \frac{\sqrt{3}}{2} = \frac{A P}{O A} = \frac{A P}{5}$

Therefore, $A P = \frac{5}{2} \sqrt{3}$

Therefore, $l (A B) = 2 \times l (A P) = 5 \sqrt{3}$

Hence, option (a).

16. CAT 2018 QA Slot 2 | Geometry - Circles CAT Question

On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is

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Answer: 24

Text Explanation :

Since points P and Q lie on the circumference of the circle drawn with BC as diameter, angles BPC and BQC are right angled triangles.

Now, area of triangle ABC = $\frac{1}{2} \times A B \times P C$ = $\frac{1}{2} \times A C \times B Q$

Therefore, BQ = $\frac{AB \times PC}{AC}$ = $\frac{30 \times 20}{25}$ = 24

Hence, 24.

17. CAT 2017 QA Slot 1 | Geometry - Circles CAT Question

Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq. cm, of the region enclosed by BPC and BQC is:

(a)
9π - 18
(b)
18
(c)
9π
(d)
9

Answer: Option B

Text Explanation :

AB = a (a = 6)
BC = a√2

BCQB is a semicircle of radius = half of BC = $\frac{a}{\sqrt{2}}$
Area of semicircle BCQB = ½ × π × ${(\frac{a}{\sqrt{2}})}^{2}$ = ¼ × π × a²

ACPBA is a quarter circle (quadrant) of radius a.
Area of BCPB = Area of sector ABPCA - Area of triangle ABC
= ¼ × π × a² - Area of triangle ABC

Now, Area of CPBQC = Area of semicircle CQB - Area of CPBC
= ¼ × π × a² - ( ¼ × π × a² - Area of triangle ABC)
= Area of triangle ABC

∴ Area of region enclosed by CPBQC = Area of ∆ABC = ½ × 6 × 6 = 18.

Hence, option (b).

18. CAT 2008 QA | Geometry - Circles CAT Question

Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the circle of the other. What is the area (in sq cm) of the intersecting region?

(a)
$\frac{π}{3} - \frac{\sqrt{3}}{4}$
(b)
$\frac{2 π}{3} + \frac{\sqrt{3}}{2}$
(c)
$\frac{4 π}{3} - \frac{\sqrt{3}}{2}$
(d)
$\frac{4 π}{3} + \frac{\sqrt{3}}{2}$
(e)
$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

Answer: Option E

Text Explanation :

Let O and P be the centres of the circles.

OR = OP = PR = 1cm

∴ ∆PRO is an equilateral triangle.

∴ m ∠ROP = 60°

∴ m ∠ROS = 120°

Now, area of the intersecting region = 2(area of sector O-RPS – area of ∆ORS)
= 2(area of sector O-RPS – area of ∆PRO) [area of ∆PRO = area of ∆ORS]

Area of sector O - RPS = $\frac{120}{360} (π)$ = $\frac{π}{3}$

Area of ∆PRO = $\frac{\sqrt{3}}{4} (1^{2})$

∴ Area of the intersecting region = $2 (\frac{π}{3})$ - $2 (\frac{\sqrt{3}}{4})$ = $\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

Hence, option (e).

19. CAT 2007 QA | Geometry - Circles CAT Question

Two circles with centres P and Q cut each other at two distinct points A and B. The circles have the same radii and neither P nor Q falls within the intersection of the circles. What is the smallest range that includes all possible values of the angle AQP in degrees?

(a)
Between 0 and 90
(b)
Between 0 and 30
(c)
Between 0 and 60
(d)
Between 0 and 75
(e)
Between 0 and 45

20. CAT 2006 QA | Geometry - Circles CAT Question

A semicircle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semicircle at D. Given that AC = 2 cm and CD = 6 cm, the area of the semicircle (in sq. cm.) will be:

(a)
32π
(b)
50π
(c)
40.5π
(d)
81π
(e)
undeterminable

21. CAT 2005 QA | Geometry - Circles CAT Question

Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is

(a)
$\frac{π}{4}$
(b)
$\frac{π}{2} - 1$
(c)
$\frac{π}{5}$
(d)
$\sqrt{2} - 1$

Answer: Option B

Text Explanation :

Let the two circles with centres P and Q intersect at M and N.

Quadrilateral PMQN is a square.

m ∠MPN = m ∠MQN = 90°

The area common to both the circles = 2(Area of sector P-MN – Area of ∆PMN)

$= 2 [(\frac{90}{360} \times π \times 1^{2}) - (\frac{1}{2} \times 1^{2})]$

$= \frac{π}{2} - 1$

Hence, option (b).

22. CAT 2005 QA | Geometry - Circles CAT Question

What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm?

(a)
1 or 7
(b)
2 or 14
(c)
3 or 21
(d)
4 or 28

23. CAT 2005 QA | Geometry - Circles CAT Question

Four points A, B, C and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

(a)
$3 \sqrt{2}$
(b)
1 + π
(c)
$\frac{4 π}{3}$
(d)
5

Answer: Option B

Text Explanation :

The ant will not go into the circles with centers B and C and radius = 1 m

The minimum distance that the ant has to traverse = the distance of the path A-H-G-D

HG = 1 m

AH = GD = $\frac{1}{4}$ × Circumference of circle = $\frac{π}{2}$

AH + GD = π m

∴ The ant must traverse (1 + π)m.

Hence, option (b).

24. CAT 2005 QA | Geometry - Circles CAT Question

In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE : EB = 1 : 2, and DF is perpendicular to MN such that NL : LM = 1 : 2. The length of DH in cm is

(a)
$2 \sqrt{2} - 1$
(b)
$\frac{(2 \sqrt{2} - 1)}{2}$
(c)
$\frac{(3 \sqrt{2} - 1)}{2}$
(d)
$\frac{(2 \sqrt{2} - 1)}{3}$

Answer: Option B

Text Explanation :

AO = OD = 1.5 cm

AE + EB = 3 cm and AE : EB = 1 : 2

∴ AE = 1 cm and EB = 2 cm

OE = AO – AE = 1.5 – 1 = 0.5 cm

Similarly, NL = 1 cm, LM = 2 cm, and OL = 0.5 cm

OEHL is a square as all its angles are right angles and OE = OL

∴ EH = HL = 0.5 cm

In ∆ODL, OD² = OL² + DL²

1.5² = 0.5² + (0.5 + DH)²

2.25 = 0.25 + 0.25 + DH + DH²

DH² + DH – 1.75 = 0

DH = $\frac{- 1 \pm \sqrt{1 - 4 (- 1.75)}}{2}$ = $\frac{2 \sqrt{2} - 1}{2}$ (DH > 0)

Hence, option (b).

25. CAT 2005 QA | Geometry - Circles CAT Question

P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR?

(a)
2r(1 + $\sqrt{3}$ )
(b)
2r(2 + $\sqrt{3}$ )
(c)
r(1 + $\sqrt{5}$ )
(d)
2r + $\sqrt{3}$

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Geometry - Circles - Previous Year CAT/MBA Questions

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