Circles, Chords, Tangents and Incircles Previous Year Questions (PYQs) for CAT: 9+ Solved Questions with Step-by-Step Solutions

Solve 9+ Circles, Chords, Tangents and Incircles previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Circles, Chords, Tangents and Incircles

CAT QA Geometry

Circles, Chords, Tangents and Incircles

A 3-step journey from measuring a circle to solving full exam geometry.

chapter PYQs

①

🧩 t1 — Chords, Arcs and Circular Segments

You learn how chord length, distance from centre, central angle, sector area, and segment area talk to each other.

2 own-course PYQs Importance: moderate Master: circle measurement

②

📐 t2 — Tangents and Circle Contact Geometry

You move from inside chords to outside touching lines and contact-based angle geometry.

2 own-course PYQs Importance: moderate

③

⭕ t3 — Cyclic Figures, Incircles and Circumcircles

You combine circles with triangles, rectangles, quadrilaterals, incircles, and circumcircles.

6 own-course PYQs Highest chapter weight

End goal: by the end of this chapter, you should see a circle question and quickly decide whether it is about measurement inside the circle, touch/contact outside the circle, or circle mixed with polygons.

Topic Hero: Chords, Arcs and Circular Segments

Selected Topic

Chords, Arcs and Circular Segments

A chord cuts the circle. The arc bends above it. The segment is the curved slice between them.

CAT skill: area + angle 2 direct PYQs Moderate frequency

One-line hook: Most chord-and-segment problems become easy once you draw the radius to the chord’s midpoint.

Circles, Chords, Tangents and Incircles: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability NAT

Two tangents drawn from a point P and a circle with center O at point Q and R. Point A and B lie on PQ and PR, repectively, such that AB is also a tangent to the same circle. If $∠ A O B = 5 0^{\circ}$ , then $∠ A P B$ , in degrees equals

Correct Answer:

Step-by-Step Solution

Let the circle be tangent to $P Q$ at $Q$ , $P R$ at $R$ , and $A B$ at $C$ . Tangents from an external point to a circle are equal in length, and the line from the center to the external point bisects the angle between the tangents. Thus, $O A$ bisects $∠ QO C$ and $O B$ bisects $∠ R O C$ . The angle $∠ A O B = ∠ A O C + ∠ B O C = \frac{1}{2} ∠ QO C + \frac{1}{2} ∠ R O C = \frac{1}{2} ∠ QO R$ . Given $∠ A O B = 5 0^{\circ}$ , we have $\frac{1}{2} ∠ QO R = 5 0^{\circ}$ , so $∠ QO R = 10 0^{\circ}$ . In the quadrilateral $P QO R$ , the angles at $Q$ and $R$ are $9 0^{\circ}$ because the radius is perpendicular to the tangent at the point of contact. The sum of angles in a quadrilateral is $36 0^{\circ}$ , so $∠ QP R + ∠ QO R = 18 0^{\circ}$ . Therefore, $∠ A P B = ∠ QP R = 18 0^{\circ} - 10 0^{\circ} = 8 0^{\circ}$ .

Question 2 · Quantitative Ability MCQ

A quadrilateral ABCD is inscribed in a circle such that AB : CD = 2 : 1 and BC : AD = 5 : 4. If AC and BD intersect at the point E, then AE : CE equals

A.
2 : 1
B.
1 : 2
C.
8 : 5
D.
5 : 8

Correct Answer:

Step-by-Step Solution

This is a cyclic-quadrilateral diagonal-ratio question. You know this because four points (A, B, C, D) lie on one circle, the two diagonals AC and BD cross at E, and you are given side ratios and asked for a diagonal-segment ratio.

Why this method applies: whenever two chords of a circle cross, the inscribed angle theorem (angles on the same arc are equal) forces two pairs of triangles at the crossing point to be similar. That similarity is what turns side ratios into diagonal-segment ratios.

Step 1 — First similar pair.

Look at triangle ABE and triangle DCE.

$∠ B A E = ∠ B A C = ∠ B D C = ∠ C D E$ (both angles stand on arc BC).

$∠ A B E = ∠ A B D = ∠ A C D = ∠ D C E$ (both angles stand on arc AD).

So $△ A B E \sim △ D C E$ , which gives:

$\frac{A E}{D E} = \frac{B E}{C E} = \frac{A B}{D C} = \frac{2}{1}$

Step 2 — Second similar pair.

Now look at triangle ADE and triangle BCE.

$∠ D A E = ∠ D A C = ∠ D B C = ∠ C B E$ (both stand on arc DC).

$∠ A D E = ∠ A D B = ∠ A C B = ∠ B C E$ (both stand on arc AB).

So $△ A D E \sim △ B C E$ , which gives:

$\frac{A E}{B E} = \frac{D E}{C E} = \frac{A D}{B C} = \frac{4}{5}$

Step 3 — Combine the two results.

From Step 1: $D E = \frac{A E}{2}$ .

From Step 2: $\frac{D E}{C E} = \frac{4}{5} \Rightarrow C E = \frac{5}{4} D E = \frac{5}{4} \cdot \frac{A E}{2} = \frac{5 A E}{8}$ .

Step 4 — Final ratio.

$\frac{A E}{C E} = \frac{A E}{\frac{5 A E}{8}} = \frac{8}{5}$

So $A E : C E = 8 : 5$ , which is option C.

Common trap: it is tempting to think $△ A B E \sim △ C D E$ directly gives $A E : C E = A B : D C = 2 : 1$ . That mixes up the two different similar-triangle pairs — $A E$ and $C E$ are NOT corresponding sides of the same similar pair, so you must combine both similarity relations as done above, not read the ratio off just one of them.

Question 3 · Quantitative Ability NAT

ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of $△ A D E$ is

Correct Answer:

Step-by-Step Solution

This is an incircle-of-a-triangle-made-inside-a-rectangle question. You can recognise the

pattern because a triangle is formed by two full sides of a rectangle and a segment to a

midpoint on the third side, and the inradius of that triangle is asked. The key is that

this triangle is right-angled because it uses two adjacent sides of the rectangle.

Step 1: Set up coordinates using the rectangle.

Let A = (0,0), B = (56,0), C = (56,45), D = (0,45), matching AB = 56 and BC = 45.

E is the midpoint of CD. Since C = (56,45) and D = (0,45), E = (28,45).

Step 2: Identify the shape of triangle ADE.

AD goes from (0,0) to (0,45): a vertical side of the rectangle, length 45.

DE goes from (0,45) to (28,45): a horizontal segment, length 28.

Since AD is vertical and DE is horizontal, they are perpendicular, so triangle ADE is

right-angled at D, with legs AD = 45 and DE = 28.

Step 3: Find the hypotenuse AE.

AE = sqrt(AD^2 + DE^2) = sqrt(45^2 + 28^2) = sqrt(2025 + 784) = sqrt(2809) = 53.

Step 4: Use the right-triangle inradius shortcut.

For a right triangle with legs a and b and hypotenuse c, the inradius is:

r = (a + b - c) / 2

r = (45 + 28 - 53) / 2 = 20 / 2 = 10

Answer: the inradius of triangle ADE is 10 cm.

Common trap: forgetting that DE is only half of CD's full opposite side (since E is a

midpoint, not an endpoint) leads to using DE = 56 instead of 28, which would give a

completely wrong triangle. Always re-check which side the midpoint actually splits before

assigning lengths.

Question 4 · Quantitative Ability MCQ

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.
$\frac{P ^{2}}{16} - R^{2}$
B.
$\frac{P ^{2}}{8} - 2 R^{2}$
C.
$\frac{P ^{2}}{2} - 2 P R$
D.
$\frac{P ^{2}}{8} - \frac{R ^{2}}{2}$

Correct Answer:

Step-by-Step Solution

This is a rectangle-in-a-circle question. The trigger is that all four vertices of the rectangle lie on the circle, which immediately means the rectangle's diagonal is a diameter of the circle — this is the key fact that connects the rectangle's dimensions to the radius.

Step 1: Recall why the diagonal equals the diameter. A rectangle's diagonal always subtends a 90° angle in a way that matches Thales' theorem: since all four vertices lie on the circle and the rectangle's angles are 90°, each diagonal must pass through the centre, making it a diameter.

$Diagonal = 2 R$

Step 2: Let the rectangle's sides be $x$ and $y$ . Since the perimeter is $P$ :

$2 (x + y) = P \Rightarrow x + y = \frac{P}{2}$

Step 3: Use the Pythagorean theorem on the diagonal:

$x^{2} + y^{2} = (2 R)^{2} = 4 R^{2}$

Step 4: We want the area, $x y$ . Use the identity:

$(x + y)^{2} = x^{2} + y^{2} + 2 x y$

Step 5: Substitute the known values:

$(\frac{P}{2})^{2} = 4 R^{2} + 2 x y$

$\frac{P ^{2}}{4} = 4 R^{2} + 2 x y$

Step 6: Solve for $x y$ :

$2 x y = \frac{P ^{2}}{4} - 4 R^{2}$

$x y = \frac{P ^{2}}{8} - 2 R^{2}$

Answer: the area of the rectangle is $\frac{P ^{2}}{8} - 2 R^{2}$ , option B.

Common trap: students often forget the factor of 2 when converting $2 (x + y) = P$ into $x + y$ , or forget to divide the final expression by 2 after isolating $2 x y$ , both of which shift the constants in the answer and lead to picking one of the other options.

Question 5 · Quantitative Ability MCQ

In a circle with center C and radius $62$ cm, PQ and SR are two parallel chords separated by one of the diameters. If $∠ P QC = 4 5^{\circ}$ , and the ratio of the perpendicular distance of PQ and SR from C is 3 : 2, then the area, in sq. cm, of the quadrilateral PQRS is

A.
$4 (3 + 14)$
B.
$4 (32 + 7)$
C.
$20 (3 + 14)$
D.
$20 (32 + 7)$

Correct Answer:

Step-by-Step Solution

This is a parallel chords on opposite sides of the centre question. You can recognise the

pattern because two chords, PQ and SR, are described as parallel and "separated by one of

the diameters" (meaning they sit on opposite sides of the centre), and the question asks

for the area of the quadrilateral PQRS formed by their endpoints. This always needs: the

distance of each chord from the centre, the length of each chord, and the distance between

the two chords.

Step 1: Use the 45 degree angle to find the distance from C to PQ.

Drop a perpendicular from C to PQ, meeting it at M. Then CM is perpendicular to PQ, so

triangle CMQ is right-angled at M, and angle CQM equals angle PQC, which is 45 degrees. In

a right triangle with a 45 degree angle, the two legs are equal, so CM = MQ.

By Pythagoras in triangle CMQ: CM^2 + MQ^2 = CQ^2 = (6*sqrt(2))^2 = 72.

Since CM = MQ, 2*CM^2 = 72, so CM^2 = 36 and CM = 6.

So the distance from C to PQ is d1 = 6.

Step 2: Use the given ratio to find the distance from C to SR.

The ratio of the two perpendicular distances is d1 : d2 = 3 : 2, so

d2 = (2/3) * 6 = 4.

Step 3: Find the half-lengths of both chords using the chord-distance formula.

Half-length of a chord = sqrt(r^2 - d^2), where r = 6*sqrt(2), so r^2 = 72.

For PQ: half-length = sqrt(72 - 36) = sqrt(36) = 6, so PQ = 12.

For SR: half-length = sqrt(72 - 16) = sqrt(56) = 2sqrt(14), so SR = 4sqrt(14).

Step 4: Find the height of trapezium PQRS.

Because PQ and SR lie on opposite sides of the centre (separated by a diameter), the

distance between the two chords equals the sum of their individual distances from the

centre: height h = d1 + d2 = 6 + 4 = 10.

Step 5: Compute the area of the trapezium.

PQRS is a trapezium with parallel sides PQ and SR and height h.

Area = (1/2) (PQ + SR) h = (1/2) (12 + 4sqrt(14)) * 10

= 5 (12 + 4sqrt(14)) = 60 + 20sqrt(14) = 20(3 + sqrt(14))

Answer: 20(3 + sqrt(14)) square cm, which is option C.

Common trap: students often add the two distances only when the chords lie on the SAME

side of the centre (where the actual gap would be d1 - d2). Here the chords are explicitly

on opposite sides of a diameter, so the distances must be added, not subtracted.

Circles, Chords, Tangents and Incircles Previous Year Questions (PYQs) for CAT: 9+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Circles, Chords, Tangents and Incircles

Circles, Chords, Tangents and Incircles

🧩 t1 — Chords, Arcs and Circular Segments

📐 t2 — Tangents and Circle Contact Geometry

⭕ t3 — Cyclic Figures, Incircles and Circumcircles

Topic Hero: Chords, Arcs and Circular Segments

Chords, Arcs and Circular Segments

Circles, Chords, Tangents and Incircles: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

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Step-by-Step Solution

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