Case Study Questions on Chapter 10 'Circles' for Class 10 with Answers

This collection of case-study questions for Chapter 10: Circles Class 10 follows CBSE and NCERT guidelines and presents short, exam-style scenarios designed to strengthen geometric reasoning, visualisation, and step-by-step problem solving. The set includes questions on identifying the correct theorems, applying logical steps, solving multi-step problems, and using circle theorems together with coordinate geometry and length calculations. Answers are provided with clear reasoning, diagram guidance, and concise steps to help learners develop procedural fluency and confidence for board examinations. The downloadable PDF includes additional practice sets for quick revision and classroom worksheets to improve speed and accuracy.

Solved Circles Case Study Questions and Answers

CBSE's case-based questions usually give you a short passage describing a real situation, sometimes with a small figure, followed by 4 - 5 questions of increasing difficulty, a couple of MCQs to check basic understanding, a short-answer question that needs a calculation, and a longer question that asks you to interpret or justify your answer.

Key Concepts You Must Know Before Attempting Case Studies

• Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

• Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

• Tangent to a Circle: A line that touches the circle at exactly one point. That point is called the point of tangency or point of contact.

• Number of Tangents from a Point:

• A point inside the circle → 0 tangents possible

• A point on the circle → exactly 1 tangent

• A point outside the circle → exactly 2 tangents

• From Theorem 10.2: If PA and PB are two tangents from an external point P to a circle with centre O, then PA = PB. Also, OP bisects the angle APB, and OA ⊥ PA, OB ⊥ PB.

• Common Tangents between Two Circles:

• Two circles that don't intersect and one is outside the other → 4 common tangents (2 direct, 2 transverse)

• Two circles that touch externally → 3 common tangents

• Two circles that intersect at two points → 2 common tangents

• Two circles that touch internally → 1 common tangent

• One circle inside another (no touching) → 0 common tangents

Case Study 1: The Well and the Tree

Read the following situation carefully and answer the questions that follow:

A farmer has a circular well in his field with centre O and radius 7 m. He wants to set up a fence from a point P, which is 25 m away from the centre of the well, to the edge of the well such that the fence touches the well at exactly one point. He plans to build two such fences from the same point P, one on each side of the well.

(i) What is the length of each fence (tangent) from point P to the well?

(ii) Are the two fences equal in length?

(iii) What angle does the fence make with the line joining P to the centre O?

(iv) If OP = 25 m and radius OA = 7 m, what is the length PA?

(v) What is the name of the triangle formed by O, P, and the point of tangency A?

Answers:

(i) and (iv) Since the fence is a tangent to the circle, OA ⊥ PA (by Theorem 10.1).

Triangle OAP is right-angled at A.

Using Pythagoras theorem:

PA² = OP² − OA²

PA² = 25² − 7²

PA² = 625 − 49 = 576

PA = 24 m

Each fence is 24 metres long.

(ii) Yes. By Theorem 10.2, the lengths of tangents drawn from an external point are always equal. So both fences are 24 m long.

(iii) By Theorem 10.1, the radius to the point of contact is always perpendicular to the tangent. So the angle between the fence and line OP (i.e., angle OAP) = 90°.

(v) Triangle OAP is a right-angled triangle, with the right angle at A.

Case Study 2: Two Circles Touching Each Other

Read the following situation carefully and answer the questions that follow:

Two circular parks in a city have centres A and B with radii 6 cm and 4 cm respectively (shown in scale model). The two circles touch each other externally at a point K. From an external point P, a common tangent PQ is drawn touching the larger circle at Q and the smaller circle at a point Y. PK is a line drawn from P through the point of tangency K.

(i) What is the distance AB between the two centres?

(ii) How many common tangents can be drawn to two circles that touch each other externally?

(iii) Since PQ is tangent to the larger circle at Q, what is the relationship between AQ and PQ?

(iv) If PA = 10 cm (distance from P to centre A), find PQ.

Answers:

(i) When two circles touch each other externally, the distance between their centres equals the sum of their radii.

AB = 6 + 4 = 10 cm

(ii) When two circles touch externally, exactly 3 common tangents can be drawn: two direct common tangents and one transverse common tangent at the point of contact.

(iii) AQ ⊥ PQ, because the radius to the point of tangency is always perpendicular to the tangent (Theorem 10.1).

(iv) In triangle PAQ, angle AQP = 90°, AQ = 6 cm (radius), PA = 10 cm.

PQ² = PA² − AQ²

PQ² = 100 − 36 = 64

PQ = 8 cm

Case Study 3: Watch Dial and Tangent Lines

Read the following situation carefully and answer the questions that follow:

A designer is making a decorative circular clock with centre O and radius 5 cm. She wants to draw two straight decorative lines from a point P outside the clock face, each line touching the clock face at exactly one point. The point P is 13 cm away from the centre O. The two points of tangency are A and B.

(i) Find the length of each tangent PA and PB.

(ii) Show that ∠OAP = 90°.

(iii) What is the length OP?

(iv) Are PA and PB equal? Which theorem confirms this?

(v) If ∠APB = 60°, find angle AOB.

Answers:

(i) In right triangle OAP (right angle at A):

PA² = OP² − OA²

PA² = 13² − 5²

PA² = 169 − 25 = 144

PA = 12 cm

Since tangents from external point are equal, PB = 12 cm as well.

(ii) OA is the radius at the point of contact A. By Theorem 10.1, OA ⊥ PA. Therefore, ∠OAP = 90°.

(iii) OP = 13 cm (given).

(iv) Yes, PA = PB = 12 cm. Theorem 10.2 confirms that the lengths of the two tangents drawn from an external point to a circle are always equal.

(v) Since OA ⊥ PA and OB ⊥ PB, the quadrilateral OAPB has angles:

∠ OAP = 90°, ∠OBP = 90°

Sum of all angles in quadrilateral = 360°

∠AOB + ∠APB = 360° − 90° − 90° = 180°

∠AOB = 180° − 60° = 120°

Questions PDF with worked-out examples for Class 10 Chapter 10: Circles, perfect for last-minute CBSE exam revision.

Class 10 Chapter 10: Circle Case Study PDF