ICSE Class 10 Maths Circles Solutions | Selina Ch 17
ICSE Class 10 Maths Circles solutions: what this page covers
ICSE Class 10 Maths Circles in Selina Concise Mathematics Chapter 17 is mainly an angle-chasing chapter: you identify the correct circle theorem, write the angle relation, and then complete the calculation using triangle or quadrilateral angle sums.
This text-only study and solutions page for the Circles chapter explains the theorem set first, then gives worked solutions for the figure-based problems covered here, followed by extra teacher-made examples, an answer index and common mistakes. Diagrams are not reproduced here, so each solution states the relevant figure condition before using it.
Concise Mathematics Selina Solutions Class 10 ICSE Chapter 17 Circles: theorem table
Most questions in this chapter are solved by choosing one theorem and combining it with a triangle angle sum. Keep this table beside you while solving the Selina Circles exercises.
| Situation in the figure | Theorem to use | How it is written in the solution |
|---|---|---|
| A chord or arc subtends an angle at the centre and at the circumference | Angle at the centre is twice the angle at the circumference on the same arc | \angle AOB=2\angle ACB |
| Two angles stand on the same chord and lie in the same segment | Angles in the same segment are equal | \angle ADB=\angle ACB |
| An angle stands on a diameter | Angle in a semicircle is a right angle | \angle APB=90^\circ |
| Four points lie on a circle | Opposite angles of a cyclic quadrilateral are supplementary | \angle A+\angle C=180^\circ |
| A radius triangle is formed, such as OA=OB | Radii of the same circle are equal, so the triangle is isosceles | \angle OAB=\angle OBA |
| Two parallel lines cut a transversal | Alternate interior angles are equal | \angle CAB=\angle ACD |
Use internal revision pages such as ICSE Class 10 solutions, Selina Concise Maths Class 10 solutions, and ICSE Class 10 question papers for connected practice after finishing this chapter.
Concept snapshot for circle angle chasing
Think of a chord as a stage and the angles standing on it as spectators looking at the same stage. If two spectators stand on the same side of the chord on the circle, they see the chord under equal angles. If the centre O looks at the same chord, the view is wider, so the central angle is twice the angle at the circumference.
This is why a problem may begin with a central angle, move to a circumference angle, and then finish inside an ordinary triangle.
Worked examples before the exercise solutions
These original examples are not copied from a textbook exercise. They show the exact style of reasoning needed for ICSE Class 10 Maths Circles.
Worked Example 1: Central angle and circumference angle
In a circle with centre O, chord AB subtends \angle AOB=112^\circ at the centre. Point C lies on the remaining circumference. Find \angle ACB.
Step 1: Use the theorem: the angle at the centre is twice the angle at the circumference on the same arc.
\angle AOB=2\angle ACB
Step 2: Substitute \angle AOB=112^\circ.
112^\circ=2\angle ACB
Step 3: Divide by 2.
\angle ACB=\frac{112^\circ}{2}=56^\circ
Final answer: \angle ACB=56^\circ.
Worked Example 2: Cyclic quadrilateral angle
In cyclic quadrilateral ABCD, \angle A=74^\circ. Find \angle C.
Step 1: Opposite angles of a cyclic quadrilateral are supplementary.
\angle A+\angle C=180^\circ
Step 2: Substitute \angle A=74^\circ.
74^\circ+\angle C=180^\circ
Step 3: Subtract 74^\circ from 180^\circ.
\angle C=180^\circ-74^\circ=106^\circ
Final answer: \angle C=106^\circ.
Worked Example 3: Diameter and triangle angle sum
AB is a diameter of a circle and P is a point on the circle. If \angle PAB=38^\circ, find \angle PBA.
Step 1: Since AB is a diameter, the angle in the semicircle is a right angle.
\angle APB=90^\circ
Step 2: Use the angle sum of triangle APB.
\angle PAB+\angle APB+\angle PBA=180^\circ
Step 3: Substitute the known values.
38^\circ+90^\circ+\angle PBA=180^\circ
Step 4: Solve for \angle PBA.
\angle PBA=180^\circ-128^\circ=52^\circ
Final answer: \angle PBA=52^\circ.
Exercise 17(A) figure-based solutions
The following worked solutions use the figure conditions stated with each question. Where a figure-based statement depends on a visible diameter or collinearity, that condition is named at the start of the solution so that the reasoning is clear.
Question 1(a): Find \angle A when \angle B=55^\circ
Step 1: In triangle OCB, OB=OC because both are radii of the same circle.
Step 2: Therefore, triangle OCB is isosceles and the base angles are equal.
\angle OBC=\angle OCB=55^\circ
Step 3: Use the angle sum of triangle OCB.
\angle BOC+55^\circ+55^\circ=180^\circ
\angle BOC=180^\circ-110^\circ=70^\circ
Step 4: The angle at the centre is twice the angle at the circumference on the same arc.
\angle BOC=2\angle A
\angle A=\frac{70^\circ}{2}=35^\circ
Final answer: \angle A=35^\circ. The correct option is 35^\circ.
Question 1(b): Find \angle P when \angle OBA=50^\circ
Step 1: In triangle OAB, OA=OB, so \triangle OAB is isosceles.
\angle OAB=\angle OBA=50^\circ
Step 2: Use the angle sum of triangle OAB.
\angle AOB+50^\circ+50^\circ=180^\circ
\angle AOB=80^\circ
Step 3: The circumference angle on the same arc is half the central angle.
\angle P=\frac{1}{2}\angle AOB=\frac{80^\circ}{2}=40^\circ
Final answer: \angle P=40^\circ. The correct option is 40^\circ.
Question 1(c): Find \angle PAB when AB=PB and \angle C=50^\circ
Step 1: Angles standing on the same chord AB in the same segment are equal.
\angle APB=\angle ACB=50^\circ
Step 2: In triangle PAB, PB=AB. Therefore, the angles opposite these equal sides are equal.
\angle PAB=\angle APB
\angle PAB=50^\circ
Final answer: \angle PAB=50^\circ. The correct option is 50^\circ.
Question 1(d): Show the relation between C, B, and D
Step 1: The figure shows AC and AD as diameters of the two intersecting circles.
Step 2: An angle in a semicircle is a right angle.
\angle ABC=90^\circ \quad \text{and} \quad \angle ABD=90^\circ
Step 3: Add the adjacent angles at B.
\angle ABC+\angle ABD=90^\circ+90^\circ=180^\circ
Step 4: Since the two adjacent angles form a straight angle, the points lie on one straight line.
Final answer: C, B, and D are collinear. The correct option is that C, B, and D are collinear.
Question 1(e): Find \angle DAB when AB\parallel DC and \angle ACD=32^\circ
Step 1: In the standard figure for this question, DC is a diameter. Therefore, the angle in the semicircle is a right angle.
\angle DAC=90^\circ
Step 2: Since AB\parallel DC, alternate interior angles are equal.
\angle CAB=\angle ACD=32^\circ
Step 3: Add the two adjacent angles at A.
\angle DAB=\angle DAC+\angle CAB
\angle DAB=90^\circ+32^\circ=122^\circ
Final answer: \angle DAB=122^\circ. The correct option is 122^\circ.
Question 2: Prove that AC is a diameter and find \angle ACB
Step 1: In triangle ABD, use the given values \angle BAD=65^\circ and \angle ABD=70^\circ.
\angle BAD+\angle ABD+\angle ADB=180^\circ
65^\circ+70^\circ+\angle ADB=180^\circ
\angle ADB=45^\circ
Step 2: The figure gives \angle BDC=45^\circ. Hence,
\angle ADC=\angle ADB+\angle BDC=45^\circ+45^\circ=90^\circ
Step 3: If an angle standing on a chord is a right angle, the chord is a diameter. Therefore, AC is a diameter.
Step 4: Angles standing on the same chord AB are equal.
\angle ACB=\angle ADB=45^\circ
Final answer: AC is a diameter and \angle ACB=45^\circ.
Question 3(i): Find a
Step 1: The figure shows an angle in a semicircle, so
\angle BAD=90^\circ
Step 2: In triangle ABD, the other given angle is \angle DBA=35^\circ.
\angle ADB+90^\circ+35^\circ=180^\circ
\angle ADB=55^\circ
Step 3: The marked angle a stands on the same chord as \angle ADB, so the two angles are equal.
Final answer: a=55^\circ.
Question 3(ii): Find b
Step 1: Let E be the point where the two chords or secants intersect. Since AC is a straight line, adjacent angles at E are supplementary.
\angle AEB+\angle BEC=180^\circ
120^\circ+\angle BEC=180^\circ
\angle BEC=60^\circ
Step 2: In triangle BEC, the given angle is \angle CBE=25^\circ.
\angle BEC+\angle CBE+\angle ECB=180^\circ
60^\circ+25^\circ+\angle ECB=180^\circ
\angle ECB=95^\circ
Step 3: Since E lies on AC, \angle ECB=\angle ACB. The angle marked b is the corresponding same-segment angle.
Final answer: b=95^\circ.
Question 3(iii): Find c
Step 1: The given circumference angle is \angle ACB=50^\circ.
Step 2: The angle at the centre on the same arc is twice the circumference angle.
\angle AOB=2\angle ACB=2\times 50^\circ=100^\circ
Step 3: Since OA=OB, triangle OAB is isosceles. Hence, \angle OAB=\angle OBA=c.
c+100^\circ+c=180^\circ
2c=80^\circ
c=40^\circ
Final answer: c=40^\circ.
Question 3(iv): Find d
Step 1: Since AB is a diameter in the figure, the angle in the semicircle is
\angle APB=90^\circ
Step 2: In triangle APB, \angle PBA=45^\circ.
\angle PAB+90^\circ+45^\circ=180^\circ
\angle PAB=45^\circ
Step 3: The angle marked d stands on the same chord as \angle PAB, so it is equal to 45^\circ.
Final answer: d=45^\circ.
Question 4: Calculate \angle CDB, \angle ABC, and \angle ACB
Step 1: Angles standing on the same chord CB are equal.
\angle CDB=\angle CAB=49^\circ
Step 2: Angles standing on the same chord AC are equal.
\angle ABC=\angle ADC=43^\circ
Step 3: Use the angle sum of triangle ABC.
\angle ABC+\angle ACB+\angle BAC=180^\circ
43^\circ+\angle ACB+49^\circ=180^\circ
\angle ACB=88^\circ
Final answer: \angle CDB=49^\circ, \angle ABC=43^\circ, and \angle ACB=88^\circ.
Question 5: Find \angle DAB+\angle ABD
Step 1: In triangle ABC, use the given angles \angle CAB=75^\circ and \angle CBA=50^\circ.
\angle ACB+75^\circ+50^\circ=180^\circ
\angle ACB=55^\circ
Step 2: Angles standing on the same chord AB are equal.
\angle ADB=\angle ACB=55^\circ
Step 3: In triangle ABD, use the angle sum.
\angle DAB+\angle ABD+\angle ADB=180^\circ
\angle DAB+\angle ABD+55^\circ=180^\circ
\angle DAB+\angle ABD=125^\circ
Final answer: \angle DAB+\angle ABD=125^\circ.
Question 6: Find \angle BDC when AOB is a diameter and \angle AOC=110^\circ
Step 1: Join AD. The angle at the centre is twice the angle at the circumference on arc AC.
\angle ADC=\frac{1}{2}\angle AOC=\frac{110^\circ}{2}=55^\circ
Step 2: Since AB is a diameter, the angle in the semicircle is a right angle.
\angle ADB=90^\circ
Step 3: From the figure, \angle BDC is the difference between \angle BDA and \angle CDA.
\angle BDC=90^\circ-55^\circ=35^\circ
Final answer: \angle BDC=35^\circ.
Question 7: Find \angle OBC when \angle AOB=60^\circ and \angle BDC=100^\circ
Step 1: The angle at the circumference on arc AB is half the central angle.
\angle ACB=\frac{1}{2}\angle AOB=\frac{60^\circ}{2}=30^\circ
Step 2: In the given figure, the angle at C used in triangle BDC is 30^\circ.
\angle DCB=30^\circ
Step 3: Use the angle sum of triangle BDC.
\angle BDC+\angle DCB+\angle CBD=180^\circ
100^\circ+30^\circ+\angle CBD=180^\circ
\angle CBD=50^\circ
Step 4: The required angle \angle OBC coincides with \angle CBD in the figure.
Final answer: \angle OBC=50^\circ.
Question 8: In cyclic quadrilateral ABCD, find \angle DBC, \angle DCB, and \angle CAB
Step 1: Given \angle DAC=27^\circ. Angles standing on the same chord DC are equal.
\angle DBC=\angle DAC=27^\circ
Step 2: Given \angle DBA=50^\circ. Angles standing on the same chord DA are equal.
\angle DCA=50^\circ
Step 3: Given \angle ADB=33^\circ. Angles standing on the same chord AB are equal.
\angle ACB=33^\circ
Step 4: Add the two parts of angle DCB.
\angle DCB=\angle DCA+\angle ACB=50^\circ+33^\circ=83^\circ
Step 5: Opposite angles of a cyclic quadrilateral are supplementary.
\angle DAB+\angle DCB=180^\circ
Step 6: Since \angle DAB=\angle DAC+\angle CAB, substitute the values.
27^\circ+\angle CAB+83^\circ=180^\circ
\angle CAB=70^\circ
Final answer: \angle DBC=27^\circ, \angle DCB=83^\circ, and \angle CAB=70^\circ.
Question 9: Find \angle DCE and \angle ABC
Step 1: CD is a diameter because it passes through the centre O. Therefore, the angle in the semicircle is
\angle CED=90^\circ
Step 2: In triangle CED, \angle CDE=40^\circ.
\angle CED+\angle CDE+\angle DCE=180^\circ
90^\circ+40^\circ+\angle DCE=180^\circ
\angle DCE=50^\circ
Step 3: In triangle BOC, \angle AOC acts as an exterior angle at O.
\angle AOC=\angle OCB+\angle OBC
Step 4: The figure gives \angle OCB=\angle DCE=50^\circ.
80^\circ=50^\circ+\angle OBC
\angle OBC=30^\circ
Step 5: Since A, O, and B are collinear, \angle ABC=\angle OBC.
Final answer: \angle DCE=50^\circ and \angle ABC=30^\circ.
Question 10: Find \angle AEB when AB\parallel CD and \angle ADC=25^\circ
Step 1: In the standard figure, CD is a diameter. Hence, angles standing on CD are right angles.
\angle CAD=90^\circ \quad \text{and} \quad \angle CBD=90^\circ
Step 2: Since AB\parallel CD, alternate interior angles are equal.
\angle BAD=\angle ADC=25^\circ
Step 3: Add the two parts of \angle BAC.
\angle BAC=\angle BAD+\angle DAC=25^\circ+90^\circ=115^\circ
Step 4: In cyclic quadrilateral ACDB, opposite angles are supplementary.
\angle CDB+\angle BAC=180^\circ
Step 5: Since \angle CDB=\angle CDA+\angle ADB,
25^\circ+\angle ADB+115^\circ=180^\circ
\angle ADB=40^\circ
Step 6: Angles standing on the same chord AB are equal.
\angle AEB=\angle ADB=40^\circ
Final answer: \angle AEB=40^\circ.
Question 11: AB is a diameter; find \angle PRB, \angle PBR, and \angle BPR
Step 1: The figure gives \angle PAB=35^\circ. Angles standing on chord PB are equal.
\angle PRB=\angle PAB=35^\circ
Step 2: Since AB is a diameter,
\angle BPA=90^\circ
Step 3: Points A, P, and Q are collinear, so \angle BPA and \angle BPQ form a linear pair.
\angle BPQ=180^\circ-90^\circ=90^\circ
Step 4: In the figure, R, B, and Q are collinear and \angle BQP=25^\circ. Therefore, \angle PBR is the exterior angle of triangle BPQ.
\angle PBR=\angle BPQ+\angle BQP=90^\circ+25^\circ=115^\circ
Step 5: In triangle BPR, use the angle sum.
\angle BPR+\angle PRB+\angle PBR=180^\circ
\angle BPR+35^\circ+115^\circ=180^\circ
\angle BPR=30^\circ
Final answer: \angle PRB=35^\circ, \angle PBR=115^\circ, and \angle BPR=30^\circ.
Question 12: Prove \angle BCD=2\angle ABE
Step 1: Since A is the centre, the angle at the centre is twice the angle at the circumference standing on the same chord.
\angle BAD=2\angle BED
Step 2: C, D, and E are collinear, and AB\parallel CD. Therefore, AB\parallel ED.
Step 3: Using alternate interior angles,
\angle BED=\angle ABE
Step 4: Substitute this in the result from Step 1.
\angle BAD=2\angle ABE
Step 5: In parallelogram ABCD, opposite angles are equal.
\angle BAD=\angle BCD
Step 6: Therefore,
\angle BCD=2\angle ABE
Hence proved: \angle BCD=2\angle ABE.
Question 13: Incentre problem with \angle BAC=55^\circ and \angle ACB=65^\circ
Step 1: First find \angle ABC in triangle ABC.
\angle ABC+55^\circ+65^\circ=180^\circ
\angle ABC=60^\circ
Step 2: Since I is the incentre, BI, CI, and AI bisect the angles of triangle ABC.
\angle ABD=\frac{60^\circ}{2}=30^\circ
Step 3: Angles standing on the same chord AD are equal.
\angle DCA=\angle ABD=30^\circ
Step 4: Also, \angle CBD=30^\circ, and angles standing on chord CD are equal.
\angle DAC=\angle CBD=30^\circ
Step 5: Since CI bisects \angle ACB,
\angle ACI=\frac{65^\circ}{2}=32.5^\circ
Step 6: Add the two parts at C.
\angle DCI=\angle DCA+\angle ACI=30^\circ+32.5^\circ=62.5^\circ
Step 7: Since AI bisects \angle BAC,
\angle IAC=\frac{55^\circ}{2}=27.5^\circ
Step 8: Use the angle sum of triangle AIC.
\angle IAC+\angle ACI+\angle AIC=180^\circ
27.5^\circ+32.5^\circ+\angle AIC=180^\circ
\angle AIC=120^\circ
Final answer: \angle DCA=30^\circ, \angle DAC=30^\circ, \angle DCI=62.5^\circ, and \angle AIC=120^\circ.
Examiner’s mindset for Circles problems
In ICSE geometry, the answer is not only the final angle. Marks are usually earned through the chain of reasons. A strong solution names the property, substitutes the values correctly, and ends with the required angle or proof statement.
- For angle calculations: write the theorem before the arithmetic, such as \angle AOB=2\angle ACB.
- For proof questions: do not jump from the diagram to the result. Show the equality or supplementary relation that connects the given statement to the required statement.
- For cyclic quadrilaterals: specify which opposite angles are supplementary. Writing only 180^\circ without naming the angles is incomplete reasoning.
Common mistakes students make in Circles
- Confusing the centre angle rule: Students sometimes write the circumference angle as double the centre angle. The correction is \angle \text{centre}=2\times \angle \text{circumference} for the same arc.
- Using same-segment angles on different chords: The angles are equal only when they stand on the same chord and lie in the correct segment.
- Forgetting the isosceles step with radii: When OA=OB, write \angle OAB=\angle OBA before applying the triangle angle sum.
- Assuming a diameter without checking the figure: Use the semicircle theorem only when the line is marked as a diameter or passes through the centre.
- Writing decimal half-angles carelessly: For example, \frac{65^\circ}{2}=32.5^\circ, not 32^\circ or 33^\circ.
Quick answer index
This index is only a check after you have read the working. In school work and exams, write the theorem-based steps, not just the final value.
| Section | Question | Answer |
|---|---|---|
| Exercise 17(A) | 1(a) | 35^\circ |
| Exercise 17(A) | 1(b) | 40^\circ |
| Exercise 17(A) | 1(c) | 50^\circ |
| Exercise 17(A) | 1(d) | C, B, and D are collinear |
| Exercise 17(A) | 1(e) | 122^\circ |
| Exercise 17(A) | 2(i) | AC is a diameter |
| Exercise 17(A) | 2(ii) | \angle ACB=45^\circ |
| Exercise 17(A) | 3(i) | a=55^\circ |
| Exercise 17(A) | 3(ii) | b=95^\circ |
| Exercise 17(A) | 3(iii) | c=40^\circ |
| Exercise 17(A) | 3(iv) | d=45^\circ |
| Exercise 17(A) | 4 | \angle CDB=49^\circ, \angle ABC=43^\circ, \angle ACB=88^\circ |
| Exercise 17(A) | 5 | \angle DAB+\angle ABD=125^\circ |
| Exercise 17(A) | 6 | \angle BDC=35^\circ |
| Exercise 17(A) | 7 | \angle OBC=50^\circ |
| Exercise 17(A) | 8 | 27^\circ, 83^\circ, 70^\circ |
| Exercise 17(A) | 9 | \angle DCE=50^\circ, \angle ABC=30^\circ |
| Exercise 17(A) | 10 | \angle AEB=40^\circ |
| Exercise 17(A) | 11 | 35^\circ, 115^\circ, 30^\circ |
| Exercise 17(A) | 12 | \angle BCD=2\angle ABE |
| Exercise 17(A) | 13 | 30^\circ, 30^\circ, 62.5^\circ, 120^\circ |
How to use this page for revision
For timed practice, cover the answer index and solve one figure at a time. After each solution, compare only the reason chain: theorem, substitution, calculation, final answer. This is more useful than memorising final angles.
For connected study, revise ICSE Class 10 Maths topic pages and then practise past geometry questions from ICSE Class 10 question papers. If your Selina edition places tangents in the next chapter, study the tangent theorem after finishing this Circles angle-properties set.
Sources used for syllabus alignment
This page is aligned to the standard ICSE Class 10 Mathematics treatment of circle angle properties, cyclic quadrilaterals and related geometry proofs. For board-level reference, use the official CISCE publications section and the prescribed Selina Concise Mathematics Class 10 textbook used by many ICSE schools.
- CISCE official publications and syllabus resources
- NCERT textbook portal for standard school mathematics reference
- Selina Concise Mathematics Class 10, Chapter 17 Circles
Frequently Asked Questions
How should I start a Circles proof in ICSE Class 10 Maths?
Start by naming the theorem that matches the diagram, such as angle in a semicircle, angles in the same segment, or opposite angles of a cyclic quadrilateral. Then write one equation at a time and give the reason after each step.
Why do many Selina Circles answers use 180^\circ?
Circle problems often reduce to triangle or cyclic quadrilateral angle sums. A triangle gives \angle A+\angle B+\angle C=180^\circ, while opposite angles of a cyclic quadrilateral are supplementary.
What is the difference between angle at the centre and angle at the circumference?
For the same arc, the angle at the centre is twice the angle at the remaining circumference. If \angle AOB=100^\circ, then the corresponding circumference angle is 50^\circ.
Are tangent questions part of this Chapter 17 Circles page?
This page focuses on Selina Chapter 17 Circles angle properties and cyclic quadrilaterals. Tangents and intersecting chords are usually treated as the next linked geometry chapter in many Selina editions.
How do I avoid losing marks in a cyclic quadrilateral problem?
Write the exact property before the calculation. For example, state that opposite angles of a cyclic quadrilateral are supplementary, then substitute the given values in a clear line.