Circles — Important Questions Bank
Chapter 5 · Ganita Manjari · Grade 9 Part I · All questions strictly from chapter content
10 Long Answer (5 marks)
Theory + Numerical + Proof
Complete Step-by-step Solutions
Part A — Short Answer Questions
15 questions · 2–3 marks each · Theory, Numerical, Application
A circle is the set of all points on a plane that are equidistant from a given fixed point on that plane.
A locus is the set of all points satisfying a given condition. Using this term: a circle is the locus of points that are equidistant from a given point.
• The fixed given point is called the centre.
• The equal distance from the centre to any point on the circle is the radius.
A chord is a line segment joining any two points on a circle.
A chord that passes through the centre is called a diameter. The diameter is the longest chord of a circle.
A diameter subtends a straight angle (180°) at the centre, since the two radii form a straight line through the centre.
Rotational symmetry: A circle has complete rotational symmetry. Rotating it by any angle about its centre produces the same circle.
Reflection symmetry: Every diameter of a circle is a line of reflection symmetry (folding along a diameter makes the two halves overlap perfectly).
Since infinitely many diameters can be drawn in a circle, a circle has infinitely many lines of reflection symmetry, each being a diameter.
Infinitely many circles can pass through two given points A and B.
The centre of any such circle must be equidistant from A and B. All such centres lie on the perpendicular bisector of AB.
The smallest circle through A and B has AB as its diameter, so its radius = AB/2 (half the distance AB). Its centre is the midpoint of AB.
Theorem 1: There is a unique circle passing through three non-collinear points.
The unique circle passing through all three vertices of a triangle is its circumcircle. Its centre — found at the intersection of the perpendicular bisectors of the sides — is the circumcentre.
For a right-angled triangle, the circumcentre lies at the midpoint of the hypotenuse.
Theorem 2: Equal chords of a circle subtend equal angles at the centre.
Theorem 3 (Converse): Chords of a circle that subtend equal angles at the centre are equal in length.
Together, these mean: two chords are equal if and only if they subtend equal angles at the centre.
Theorem 4: The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord. So CM ⊥ AB.
Theorem 5 (Converse): The perpendicular from the centre of a circle to a chord bisects the chord.
These two theorems together say: the line from the centre to a chord is perpendicular to it if and only if it meets the chord at its midpoint.
Theorem 6: Chords of a circle having the same length are all at the same distance from the centre.
Theorem 7: Chords of a circle that are equidistant from the centre have equal length.
Together: two chords are equal if and only if they are equidistant from the centre.
The longer chord is closer to the centre.
Theorem 8: Let AB and DE be two chords of a circle with centre C. If AB > DE, then the perpendicular distance from C to AB is less than the perpendicular distance from C to DE.
Practical example: A diameter (the longest chord) passes through the centre, so its distance from the centre is 0 — it is the closest possible chord. As a chord moves away from the centre, its length decreases.
An arc is a connected portion of a circle, defined by two endpoints on the circle and the curve connecting them.
Two points on a circle divide it into two arcs. The larger arc is the major arc and the smaller arc is the minor arc.
An arc is classified by the angle it subtends at the centre: if the angle is less than 180°, it is a minor arc; if greater than 180°, it is a major arc.
∠ACB = 90°
Corollary of Theorem 9: The angle subtended by a diameter at any point on the circle is 90°.
Reason: AB is a diameter, so the arc (not containing C) subtends a straight angle (180°) at the centre. By Theorem 9, the angle at C = 180° ÷ 2 = 90°.
A quadrilateral whose four vertices all lie on a circle is called a cyclic quadrilateral.
Theorem 11: The sum of two opposite angles of a cyclic quadrilateral is 180°.
Part B — Long Answer Questions
10 questions · 5 marks each · Proof, Word Problems, Numerical
To Prove: ∠ACB = ∠DCE.
- In △CAB and △CDE:
- CA = CD = r (both are radii of the circle)
- CB = CE = r (both are radii of the circle)
- AB = DE (given)
- By the SSS congruence rule: △CAB ≅ △CDE
- Therefore, corresponding angles are equal: ∠ACB = ∠DCE
To Prove: CM ⊥ AB.
- CA = CB = r (radii), so △CAB is an isosceles triangle with CA = CB.
- Therefore ∠CAB = ∠CBA (base angles of isosceles triangle).
- M is the midpoint of AB, so AM = BM.
- In △CMA and △CMB: CA = CB (radii), AM = BM (M is midpoint), CM = CM (common).
- By SAS congruence: △CMA ≅ △CMB.
- So ∠CMA = ∠CMB (corresponding angles).
- But ∠CMA + ∠CMB = 180° (angles on a straight line AB).
- Therefore ∠CMA = ∠CMB = 90°.
To Prove: ∠BCA = 2∠BDA.
- Join D to C and extend DC to meet the circle at E on arc AFB.
- In △DCB: CB = CD (both radii), so △DCB is isosceles.
- Therefore ∠CBD = ∠CDB.
- ∠BCE is an exterior angle of △BCD. By the Exterior Angle Theorem: ∠BCE = ∠CBD + ∠CDB = 2∠BDC.
- In △ADC: CA = CD (radii), so △ADC is isosceles.
- Therefore ∠CAD = ∠CDA.
- ∠ACE is the exterior angle of △ADC: ∠ACE = ∠CAD + ∠CDA = 2∠CDA.
- Now: ∠BCA = ∠BCE + ∠ECA and ∠BDA = ∠BDC + ∠CDA.
- Therefore: ∠BCA = 2∠BDC + 2∠CDA = 2(∠BDC + ∠CDA) = 2∠BDA.
To Prove: ∠BAD + ∠BCD = 180°.
- Consider arc BCD. Point A is outside this arc, on the circle.
- By Theorem 9: ∠BAD = ½ × (angle subtended by arc BCD at centre O).
- The angle subtended by arc BCD at O (going B→C→D) is the reflex angle BOD.
- So: ∠BAD = ½ × (reflex ∠BOD).
- Consider arc BAD. Point C is outside this arc, on the circle.
- ∠BCD = ½ × ∠BOD (the non-reflex angle BOD, going B→A→D).
- Now: reflex ∠BOD + ∠BOD = 360° (complete angle at O).
- So: ∠BAD + ∠BCD = ½ × reflex ∠BOD + ½ × ∠BOD = ½ × 360° = 180°.
Proof:
- Let △ABC have ∠A = 90°. So BC is the hypotenuse.
- The circumcircle passes through B and C, with BC as a chord.
- By the corollary of Theorem 9: the angle in a semicircle = 90°. So A lies on the circle where BC is the diameter.
- If BC is the diameter, then the centre of the circumcircle is the midpoint of BC.
Calculation:
Proof:
- By Theorem 11: ∠ADC + ∠ABC = 180° (opposite angles of cyclic quadrilateral).
- ∠CDE + ∠ADC = 180° (∠CDE and ∠ADC form a linear pair, E is on the extension of CD).
- From both equations: ∠CDE = 180° − ∠ADC = ∠ABC.
Verification with ∠A = 75°, ∠B = 110°:
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