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Class 9 Maths • Chapter 9 • NCERT Core

Circles

Vardaan Learning Institute  |  School-Exam Focused Notes

🔵 1. Terms Related to a Circle

• Circle: The collection of all points in a plane, which are at a fixed distance from a fixed point in the plane.
• Chord: A line segment joining any two points on the circle.
• Diameter: The longest chord of the circle, passing through the centre.
• Arc: A piece of a circle between two points (Minor Arc and Major Arc).
• Segment: Region between a chord and either of its arcs (Minor Segment, Major Segment).
• Sector: Region between an arc and the two radii joining the centre to the end points of the arc.
O A B Minor Segment Major Segment Chord AB Minor Sector Major Sector O

📏 2. Angle Subtended by a Chord at a Point

Theorem 9.1: Equal chords of a circle subtend equal angles at the centre.
Converse (Theorem 9.2): If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.
Proof uses SAS congruence between the two triangles formed by radii and chords.

⛏️ 3. Perpendicular from the Centre to a Chord

📐 Theorem 9.3 & 9.4 (Very Important for Numericals)
Theorem 9.3: The perpendicular from the centre of a circle to a chord bisects the chord.
If $OM \perp AB$, then $AM = MB$.

Theorem 9.4 (Converse): The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
If $AM = MB$, then $OM \perp AB$.

Application: To find radius, chord length, or distance to centre, draw the perpendicular, form a right triangle ($OMA$), and use the Pythagorean theorem: $OA^2 = OM^2 + AM^2$.

⭕ 4. Circle through Three Points & Equal Chords

Theorem 9.5: There is one and only one circle passing through three given non-collinear points.

Theorem 9.6: Equal chords of a circle (or of congruent circles) are equidistant from the centre.
Theorem 9.7 (Converse): Chords equidistant from the centre of a circle are equal in length.

📐 5. Angles Subtended by an Arc

📐 Theorem 9.8: Angle at Centre is Double Angle at Circumference
Statement: The angle subtended by an arc at the centre is double the angle subtended by it at any remaining part of the circle.
$\angle AOB = 2 \times \angle ACB$
(This is one of the most frequently used theorems in Angle Chasing problems).
Theorem 9.9: Angles in the same segment of a circle are equal. ($\angle ACB = \angle ADB$ if they stand on same arc AB).
Theorem (Angle in Semicircle): The angle in a semicircle is a right angle (90°).
Since straight angle is 180° at centre, the angle at circumference is $180/2 = 90°$.
A B C D O (2x) x x

⏹️ 6. Cyclic Quadrilaterals

A quadrilateral whose all four vertices lie on a circle is called a cyclic quadrilateral.

Theorem 9.11: The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
$\angle A + \angle C = 180°$ and $\angle B + \angle D = 180°$.

Theorem 9.12 (Converse): If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.
✏️ Practice Questions — Circles (28 Questions)

Section A — 1 Mark / MCQs Easy

Q1. The longest chord of a circle is known as ________.
Q2. The region between a chord and its corresponding arc is called ________.
Q3. Angle in a semicircle is ________.
Q4. If $\angle AOB = 120°$ at the centre, what is the angle subtended by the arc at the remaining circle?
Q5. ABCD is a cyclic quadrilateral. If $\angle A = 70°$, find $\angle C$.
Q6. True/False: Equal chords subtend equal angles at the centre.
Q7. Distance of chord from centre = 3cm, radius = 5cm. Find chord length.
Q8. Two circles intersect at 2 points. Their centres lie on the _______ of the common chord.

Section B — Core Problems Medium

Q9. In a circle with radius 5cm, two parallel chords 6cm and 8cm are on opposite sides of the centre. Find distance between them.
Q10. Two chords AB and CD of a circle intersect inside. If AB=10, CD=10, and their distances from O are 'x' and 'y', relation between x and y?
Q11. A, B and C are three points on a circle with centre O. $\angle BOC = 30°$ and $\angle AOB = 60°$. If D is a point on the circle other than arc ABC, find $\angle ADC$.
Q12. Prove that a cyclic parallelogram is a rectangle.
Q13. O is centre. $\angle OBC = 30°$, $\angle OCB = 30°$. Find $\angle BOC$ and then $\angle BAC$.
Q14. In cyclic quad ABCD, AB || DC. If $\angle ADC = 65°$, find $\angle ABC$.

Section C — NCERT Proofs & Logic Medium

  1. Q15. Prove that equal chords of a circle subtend equal angles at the centre.
  2. Q16. Prove that a line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
  3. Q17. If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.
  4. Q18. A circular park of radius 20m. Three boys are sitting at equal distances from each other on the boundary. Find the distance between each string. (Equilateral triangle in circle).
  5. Q19. Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any remaining part.
  6. Q20. If diagonals of a cyclic quadrilateral are diameters of the circle, prove it's a rectangle.

Section D — Board Level / Complex Numericals Hard

  1. Q21. In a circle of radius 5cm, AB and AC are two chords of 6cm each. Find length of chord BC.
  2. Q22. A, B, C, D are four points on a circle. AC and BD intersect at E. $\angle BEC = 130°$ and $\angle ECD = 20°$. Find $\angle BAC$.
  3. Q23. Non-parallel sides of a trapezium are equal. Prove it is cyclic.
  4. Q24. Two circles intersect at points A and B. AD and AC are diameters to the two circles. Prove B lies on line segment DC.
  5. Q25. Prove that opposite angles of a cyclic quadrilateral are supplementary.
  6. Q26. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.
  7. Q27. Two congruent circles intersect at A and B. A line passing through A meets the circles at P and Q. Prove $BP = BQ$. (Use Thm 9.9)
  8. Q28. In cyclic quad ABCD, $\angle DBA = 50°$ and $\angle ADB = 33°$. Find $\angle BCD$. (Hint: use sum of angles in $\triangle ABD$ then cyclic prop).
✅ Quick Numerical Checks: Q4: 60° | Q5: 110° | Q7: 8cm (half is 4) | Q9: 7cm (4+3) | Q11: Total at O is 90°, so at D is 45° | Q12: Opp angles sum to 180, and equal in ||gram -> 90. | Q18: $20\sqrt{3}$ meters. | Q22: $\angle CED = 50$, $\angle CDE = 110$. By same arc, $\angle BAC = \angle BDC = 110$. Wait: $\triangle CDE$ angle sum is 180 -> $50+20+\angle CDE = 180 \implies \angle CDE = 110°$. Hence $BAC = 110°$.