Circles Notes: Class 9 Maths Chapter 10
Class 9 Maths Chapter 10 Circles covers the parts of a circle, the equal chords theorem, angle subtended by an arc, and properties of a cyclic quadrilateral.
Overview of Topics & Subtopics (Click to Jump)
1. Circles and Chords
Short Note:
• Equal chords of a circle subtend equal angles at the centre, and conversely.
• The perpendicular from the centre of a circle to a chord bisects the chord, and the line joining the centre to the mid-point of a chord is perpendicular to the chord.
• Equal chords of a circle (or of congruent circles) are equidistant from the centre, and conversely.
2. Angle Subtended by an Arc
Short Note:
• The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
• Angles in the same segment of a circle are equal.
• The angle in a semicircle is a right angle.
3. Cyclic Quadrilaterals
Short Note:
• A quadrilateral whose all four vertices lie on a circle is a cyclic quadrilateral.
• The sum of either pair of opposite angles of a cyclic quadrilateral is 180°, and conversely, if the sum of a pair of opposite angles of a quadrilateral is 180°, it is cyclic.
NCERT Textbook Exercise Questions
Two circles of radii 5 cm and 3 cm intersect at two points, and the distance between their centres is 4 cm. Find the length of the common chord.
Answer: Let the common chord be AB, intersecting the line of centres at M. Using the two right triangles formed and solving simultaneously (Pythagoras theorem in each circle), the length of the common chord AB works out to 6 cm.
If a line intersects two concentric circles with centre O at points A, B, C, D, prove that AB = CD.
Answer: Draw OM ⊥ AD. Since the perpendicular from the centre bisects the chord in each circle, M is the mid-point of AD as well as of BC. So AM = MD and BM = MC, subtracting gives AM − BM = MD − MC, i.e., AB = CD.
In the given figure, ∠PQR = 100°, where P, Q, R are points on a circle with centre O. Find ∠OPR.
Answer: The reflex angle POR = 2 × ∠PQR = 200°, so the (non-reflex) ∠POR = 360° − 200° = 160°. Since OP = OR (radii), ΔOPR is isosceles, so ∠OPR = ∠ORP = (180° − 160°)/2 = 10°.
ABCD is a cyclic quadrilateral in which AC and BD are diameters of the circle. Prove that ABCD is a rectangle.
Answer: Since AC is a diameter, ∠ABC = ∠ADC = 90° (angle in a semicircle). Since BD is a diameter, ∠BAD = ∠BCD = 90°. All four angles of ABCD are 90°, so ABCD is a rectangle.
