CIRCLE

1. NAME = YOGESH KATYAL CLASS = 9C ROLL NO = 30  SUBJECT = MATHS TOPIC ON NEXT PAGE

2. CIRCLE

3. CIRCLE DEFINITION CIRCLE : The set of points coplanar points equidistant from give point.   : He given point is called the centre of the circle . The distance from the centre to the circle is called the radius .  RADIUS CENTER

4. DEFINITION Chord : The segment whose Diameter : a chord that contain the center of the circle . Tangent : A is a line or line segment that touches a circle at one point only . ARC Tangent chord pair of tangency : The point where the tangent line intersects the circle diameter Secant : a line that intersects a curve at a minimum of two distinct points is called secant secant

5. Theorems and theorem 1 “Two equal chords of a circle subtend equal angles at the centre of the circle. Proof: Given, in ∆AOB and ∆POQ, AB = PQ (Equal Chords) …………..(1) OA = OB= OP=OQ (Radii of the circle) ……..(2) From eq 1 and 2, we get; ∆AOB ≅ ∆POQ (SSS Axiom of congruency) Therefore, by CPCT (corresponding parts of congruent triangles), we get; ∠AOB = ∠POQ Hence, Proved.

6. Converse of Theorem 1 “If two angles subtended at the centre by two chords are equal, then the chords are of equal length.” Proof: Given, in ∆AOB and ∆POQ, ∠AOB = ∠POQ (Equal angle subtended at center O) ..…… (1) OA = OB = OP = OQ (Radii of the same circle) ……………(2) From eq. 1 and 2, we get; ∆AOB ≅ ∆POQ (SAS Axiom of congruency) Hence, AB = PQ (By CPCT)

7. THEOREM 2 “The perpendicular to a chord bisects the chord if drawn from the centre of the circle.” Proof: Given, in ∆AOD and ∆BOD, ∠ADO = ∠BDO = 90° (OD ⊥ AB) ………(1) OA = OB (Radii of the circle) ……….(2) OD = OD (Common side) ………….(3) From eq. (1), (2) and (3), we get; ∆AOB ≅ ∆POQ (R.H.S Axiom of congruency) Hence, AD = DB (By CPCT)

8. Converse of theorem 2 “A straight line passing through the Centre of a circle to bisect a chord is perpendicular to the chord.” Proof: Given, in ∆AOD and ∆BOD, AD = DB (OD bisects AB) ………….(1) OA = OB (Radii of the circle) ………….(2) OD = OD (Common side) …………..(3) From eq. 1, 2 and 3, we get; ∆AOB ≅ ∆POQ (By SSS Axiom of congruency) Hence, ∠ADO = ∠BDO = 90° (By CPCT)

9. THEOREM 3 “Equal chords of a circle are equidistant (equal distance) from the Centre of the circle.” Construction: Join OB and OD Proof: Given, In ∆OPB and ∆OQD BP = 1/2 AB (Perpendicular to a chord bisects it) ……..(1) DQ = 1/2 CD (Perpendicular to a chord bisects it) ……..(2) AB = CD (Given) BP = DQ (from eq 1 and 2) OB = OD (Radii of the same circle) ∠OPB = ∠OQD = 90° (OP ⊥ AB and OQ ⊥ CD) ∆OPB ≅ ∆OQD ( By R.H.S Axiom of Congruency) Hence, OP = OQ ( By CPCT)

10. Converse of theorem 3 “Chords of a circle, which are at equal distances from the centre are equal in length, is also true.” Proof: Given, in ∆OPB and ∆OQD, OP = OQ ………….(1) ∠OPB = ∠OQD = 90° ………..(2) OB = OD (Radii of the same circle) ………..(3) Therefore, from eq. 1, 2 and 3, we get; ∆OPB ≅ ∆OQD (By R.H.S Axiom of Congruency) BP = DQ ( By CPCT) 1/2 AB = 1/2 CD (Perpendicular from center bisects the chord) Hence, AB = CD

11. THEOREM 4 “Measure of angles subtended to any point on the circumference of the circle from the same arc is equal to half of the angle subtended at the center by the same arc.” From the above figure, ∠AOB = 2∠APB Construction: Join PD passing through centre O Proof: In ∆AOP, OA = OP (Radii of the same circle) ………..(1) ∠OAP = ∠OPA (Angles opposite to equal sides of a triangle) …………(2) ∠AOD = ∠OAP + ∠OPA (Exterior Angle Property of the triangle) …………(3) Hence, from eq. 2 and 3 we get; ∠AOD = 2∠OPA ………….(4) Similarly in ∆BOP, Exterior angle, ∠BOD = 2 ∠OPB …………(5) ∠AOB = ∠AOD + ∠BOD From eq. 4 and 5, we get; ⇒∠AOB = 2∠OPA + 2∠OPB ⇒∠AOB = 2(∠OPA + ∠OPB) ⇒∠AOB = 2∠APB Hence, proved.

12. THEOREM 5 “The opposite angles in a cyclic quadrilateral are supplementary.” Proof: Suppose, for arc ABC, ∠AOC = 2∠ABC = 2α (Theorem 4) …………….(1) Consider for arc ADC, Reflex ∠AOC = 2 ∠ADC = 2β (Theorem 4) …………..(2) ∠AOC + Reflex ∠AOC = 360° From eq. 1 and 2, we get; ⇒ 2 ∠ABC + 2∠ADC = 360° ⇒ 2α + 2β = 360° ⇒α + β = 180°