Page 4 - Lesson note-7 Extra Questions Ch.10 Circle

As Central angle subtended by arc AC = ∠COA
then ∠COA = 2 x ∠ABC = 2 x 30⁰= 60⁰...(1)
In triangle OCA,
OC = OA [Radius of same circle)
∠OCA = ∠CAO ... (2) [Angle opposite to equal sides]
In triangle COA,
∠OCA + ∠CAO + ∠COA = 180⁰
From (1) and (2), we get
2∠CAO + 60⁰= 180⁰
∠CAO = 60⁰

Question 8. Prove that the line of centers of two intersecting circles subtends equal
angles at the two points of intersection.

Sol. Given : Two intersecting circles, in which OO′ is the line of centers and A and B are
two points of intersection.
To prove: ∠OAO′ = ∠OBO′
Construction: Join AO, BO, AO′ and BO′.
Proof: In ∆AOO′ and ∆BOO′, we have
AO = BO [Radii of the same circle]
AO′ = BO′ [Radii of the same circle]
OO′ = OO′ [Common]
∴ ∆AOO′ ≅ ∆BOO′ [SSS axiom]
⇒ ∠OAO′ = ∠OBO′ [CPCT]
Proved.

Question 9. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle
are parallel to each other and are on opposite sides of its centre. If the distance
between AB and CD is 6 cm, find the radius of the circle.