NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.4 – Part 2

Circle With Center O and Equal Chords

Circle with center O and chords AB and CD are of equal length

Q-3 If two equal chords of a circle intersect within the circle; prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Solution:

Circle c Circle c: Circle through B with center O Angle ? Angle ?: Angle between O, N, O' Angle ? Angle ?: Angle between O, N, O' Angle ? Angle ?: Angle between S, M, S' Angle ? Angle ?: Angle between S, M, S' Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [C, D] Segment h Segment h: Segment [P, Q] Segment i Segment i: Segment [O, M] Segment j Segment j: Segment [O, N] Point O O = (-0.68, 2.98) Point O O = (-0.68, 2.98) Point O O = (-0.68, 2.98) Point B B = (0.1, 1.14) Point B B = (0.1, 1.14) Point B B = (0.1, 1.14) Point A Point A: Point on c Point A Point A: Point on c Point A Point A: Point on c Point D Point D: Point on c Point D Point D: Point on c Point D Point D: Point on c Point C Point C: Point on c Point C Point C: Point on c Point C Point C: Point on c Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point M Point M: Point on h Point M Point M: Point on h Point M Point M: Point on h Point N Point N: Point on f Point N Point N: Point on f Point N Point N: Point on f Point S Point S: Intersection point of f, h Point S Point S: Intersection point of f, h Point S Point S: Intersection point of f, h in Circle With Center O With AB and PQ Are Two Chords

Circle and its chord AB and PQ intersecting at S,CD is the diameter.

Given,

To prove,

Now, Construction,

In Δ O S M and Δ O S N ,

O M = O N (Equal chords are equidistant from the centre)

O S = O S (Common line)

∠ O M S = ∠ O N S ( 90 ° by construction)

Therefore, Δ O S M ≅ Δ O S N (By Right Angle Hypotenuse congruence condition)

Therefore,

Q-4 If a line intersects two concentric circles (circles with the same centre) with centre O. A common chord intersects the circles at A, B, C and D, prove that AB = CD (see Fig.).

Circle c Circle c: Circle through B_1 with center O Circle d Circle d: Circle through C with center O Segment f Segment f: Segment [A, D] Point O O = (-0.14, 2.28) Point O O = (-0.14, 2.28) Point O O = (-0.14, 2.28) Point C C = (1.5, 2.56) Point C C = (1.5, 2.56) Point C C = (1.5, 2.56) Point B Point B: Point on d Point B Point B: Point on d Point B Point B: Point on d Point A Point A: Point on c Point A Point A: Point on c Point A Point A: Point on c Point D Point D: Point on c Point D Point D: Point on c Point D Point D: Point on c Two Concentric Circles With Centre O at a,B,C and D

Two concentric circles with centre O and common chord intersecting at A,B,C and D

Solution:

Given,

Two concentric circles, with centre at O and common chord intersecting at A,B,C and D

Contruction, draw a line OM ⊥ AD, intersecting AD at M.

Circle c Circle c: Circle through B_1 with center O Circle d Circle d: Circle through C with center O Angle ? Angle ?: Angle between O, M, O' Angle ? Angle ?: Angle between O, M, O' Segment f Segment f: Segment [A, D] Segment g Segment g: Segment [O, M] Point O O = (-0.14, 2.28) Point O O = (-0.14, 2.28) Point O O = (-0.14, 2.28) Point C C = (1.54, 2.42) Point C C = (1.54, 2.42) Point C C = (1.54, 2.42) Point B Point B: Point on d Point B Point B: Point on d Point B Point B: Point on d Point A Point A: Point on c Point A Point A: Point on c Point A Point A: Point on c Point D Point D: Point on c Point D Point D: Point on c Point D Point D: Point on c Point M Point M: Point on f Point M Point M: Point on f Point M Point M: Point on f Two Concentric Circle With Centre at O and Common Chord Intersecting at a,B,C and D

Two concentric circle with centre O and common chord intersecting at A,B,C and D also OM ⊥ AD. OM bisects AD as OM ⊥ AD

Now, since O M ⊥ A D and is drawn from center O

Since O is also the center of inner circle, therefore OM also bisects BC as O M ⊥ B C .

From (1) and (2),