NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.6 – Part 1

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Theorems of the circle

Theorems of circle

Theorems of Circle

Theorems of circle

Q-1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Solution:

Circle c Circle c: Circle through B with center P Circle d Circle d: Circle through D with center R Segment f Segment f: Segment [P, Q] Segment g Segment g: Segment [Q, R] Segment h Segment h: Segment [R, S] Segment i Segment i: Segment [P, S] Segment j Segment j: Segment [Q, S] Segment k Segment k: Segment [P, R] Point P P = (0.4, 2.58) Point P P = (0.4, 2.58) Point P P = (0.4, 2.58) Point R R = (3.28, 2.7) Point R R = (3.28, 2.7) Point R R = (3.28, 2.7) Point Q Point Q: Intersection point of c, d Point Q Point Q: Intersection point of c, d Point Q Point Q: Intersection point of c, d Point S Point S: Intersection point of c, d Point S Point S: Intersection point of c, d Point S Point S: Intersection point of c, d

Two Circles With Centres P and R

Two circles with centre P and R, intersecting each other at Q and S

Given

  • Two circles with centres P and R, which intersect each other at Q and S

Construction

  • Join PQ, PS, SR, QR.

In Equation and Equation

  • Equation (radius of the same circle)

  • Equation (radius of the same circle)

  • Equation (common line)

  • Therefore by Side-Side-Side criterion of congruence, Equation

Therefore, Equation (corresponding parts of congruent triangles)

Q-2 Two chords PQ and RS of lengths 5cm and 11cm respectively of a circle are parallel to each and are on opposite sides of its centre. If the distance between PQ and RS is 6cm, find the radius of the circle

Solution:

Circle c Circle c: Circle through B_1 with center O Segment f Segment f: Segment [R, S] Segment g Segment g: Segment [A, B] Segment h Segment h: Segment [P, Q] Segment i Segment i: Segment [R, O] Segment j Segment j: Segment [O, P] Point O O = (0.6, 2.24) Point O O = (0.6, 2.24) Point O O = (0.6, 2.24) Point R Point R: Point on c Point R Point R: Point on c Point R Point R: Point on c Point S Point S: Point on c Point S Point S: Point on c Point S Point S: Point on c Point A Point A: Point on f Point A Point A: Point on f Point A Point A: Point on f Point B B = (0.58, 0.84) Point B B = (0.58, 0.84) Point B B = (0.58, 0.84) 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 r cm text1 = "r cm" r cm text1_1 = "r cm" (6-x)cm text2 = "(6-x)cm" x cm text3 = "x cm"

O Is the Centre of Circle

O be the centre of circle and radius be r cm also OB⊥PQ and OA⊥RS, AB=6cm

Given,

  • O is the centre of the circle

  • And its radius be r cm

  • Equation and Equation

  • Equation

Construction,

  • Join OP and OR

  • Since both P and R are on the circle and O is the center, Equation

Equation and Equation and Equation , Therefore, B, O, A are collinear.

Let Equation , since Equation . Therefore, Equation

Since, the perpendicular from the centre to a chord of the circle bisects the chord, therefore,

  • Equation ( Equation )

  • Equation ( Equation )

In right Equation

  • Equation

  • Equation

  • Equation ………….equation (1)

In right Equation

  • Equation

  • Equation

  • Equation

  • Equation …………equation (2)

From equation (1) and (2)

  • Equation

  • Equation

  • Equation

  • Equation

Now, putting Equation in equation (1)

  • Equation

  • Equation

  • Equation

  • Equation

  • Equation

So, the radius of the circle is Equation .

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