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 PQR and PSR

  • PQ=PS (radius of the same circle)

  • QR=SR (radius of the same circle)

  • PR=PR (common line)

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

Therefore, PQR=PSR (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

  • OBPQ and OARS

  • RSPQ.

Construction,

  • Join OP and OR

  • Since both P and R are on the circle and O is the center, OP=OR=r

OBPQ and OARS and RSPQ , Therefore, B, O, A are collinear.

Let OB=x , since AB=6cm . Therefore, OA=(6x)cm.

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

  • PB=BQ=2.5cm ( PQ=5cm )

  • RA=AS=5.5cm. ( RS=11cm )

In right OPB

  • OP2=OB2+PB2

  • r2=x2+(2.5)2

  • r2=x2+6.25 ………….equation (1)

In right ORA

  • OR2=RA2+AO2

  • r2=(5.5)2+(6x)2

  • r2=30.25+3612x+x2

  • r2=66.2512x+x2 …………equation (2)

From equation (1) and (2)

  • x2+6.25=66.2512x+x2

  • 12x=66.256.25

  • 12x=60

  • x=5

Now, putting x=5 in equation (1)

  • r2=x2+6.25

  • r2=(5)2+6.25

  • r2=25+6.25

  • r2=31.25

  • r=5.59

So, the radius of the circle is 5.59cm .

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