NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.6 – Part 5

Understanding tangent, concentric and congruent circles

Understanding Tangent, Concentric and Congruent Circles

Understanding tangent, concentric and congruent circles

Q-9 Two congruent circles intersect each other at point P and Q. Through P any line segment APB is drawn so that A, B lie on the two circles. Prove that QA=QB.

Solution:

Circle c Circle c: Circle through B_1 with center O_2 Circle d Circle d: Circle through D_1 with center O_1 Segment f Segment f: Segment [A, B] Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [A, Q] Segment h Segment h: Segment [B, Q] Segment i Segment i: Segment [P, Q] Point O_2 O_2 = (-2.02, 2.57) Point O_2 O_2 = (-2.02, 2.57) Point O_2 O_2 = (-2.02, 2.57) Point O_2 O_2 = (-2.02, 2.57) Point O_1 O_1 = (0.62, 2.62) Point O_1 O_1 = (0.62, 2.62) Point O_1 O_1 = (0.62, 2.62) Point O_1 O_1 = (0.62, 2.62) Point P Point P: Intersection point of c, d Point P Point P: Intersection point of c, d Point P Point P: Intersection point of c, d 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 A Point A: Point on c Point A Point A: Point on c Point A Point A: Point on c Point B Point B: Point on d Point B Point B: Point on d Point B Point B: Point on d Point C Point C: Point on c Point C Point C: Point on c Point C Point C: Point on c Point D Point D: Point on d Point D Point D: Point on d Point D Point D: Point on d

Two Congruent Circles With Equal Radius

Two congruent circles intersecting each other at point P and Q

Given,

  • Two congruent circles intersecting each other at point P and Q.

  • PQ is a common chord of these circles.

Since PQ is common chord

  • arc(PCQ)=arc(PDQ)

  • Angle subtended by arcs of equal lengths on congruence circles must be equal, therefore, QAP=QBP

  • Therefore in isosceles triangle AQB, QA=QB

Q-10 In any triangle PQR, if the angle bisector of P and perpendicular bisector of QR intersect, prove that they intersect on the circumcircle of the circumcircle of triangle PQR.

Solution:

Circle c Circle c: Circle through A with center O Angle ? Angle ?: Angle between Q, P, A Angle ? Angle ?: Angle between Q, P, A Angle ? Angle ?: Angle between A, P, R Angle ? Angle ?: Angle between A, P, R Segment f Segment f: Segment [Q, R] Segment g Segment g: Segment [Q, P] Segment h Segment h: Segment [P, R] Segment i Segment i: Segment [P, A] Segment j Segment j: Segment [R, A] Segment k Segment k: Segment [Q, A] Segment l Segment l: Segment [A, B] Point O O = (0.44, 2.62) Point O O = (0.44, 2.62) Point O O = (0.44, 2.62) Point A A = (0.46, 0.7) Point A A = (0.46, 0.7) Point A A = (0.46, 0.7) Point Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point R Point R: Point on c Point R Point R: Point on c Point R Point R: 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 B Point B: Point on f Point B Point B: Point on f Point B Point B: Point on f

Triangle PQR and Its Circumcircle With Center O

Triangle PQR and its circumcircle with center O, bisector of ∠P and perpendicular bisector of QR intersect at A.

Given,

  • PQR is a triangle inscribed in a circle with centre at O, A is a point on the circle

  • PA is the internal bisector of QPR and B is the mid-point of QR

Prove:

BA is the right bisector of QR

In QBAandRBA,QA=AR , since both the arcs QA and AR subtend equal angle at the circumference ( QPA=RPA) .

Therefore in triangle QAR,

  • QPA=RPA

  • QB=BR (Given)

  • BA=BA (common line)

Therefore, by SSS criterion of congruence, QBARBA

Hence, QBA=BAR (corresponding parts of congruent triangles)

Now,

  • QBA+BAR=180°

  • QBA+QBA=180° ( QBA=BAR )

  • 2QBA=180°

  • QBA=CDE=90°

Therefore, BA is the right bisector of QR.

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