NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.3 – Part 2

Conditions of congruence of triangles

Congruence Conditions

Conditions of congruence of triangles

Q-3 If two circles intersect at two points; prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

Circle c Circle c: Circle through B_1 with center O Circle d Circle d: Circle through D with center O' Segment f Segment f: Segment [O, A] Segment g Segment g: Segment [A, O'] Segment h Segment h: Segment [O', B] Segment i Segment i: Segment [B, O] Segment j Segment j: Segment [A, B] Segment k Segment k: Segment [O, O'] Point O O = (-1.12, 2.9) Point O O = (-1.12, 2.9) Point O O = (-1.12, 2.9) Point O' O' = (1.42, 2.98) Point O' O' = (1.42, 2.98) Point O' O' = (1.42, 2.98) Point A Point A: Intersection point of c, d Point A Point A: Intersection point of c, d Point A Point A: Intersection point of c, d Point B Point B: Intersection point of c, d Point B Point B: Intersection point of c, d Point B Point B: Intersection point of c, d Point C Point C: Intersection point of j, k Point C Point C: Intersection point of j, k Point C Point C: Intersection point of j, k

Circles Interact Each Other at a and B

Circles intersect at A and B, OO’ is perpendicular bisector of AB

  • A and B are the two points at the intersection of two circles.

  • To prove, AB is bisector and OO’ is perpendicular to OO’.

In ΔAOO' and ΔBOO' ,

  • OA=OB (Radius)

  • OO'=OO' (Common line)

  • O'A=OB (Radius)

  • ΔAOO'ΔBOO' (Side-Side-Side congruence condition)

Thus, AOO'=BOO' (by corresponding parts of congruent triangle)

In ΔAOC and ΔBOC,

  • OA=OB (Radius)

  • AOC=BOC

  • OC=OC (Common line)

  • ΔAOCΔBOC (Side-Angle-Side congruence condition)

Thus, ACO=BCO (by corresponding parts of congruent triangles)

Also,

  • ACO+BCO=180°

  • ACO+ACO=180° ( ACO=BCO )

  • 2ACO=180°

  • ACO=BCO=(180°12)=90°

Hence, OO' is perpendicular to AB. Since ΔAOCΔBOC , therefore AC = CB, i.e. C is the midpoint of AB. Therefore, OO' is perpendicular bisector of AB.

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