Q1 of 15 Page 82

Prove that the line joining the centres of two intersecting circles is the perpendicular bisector of the line joining the points of intersection.


O is the centre of the bigger circle and P is the centre of the smaller circle. DE is the line joining the points of intersection. We have to prove that, OP is the perpendicular bisector of DE.


In ΔODP and ΔOEP we have,


OD = OE [radius of same circle]


PD = PE [radius of same circle]


OP is the common side.


∴ΔODP≅ΔOEP [SSS congruency]


∴ ∠DOP = ∠EOP


∴ ∠DOG = ∠EOG ……. (i)


Now in ΔODG and ΔOEG we have,


OD = OE [radius of same circle]


∠DOG = ∠EOG from (i)


OG is the common side.


∴ΔODG≅ΔOEG [SAS congruency]


∴ DG = EG


∴ ∠OGD = ∠OGE


We know, ∴ ∠OGD + ∠OGE = 180°


∴ ∠OGD = ∠OGE = 90°


∴ DG = EG and ∠OGD = ∠OGE = 90°


∴ OP is perpendicular bisector of DE


More from this chapter

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2

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Prove that the parts of the line between the circles on either side are equal.

 3

A chord and the diameter through one of its ends are drawn in a circle. A chord of the same inclination is drawn on the other side of the diameter.


Prove that the chords are of the same length.

 4

Prove that the angle made by two equal chords drawn from a point on the circle is bisected by the diameter through that point.

 5

Draw a square and a circle through all four vertices. Draw diameters parallel to the sides of the square and draw a polygon joining the end points of these diameters and the vertices of the square.


Prove that this polygon is a regular octagon.