# If two circles intersect at two points,prove that their centres lie on the perpendicular bisector of the common chord.

2
by pluffy

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by pluffy

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2 2circles intersect at a and b

T.P

the line formed by their centers(o and p) is the perpendicular bisector of chord ab

PROOF

con.tri. oap and obp

oa=ob (radii of same circle)

pb=pa

ap=pa (common)

therefore, oap is cong. to obp by SSS

let m the point of intersection of ab and op

join om

oa=ob

aom=bom (proved by CPCT)

om=om (common)

therefore tri. aom is cong. to bom by sas

hence,

am=bm

amo=bmo

now ang. amo+bmo=180 (linear pair)

as amo=bmo,

2 amo=180

amo=90

hence, bmo=90

lly we can prove it for the other two.

i.e all angles formed r 90

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See diagram.

AB is the line joining intersections of the circles with centers O and P. Let OC be the perpendicular onto AB from O.

OA = OB as OA and OB are radii of circle with center O. angle OCA = angle OCB = 90 deg. OC is a common side. Hence two triangles OCA and OCB are congruent. Hence AC = CB. So O lies on perpendicular bisector of AB.

PA = PB as they are radii of circle with center P. draw a perpendicular from P onto AB. Let it meet at C'. Now between the two triangles AC'P and BC'P, PC' is common side, angle AC'P = angle BC'P = 90 deg. Hence the two triangles are congruent. Hence AC' = BC'. SO P lies on the perpendicular bisector of AB.

AB is the line joining intersections of the circles with centers O and P. Let OC be the perpendicular onto AB from O.

OA = OB as OA and OB are radii of circle with center O. angle OCA = angle OCB = 90 deg. OC is a common side. Hence two triangles OCA and OCB are congruent. Hence AC = CB. So O lies on perpendicular bisector of AB.

PA = PB as they are radii of circle with center P. draw a perpendicular from P onto AB. Let it meet at C'. Now between the two triangles AC'P and BC'P, PC' is common side, angle AC'P = angle BC'P = 90 deg. Hence the two triangles are congruent. Hence AC' = BC'. SO P lies on the perpendicular bisector of AB.