Question

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

Answer 1

given
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
HENCE THE PROOF..

Answer 2

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.