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Analytical Geometry
Comments (5)
Let P be a complex number z.
A and B be z1 and z2. PA=|z-z1| , PB=|z-z2|
PA/PB=K
i.e |z-z1|=K|z-z2| ...
if K=1 then |z-z1|=|z-z2| (use z=x+iy if you dont get my explanation)
which is the equation of perpendicular bisector of the line joining the points A and B.
orher wise it is a circle.
hello dear
nice kishore
well according to the given condition in the question..that the point lies on the circle
PA = k PB
k can not be equal to 1 bcz .otherwise in this it will the perpendicular bisector of the line joing the point A and B
for the equation to be circle:
PA = k PB
putting in the algebric from we get:
x2 (1-k) + y2(1-k) - x (2x1 + 2kx2) - y (2y1 + 2ky2) + (x12 + y12 -kx22 - k y22) = 0
hence k can not be equal to 1 for the above equation to be circle
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