The two points A and B in a plane are such that for all points P lies on circle satisfied PAPB=k, then k will not be
equal to
1
Let A = (a, 0), B(-a, 0), P = (α,β)
∴PA2PB2=k2⇒(α−a)2+β2=k2[(α+a)2+β2]
∴ Locus is (x2+y2)(1−k2)−2a(1+k2)x+(1−k2)a2=0
This is a circle for k ≠ 1.