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(B) Let $A = (a, 0)$,$B = (-a, 0)$,and $P = (x, y)$.Given the condition $\frac{PA}{PB} = k$,we have $\frac{PA^2}{PB^2} = k^2$.Substituting the coordin

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(B) Let $A = (a, 0)$,$B = (-a, 0)$,and $P = (x, y)$.Given the condition $\frac{PA}{PB} = k$,we have $\frac{PA^2}{PB^2} = k^2$.Substituting the coordinates,we get $\frac{(x - a)^2 + y^2}{(x + a)^2 + y^2} = k^2$.$(x - a)^2 + y^2 = k^2((x + a)^2 + y^2)$.$(x^2 - 2ax + a^2 + y^2) = k^2(x^2 + 2ax + a^2 + y^2)$.$(x^2 + y^2)(1 - k^2) - 2ax(1 + k^2) + a^2(1 - k^2) = 0$.If $k = 1$,the equation becomes $-4ax = 0$,which represents a straight line (the perpendicular bisector of $AB$),not a circle.Therefore,for the locus to be a circle,$k$ must not be equal to $1$.

The two points $A$ and $B$ in a plane are such that for all points $P$ lying on a circle satisfying $\frac{PA}{PB} = k$,then $k$ will not be equal to:

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