Let A = (a, 0) and B = (-a, 0) be two fixed p | vedclass

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(B) Let $P = (x, y)$. The condition $PA = nPB$ implies $PA^2 = n^2 PB^2$.$(x - a)^2 + y^2 = n^2((x + a)^2 + y^2)$.$x^2 - 2ax + a^2 + y^2 = n^2(x^2 + 2

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$A$ lies outside the circle and $B$ lies inside the circle.

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(B) Let $P = (x, y)$. The condition $PA = nPB$ implies $PA^2 = n^2 PB^2$.$(x - a)^2 + y^2 = n^2((x + a)^2 + y^2)$.$x^2 - 2ax + a^2 + y^2 = n^2(x^2 + 2ax + a^2 + y^2)$.$(n^2 - 1)x^2 + (n^2 - 1)y^2 + 2ax(n^2 + 1) + a^2(n^2 - 1) = 0$.Dividing by $(n^2 - 1)$,we get $x^2 + y^2 + 2ax \frac{n^2 + 1}{n^2 - 1} + a^2 = 0$.This is the equation of a circle with center $C = (-a \frac{n^2 + 1}{n^2 - 1}, 0)$ and radius $r = |\frac{2an}{n^2 - 1}|$.Since $a  0$. Then the center is $C = (k \frac{n^2 + 1}{n^2 - 1}, 0)$.For $n > 1$,the circle encloses the point $B(-a, 0)$ because $B$ is closer to the center relative to the radius,while $A(a, 0)$ lies outside.

Let $A = (a, 0)$ and $B = (-a, 0)$ be two fixed points. For $a \in (-\infty, 0)$,point $P$ moves in the plane such that $PA = nPB$ $(n \neq 0, n \neq 1)$. If $n > 1$,then which of the following is true regarding the positions of $A$ and $B$ with respect to the locus of $P$?

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