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

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(A) Given $PA = nPB$. Squaring both sides,we get $PA^2 = n^2 PB^2$.$(x - a)^2 + y^2 = n^2((x + a)^2 + y^2)$.$x^2 - 2ax + a^2 + y^2 = n^2(x^2 + 2ax + a

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

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(A) Given $PA = nPB$. Squaring both sides,we get $PA^2 = n^2 PB^2$.$(x - a)^2 + y^2 = n^2((x + a)^2 + y^2)$.$x^2 - 2ax + a^2 + y^2 = n^2(x^2 + 2ax + a^2 + y^2)$.$(1 - n^2)x^2 - 2a(1 + n^2)x + (1 - n^2)y^2 + a^2(1 - n^2) = 0$.Dividing by $(1 - n^2)$,we get $x^2 - 2a \frac{1 + n^2}{1 - n^2} x + y^2 + a^2 = 0$.This is the equation of a circle with center $C = (a \frac{1 + n^2}{1 - n^2}, 0)$ and radius $r = \sqrt{(a \frac{1 + n^2}{1 - n^2})^2 - a^2} = |a| \sqrt{\frac{(1 + n^2)^2 - (1 - n^2)^2}{(1 - n^2)^2}} = |a| \frac{2n}{1 - n^2}$.Since $a  0$. Then $A = (-k, 0)$ and $B = (k, 0)$.The center is $C = (-k \frac{1 + n^2}{1 - n^2}, 0)$.Since $0  1$,so the center $C$ is at a distance greater than $k$ from the origin in the negative direction.Thus,$A$ is closer to the center than $B$,meaning $A$ lies inside the circle and $B$ lies outside.

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

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