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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Never passes through $A$ and $B$.

Vedclass · Answer

(B) Let $P = (x, y)$. The condition $PA = nPB$ implies $PA^2 = n^2 PB^2$.$(x - a)^2 + y^2 = n^2((x + a)^2 + y^2)$.$x^2 - 2ax + a^2 + y^2 = n^2(x^2 + 2ax + a^2 + y^2)$.$(1 - n^2)x^2 + (1 - n^2)y^2 - 2ax(1 + n^2) + a^2(1 - n^2) = 0$.Dividing by $(1 - n^2)$ (since $n 
eq 1$):$x^2 + y^2 - 2ax \frac{1 + n^2}{1 - n^2} + a^2 = 0$.This is the equation of a circle.If the circle passes through $A(a, 0)$,then $a^2 + 0 - 2a^2 \frac{1 + n^2}{1 - n^2} + a^2 = 0$,which implies $2a^2 = 2a^2 \frac{1 + n^2}{1 - n^2}$,so $1 = \frac{1 + n^2}{1 - n^2}$,which means $1 - n^2 = 1 + n^2$,so $n = 0$,which contradicts $n 
eq 0$.Similarly,it does not pass through $B(-a, 0)$.

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 $PA = nPB$ $(n \neq 0, 1)$. If the locus of $P$ is a circle,then the circle:

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