Question

If locus of $P$ is a circle, then the circle

• A. passes through $A$ and $B$
• B. never passes through $A$ and $B$
• C. passes through $A$ but does not pass through $B$
• D. passes through $B$ but does not pass through $A$

Answer 1

Answer: (B)

never passes through $A$ and $B$

$\because PA=n\left(PB\right),$
Let $P\equiv (x,y)$
${\left(PA\right)}^2=n^2{\left(PB\right)}^2$
$\Rightarrow {(x-a)}^2+y^2=n^2({(x+a)}^2+y^2)$ ...............$\left(1\right)$
Expanding, we get$\Rightarrow x^2+a^2-2ax+y^2=n^2(x^2+a^2+2ax+y^2)$
Taking common, we get$\Rightarrow (n^2-1)x^2+(n^2-1)y^2+(n^2+1)2ax+(n^2-1)a^2=0$
Dividing the above equation by $(n^2-1)$ we get
$\Rightarrow x^2+y^2+\left(\frac{n^2+1}{n^2-1}\right)2ax+a^2=0$ .........$\left(2\right)$
Let $S=x^2+y^2+\left(\frac{n^2+1}{n^2-1}\right)2ax+a^2=0$
Put $x=a$
$\therefore S_A=a^2+0+\left(\frac{n^2+1}{n^2-1}\right)2a^2+a^2=0$
$=2a^2(1+\frac{n^2+1}{n^2-1})$
$=\frac{4a^2{n}^2}{n^2-1}>0\ne 0$ for$n>1$
$=\frac{4a^2{n}^2}{n^2-1}<0\ne 0$ for$0<n<1$
Put $x=-a$ then $S_B=a^2+0+\left(\frac{n^2+1}{n^2-1}\right)(-2a^2)+a^2=0$
$=2a^2(1-\frac{n^2+1}{n^2-1})$
$=-\frac{4a^2}{n^2-1}<0\ne 0$ for $n>1$
$=-\frac{4a^2}{n^2-1}>0\ne 0$ for $0<n<1$
Thus, $S_A\ne 0,S_B\ne 0$
Hence, the circle never passes through $A$ and $B$