Question

Let $A=\left( a,0 \right)$ and $B=\left( -a,0 \right)$ be two fixed points $\forall a\in \left( -\infty ,0 \right)$ and P moves on a plane such that $PA=nPB$ $\left( n\ne 0 \right).$If $0 < n < 1$,then
a) ‘A’ lies inside the circle and ‘B’ lies outside the circle.
b) ‘A’ lies outside the circle and ‘B’ lies inside the circle.
c) both ‘A’ and ‘B’ lie on the circle.
d) both ‘A’ and ‘B’ lie inside the circle.

Answer 1

We start solving the problem by assuming the coordinates for the point representing the locus. We then use the fact that the distance between the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ and make necessary arrangements to get the equation of the required locus. We then substitute the points A and B to check whether the given points are inside or outside of the circle.

Complete step by step answer:
According to the problem, we have given the points $A=\left( a,0 \right)$ and $B=\left( -a,0 \right)$ are points and the point P moves on a plane such that $PA=nPB$, $\left( n\ne 0 \right)$ and the values of n lies in the interval $0 < n < 1$.
let us assume the point P as $P=\left( h,k \right)$.
According to the problem, we have been given the relation as $PA=nPB$.
We know that the distance between the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$.
So, we get $\sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}=n\left[ \sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}} \right]$.
On squaring both sides, we get as follows
$\Rightarrow {{\left( h-a \right)}^{2}}+{{k}^{2}}={{n}^{2}}\left[ {{\left( h+a \right)}^{2}}+{{k}^{2}} \right]$.
$\Rightarrow {{h}^{2}}+{{a}^{2}}-2ha+{{k}^{2}}={{n}^{2}}\left( {{h}^{2}}+{{a}^{2}}+2ha+{{k}^{2}} \right)$.
Now let us find the locus of $P=\left( h,k \right)$.
To find the locus of $P=\left( h,k \right)$ we need to replace $\left( h,k \right)$ with $\left( x,y \right)$.
Therefore, locus of $P=\left( h,k \right)$ is
$\Rightarrow {{x}^{2}}+{{a}^{2}}-2xa+{{y}^{2}}={{n}^{2}}\left( {{x}^{2}}+{{a}^{2}}+2xa+{{y}^{2}} \right)$.
$\Rightarrow {{x}^{2}}-{{n}^{2}}{{x}^{2}}+{{a}^{2}}-{{n}^{2}}{{a}^{2}}-2xa-{{n}^{2}}(2xa)+{{y}^{2}}-{{n}^{2}}{{y}^{2}}=0$.
On simplifying we get, ${{x}^{2}}(1-{{n}^{2}})+{{a}^{2}}(1-{{n}^{2}})-2xa(1+{{n}^{2}})+{{y}^{2}}(1-{{n}^{2}})=0$.
Let us take the $(1-{{n}^{2}})$term as common then we get
$\Rightarrow {{x}^{2}}+{{y}^{2}}-2xa\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}=0$---(1).
So, we have found the locus of the point $P=\left( h,k \right)$ as a circle with equation$\Rightarrow {{x}^{2}}+{{y}^{2}}-2xa\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}=0$.
Let us substitute the point $A=\left( a,0 \right)$ in equation (1) Let us assume the value be c.
$\Rightarrow c={{a}^{2}}+{{0}^{2}}-2\left( a \right)a\dfrac{\left( 1+{{n}^{2}} \right)}{\left( 1-{{n}^{2}} \right)}+{{a}^{2}}$.
$\Rightarrow c={{a}^{2}}-2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}$.
$\Rightarrow c=2{{a}^{2}}-2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)$.
$\Rightarrow c=2{{a}^{2}}\left( 1-\dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)$.
$\Rightarrow c=2{{a}^{2}}\left( \dfrac{1-{{n}^{2}}-1-{{n}^{2}}}{1-{{n}^{2}}} \right)$.
$\Rightarrow c=2{{a}^{2}}\left( \dfrac{-2{{n}^{2}}}{1-{{n}^{2}}} \right)$.
$\Rightarrow c=\dfrac{-4{{a}^{2}}{{n}^{2}}}{1-{{n}^{2}}}$.
Since $0 < n < 1$, we get $0 < 1-{{n}^{2}} < 1$ and ${{n}^{2}} > 0$. So, this makes $\dfrac{{{n}^{2}}}{1-{{n}^{2}}} > 0$ and $\dfrac{-{{n}^{2}}}{1-{{n}^{2}}} < 0$.
So, we get $c=\dfrac{-4{{a}^{2}}{{n}^{2}}}{1-{{n}^{2}}} < 0$.
Therefore, the point $A=\left( a,0 \right)$ lies inside the circle.
Let us substitute the point $B=\left( -a,0 \right)$ in equation (1)
$\Rightarrow d={{\left( -a \right)}^{2}}+{{0}^{2}}-2\left( -a \right)a\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}$.
$\Rightarrow d={{a}^{2}}+2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}$.
$\Rightarrow d=2{{a}^{2}}+2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)$.
$\Rightarrow d=2{{a}^{2}}\left( 1+\dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)$.
$\Rightarrow d=2{{a}^{2}}\left( \dfrac{1-{{n}^{2}}+1+{{n}^{2}}}{1-{{n}^{2}}} \right)$.
$\Rightarrow d=2{{a}^{2}}\left( \dfrac{2}{1-{{n}^{2}}} \right)$.
$\Rightarrow d=\dfrac{4{{a}^{2}}}{1-{{n}^{2}}} > 0$.
Therefore, the point $B=\left( -a,0 \right)$ lies outside the circle.

So, the correct answer is “Option a”.

Note: We should not confuse the condition of the points lying inside and outside of the circle. We should not make calculation mistakes while solving these problems. Whenever we get this type of problem, we should first try to find the equation of the locus of all points. We should know the general equation of all conics before solving this problem.