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 $n>1$,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’ lies on the circle.
d) both ‘A’ and ‘B’ lies inside the circle.
Answer
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Hint:We start solving the problem by assigning the co-ordinates for the point to represent the locus satisfying the condition $PA=nPB$. 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 $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ for PA and PB. We then make necessary arrangements to get the equation of the required locus. We then substitute the points A and B in the obtained equation of locus to check whether those points are inside or outside of the circle.
Complete step by step answer:
According to the problem, we have given two points \[A=\left( a,0 \right)\] and \[B=\left( -a,0 \right)\] 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 $n>1$.
let us assume the point P as $P=\left( h,k \right)$ to represent all the points present in that locus.
According to the problem, we have been given the relation to $PA=nPB$.
Let us use the distance between the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ to the given relation.
We get $\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=n\left[ \sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}} \right]$. Let us square this on both sides.
${{\left( h+a \right)}^{2}}+{{k}^{2}}={{n}^{2}}\left[ {{\left( h-a \right)}^{2}}+{{k}^{2}} \right]$.
${{h}^{2}}+{{a}^{2}}+2ha+{{k}^{2}}={{n}^{2}}\left( {{h}^{2}}+{{a}^{2}}-2ha+{{k}^{2}} \right)$.
Now let us find the locus of $P=\left( h,k \right)$.
To find the locus of the point $P=\left( h,k \right)$, we replace $\left( h,k \right)$ with $\left( x,y \right)$.
Therefore, locus of the point $P=\left( h,k \right)$ is ${{x}^{2}}+{{a}^{2}}+2xa+{{y}^{2}}={{n}^{2}}\left( {{x}^{2}}+{{a}^{2}}-2xa+{{y}^{2}} \right)$.
\[{{x}^{2}}-{{n}^{2}}{{x}^{2}}+{{a}^{2}}-{{n}^{2}}{{a}^{2}}+2xa+{{n}^{2}}(2xa)+{{y}^{2}}-{{n}^{2}}{{y}^{2}}=0\].
\[{{x}^{2}}(1-{{n}^{2}})+{{a}^{2}}(1-{{n}^{2}})+2xa(1+{{n}^{2}})+{{y}^{2}}(1-{{n}^{2}})=0\].
\[{{x}^{2}}+{{y}^{2}}+2xa\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}=0\]---(1).
So, we have found the locus of the point $P=\left( h,k \right)$ as a circle with equation\[{{x}^{2}}+{{y}^{2}}+2xa\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}=0\].
Let us substitute the point $A=\left( a,0 \right)$ in equation (1) and let us assume the value be p.
\[p={{a}^{2}}+{{0}^{2}}+2\left( a \right)a\dfrac{\left( 1+{{n}^{2}} \right)}{\left( 1-{{n}^{2}} \right)}+{{a}^{2}}\].
\[p={{a}^{2}}+2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}\].
\[p=2{{a}^{2}}+2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[p=2{{a}^{2}}\left( 1+\dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[p=2{{a}^{2}}\left( \dfrac{1-{{n}^{2}}+1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[p=2{{a}^{2}}\left( \dfrac{1+1}{1-{{n}^{2}}} \right)\].
\[p=\dfrac{4{{a}^{2}}}{1-{{n}^{2}}}\].
According to the problem, we have $n>1$ so, we get $1-{{n}^{2}}<0$. Which makes \[\dfrac{1}{1-{{n}^{2}}}<0\].
So, we get \[p=\dfrac{4{{a}^{2}}}{1-{{n}^{2}}}<0\].
So, we have found that the point $A=\left( a,0 \right)$ lies inside the circle.
Let us substitute the point \[B=\left( -a,0 \right)\] in equation (1)
\[d={{\left( -a \right)}^{2}}+{{0}^{2}}+2\left( -a \right)a\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}\].
\[d={{a}^{2}}-2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}\].
\[d=2{{a}^{2}}-2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[d=2{{a}^{2}}\left( 1-\dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[d=2{{a}^{2}}\left( \dfrac{1-{{n}^{2}}-1-{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[d=2{{a}^{2}}\left( \dfrac{-2{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[d=\dfrac{-4{{a}^{2}}{{n}^{2}}}{1-{{n}^{2}}}>0\].
Therefore, the point \[B=\left( -a,0 \right)\] lies outside the circle
Hence, the correct option is (a)
Note:
We should not get confused while mentioning the points whether they are lying inside or outside of the circle. We should not make calculation mistakes while solving this problem. Whenever we get this type of problem, we should first try to find the equation of the locus of all points. We should not confuse with the equation of conics while solving this type of problem.
Complete step by step answer:
According to the problem, we have given two points \[A=\left( a,0 \right)\] and \[B=\left( -a,0 \right)\] 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 $n>1$.
let us assume the point P as $P=\left( h,k \right)$ to represent all the points present in that locus.
According to the problem, we have been given the relation to $PA=nPB$.
Let us use the distance between the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ to the given relation.
We get $\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=n\left[ \sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}} \right]$. Let us square this on both sides.
${{\left( h+a \right)}^{2}}+{{k}^{2}}={{n}^{2}}\left[ {{\left( h-a \right)}^{2}}+{{k}^{2}} \right]$.
${{h}^{2}}+{{a}^{2}}+2ha+{{k}^{2}}={{n}^{2}}\left( {{h}^{2}}+{{a}^{2}}-2ha+{{k}^{2}} \right)$.
Now let us find the locus of $P=\left( h,k \right)$.
To find the locus of the point $P=\left( h,k \right)$, we replace $\left( h,k \right)$ with $\left( x,y \right)$.
Therefore, locus of the point $P=\left( h,k \right)$ is ${{x}^{2}}+{{a}^{2}}+2xa+{{y}^{2}}={{n}^{2}}\left( {{x}^{2}}+{{a}^{2}}-2xa+{{y}^{2}} \right)$.
\[{{x}^{2}}-{{n}^{2}}{{x}^{2}}+{{a}^{2}}-{{n}^{2}}{{a}^{2}}+2xa+{{n}^{2}}(2xa)+{{y}^{2}}-{{n}^{2}}{{y}^{2}}=0\].
\[{{x}^{2}}(1-{{n}^{2}})+{{a}^{2}}(1-{{n}^{2}})+2xa(1+{{n}^{2}})+{{y}^{2}}(1-{{n}^{2}})=0\].
\[{{x}^{2}}+{{y}^{2}}+2xa\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}=0\]---(1).
So, we have found the locus of the point $P=\left( h,k \right)$ as a circle with equation\[{{x}^{2}}+{{y}^{2}}+2xa\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}=0\].
Let us substitute the point $A=\left( a,0 \right)$ in equation (1) and let us assume the value be p.
\[p={{a}^{2}}+{{0}^{2}}+2\left( a \right)a\dfrac{\left( 1+{{n}^{2}} \right)}{\left( 1-{{n}^{2}} \right)}+{{a}^{2}}\].
\[p={{a}^{2}}+2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}\].
\[p=2{{a}^{2}}+2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[p=2{{a}^{2}}\left( 1+\dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[p=2{{a}^{2}}\left( \dfrac{1-{{n}^{2}}+1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[p=2{{a}^{2}}\left( \dfrac{1+1}{1-{{n}^{2}}} \right)\].
\[p=\dfrac{4{{a}^{2}}}{1-{{n}^{2}}}\].
According to the problem, we have $n>1$ so, we get $1-{{n}^{2}}<0$. Which makes \[\dfrac{1}{1-{{n}^{2}}}<0\].
So, we get \[p=\dfrac{4{{a}^{2}}}{1-{{n}^{2}}}<0\].
So, we have found that the point $A=\left( a,0 \right)$ lies inside the circle.
Let us substitute the point \[B=\left( -a,0 \right)\] in equation (1)
\[d={{\left( -a \right)}^{2}}+{{0}^{2}}+2\left( -a \right)a\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}\].
\[d={{a}^{2}}-2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)+{{a}^{2}}\].
\[d=2{{a}^{2}}-2{{a}^{2}}\left( \dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[d=2{{a}^{2}}\left( 1-\dfrac{1+{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[d=2{{a}^{2}}\left( \dfrac{1-{{n}^{2}}-1-{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[d=2{{a}^{2}}\left( \dfrac{-2{{n}^{2}}}{1-{{n}^{2}}} \right)\].
\[d=\dfrac{-4{{a}^{2}}{{n}^{2}}}{1-{{n}^{2}}}>0\].
Therefore, the point \[B=\left( -a,0 \right)\] lies outside the circle
Hence, the correct option is (a)
Note:
We should not get confused while mentioning the points whether they are lying inside or outside of the circle. We should not make calculation mistakes while solving this problem. Whenever we get this type of problem, we should first try to find the equation of the locus of all points. We should not confuse with the equation of conics while solving this type of problem.
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