Let T be the line passing through the points P(-2, 7) and Q(2, -5). Let ${{F}_{1}}$ be the set of all pair of circles $\left( {{S}_{1}},{{S}_{2}} \right)$ such that T is the tangent to ${{S}_{1}}$ at P and tangent to ${{S}_{2}}$at Q, and also such that ${{S}_{1}}$ and ${{S}_{2}}$ touch each other at point, say M. Let ${{E}_{1}}$ be the set representing the locus of M as the pair $\left( {{S}_{1}},{{S}_{2}} \right)$ varies in ${{F}_{1}}$. Let the set of all straight line segments joining a pair of distinct points of ${{E}_{1}}$ and passing through the point R(1, 1) be ${{F}_{2}}$. Let ${{E}_{2}}$ be the set of midpoints of the line segments in the set ${{F}_{2}}$. Then, which one of the following statements is/are true?
(a) The point (-2, 7) lies in${{E}_{1}}$
(b) The point $\left( \dfrac{4}{5},\dfrac{7}{5} \right)$does not lie in ${{E}_{2}}$
(c) The point $\left( \dfrac{1}{2},1 \right)$ lies in ${{E}_{2}}$
(d) The point $\left( 0,\dfrac{3}{2} \right)$ does not lie in${{E}_{1}}$
Answer
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Hint: First, before proceeding with this, we must draw a general diagram of the given condition so that we can understand the question easily by considering ${{C}_{1}},{{r}_{1}}$ being the center and radius of the first circle and ${{C}_{2}},{{r}_{2}}$ being the center and radius of the second circle. Then, by using the angle sum property of the triangle which gives that sum of all the angles of the triangle is ${{180}^{\circ }}$ for triangle PMQ. Then, we get the angle $\angle PMQ$ as ${{90}^{\circ }}$ which means that triangle PMQ forms a semicircle with center at M and diameter points as P and Q and by using the equation of circle by diameter points as we get the desired result.
Complete step-by-step solution:
In this question, we are supposed to find the correct statements when T is the line passing through the points P(-2, 7) and Q(2, -5). Let ${{F}_{1}}$ be the set of all pair of circles $\left( {{S}_{1}},{{S}_{2}} \right)$ such that T is the tangent to ${{S}_{1}}$ at P and tangent to ${{S}_{2}}$ at Q, and also such that ${{S}_{1}}$ and ${{S}_{2}}$ touch each other at point, say M. Let ${{E}_{1}}$ be the set representing the locus of M as the pair $\left( {{S}_{1}},{{S}_{2}} \right)$varies in ${{F}_{1}}$. Let the set of all straight line segments joining a pair of distinct points of ${{E}_{1}}$ and passing through the point R(1, 1) be ${{F}_{2}}$. Let ${{E}_{2}}$ be the set of midpoints of the line segments in the set ${{F}_{2}}$.
So, before proceeding with this, we must draw a general diagram of the given condition so that we can understand the question easily by considering ${{C}_{1}},{{r}_{1}}$ being the center and radius of the first circle and ${{C}_{2}},{{r}_{2}}$ being the center and radius of the second circle as:
Also, we have considered the angle made by PM line in the triangle $PM{{C}_{1}}$ as $\theta $ and similarly angle made by QM in the triangle $QM{{C}_{2}}$ be $\phi $.
Then, we can see that in triangle $PM{{C}_{1}}$, sides ${{C}_{1}}M$ and $P{{C}_{1}}$ are the radius of the circle which gives us:
$\angle PM{{C}_{1}}=\theta $
Similarly, we can see that in triangle $QM{{C}_{2}}$, sides ${{C}_{2}}M$ and $Q{{C}_{2}}$ are the radius of the circle which gives us:
$\angle QM{{C}_{2}}=\phi $
Then, we can see clearly that ${{C}_{1}}{{C}_{2}}$ is the straight line which makes the total angle of ${{180}^{\circ }}$.
So, by using this, we get the value of $\angle PMQ$ as:
$\begin{align}
& \angle PMQ+\angle PM{{C}_{1}}+\angle QM{{C}_{2}}={{180}^{\circ }} \\
& \Rightarrow \angle PMQ+\theta +\phi ={{180}^{\circ }} \\
& \Rightarrow \angle PMQ={{180}^{\circ }}-\theta -\phi \\
\end{align}$
Also, we need to get the other two angles of the triangle PMQ by using the figure as:
$\begin{align}
& \theta +\angle MPQ={{90}^{\circ }} \\
& \Rightarrow \angle MPQ={{90}^{\circ }}-\theta \\
\end{align}$
Similarly, for other angle also by using the figure, we get:
$\begin{align}
& \phi +\angle MQP={{90}^{\circ }} \\
& \Rightarrow \angle MQP={{90}^{\circ }}-\phi \\
\end{align}$
Now, by using the angle sum property of the triangle which gives that sum of all the angles of the triangle is ${{180}^{\circ }}$.
So, by applying the same property to triangle PMQ, we get:
$\begin{align}
& \angle PMQ+\angle PQM+\angle MPQ={{180}^{\circ }} \\
& \Rightarrow {{180}^{\circ }}-\theta -\phi +{{90}^{\circ }}-\phi +{{90}^{\circ }}-\theta ={{180}^{\circ }} \\
& \Rightarrow 2\left( \theta +\phi \right)={{180}^{\circ }} \\
& \Rightarrow \left( \theta +\phi \right)={{90}^{\circ }} \\
\end{align}$
So, we get the angle $\angle PMQ$ as ${{90}^{\circ }}$ which means that triangle PMQ forms a semicircle with centre at M and diameter points as P and Q.
Now, we also know that the equation of the circle with diameter points $\left( {{x}_{1}},{{x}_{2}} \right)$ and $\left( {{y}_{1}},{{y}_{2}} \right)$ is given by :
$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$
So, by using the diameter points P(-2, 7) and Q(2, -5), we get the equation of circle as:
$\begin{align}
& \left( x+2 \right)\left( x-2 \right)+\left( y-7 \right)\left( y+5 \right)=0 \\
& \Rightarrow {{x}^{2}}-4+{{y}^{2}}-2y-35=0 \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}-2y-39=0 \\
\end{align}$
So, we get the locus of ${{E}_{1}}$ by the expression as ${{x}^{2}}+{{y}^{2}}-2y-39=0$.
Now, ${{E}_{2}}$ be the mid points of the points given in the question (1, 1) and the above equation centre (0, 1) and we get the locus of ${{E}_{2}}$ as:
$\begin{align}
& \left( x-1 \right)\left( x-0 \right)+\left( y-1 \right)\left( y-1 \right)=0 \\
& \Rightarrow {{x}^{2}}-x+{{y}^{2}}-2y+1=0 \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}-x-2y+1=0 \\
\end{align}$
Hence, options (a) and (d) are correct.
Note: Now, to solve these types of questions we need to know some of the basics of the identities beforehand to solve the expression easily. So, the required expressions are as follows:
$\begin{align}
& {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
& \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} \\
\end{align}$
Also we must know the angle sum property of the triangle which states that sum of all the angle of a triangle is ${{180}^{\circ }}$ and straight line makes a total angle of ${{180}^{\circ }}.
Complete step-by-step solution:
In this question, we are supposed to find the correct statements when T is the line passing through the points P(-2, 7) and Q(2, -5). Let ${{F}_{1}}$ be the set of all pair of circles $\left( {{S}_{1}},{{S}_{2}} \right)$ such that T is the tangent to ${{S}_{1}}$ at P and tangent to ${{S}_{2}}$ at Q, and also such that ${{S}_{1}}$ and ${{S}_{2}}$ touch each other at point, say M. Let ${{E}_{1}}$ be the set representing the locus of M as the pair $\left( {{S}_{1}},{{S}_{2}} \right)$varies in ${{F}_{1}}$. Let the set of all straight line segments joining a pair of distinct points of ${{E}_{1}}$ and passing through the point R(1, 1) be ${{F}_{2}}$. Let ${{E}_{2}}$ be the set of midpoints of the line segments in the set ${{F}_{2}}$.
So, before proceeding with this, we must draw a general diagram of the given condition so that we can understand the question easily by considering ${{C}_{1}},{{r}_{1}}$ being the center and radius of the first circle and ${{C}_{2}},{{r}_{2}}$ being the center and radius of the second circle as:
Also, we have considered the angle made by PM line in the triangle $PM{{C}_{1}}$ as $\theta $ and similarly angle made by QM in the triangle $QM{{C}_{2}}$ be $\phi $.
Then, we can see that in triangle $PM{{C}_{1}}$, sides ${{C}_{1}}M$ and $P{{C}_{1}}$ are the radius of the circle which gives us:
$\angle PM{{C}_{1}}=\theta $
Similarly, we can see that in triangle $QM{{C}_{2}}$, sides ${{C}_{2}}M$ and $Q{{C}_{2}}$ are the radius of the circle which gives us:
$\angle QM{{C}_{2}}=\phi $
Then, we can see clearly that ${{C}_{1}}{{C}_{2}}$ is the straight line which makes the total angle of ${{180}^{\circ }}$.
So, by using this, we get the value of $\angle PMQ$ as:
$\begin{align}
& \angle PMQ+\angle PM{{C}_{1}}+\angle QM{{C}_{2}}={{180}^{\circ }} \\
& \Rightarrow \angle PMQ+\theta +\phi ={{180}^{\circ }} \\
& \Rightarrow \angle PMQ={{180}^{\circ }}-\theta -\phi \\
\end{align}$
Also, we need to get the other two angles of the triangle PMQ by using the figure as:
$\begin{align}
& \theta +\angle MPQ={{90}^{\circ }} \\
& \Rightarrow \angle MPQ={{90}^{\circ }}-\theta \\
\end{align}$
Similarly, for other angle also by using the figure, we get:
$\begin{align}
& \phi +\angle MQP={{90}^{\circ }} \\
& \Rightarrow \angle MQP={{90}^{\circ }}-\phi \\
\end{align}$
Now, by using the angle sum property of the triangle which gives that sum of all the angles of the triangle is ${{180}^{\circ }}$.
So, by applying the same property to triangle PMQ, we get:
$\begin{align}
& \angle PMQ+\angle PQM+\angle MPQ={{180}^{\circ }} \\
& \Rightarrow {{180}^{\circ }}-\theta -\phi +{{90}^{\circ }}-\phi +{{90}^{\circ }}-\theta ={{180}^{\circ }} \\
& \Rightarrow 2\left( \theta +\phi \right)={{180}^{\circ }} \\
& \Rightarrow \left( \theta +\phi \right)={{90}^{\circ }} \\
\end{align}$
So, we get the angle $\angle PMQ$ as ${{90}^{\circ }}$ which means that triangle PMQ forms a semicircle with centre at M and diameter points as P and Q.
Now, we also know that the equation of the circle with diameter points $\left( {{x}_{1}},{{x}_{2}} \right)$ and $\left( {{y}_{1}},{{y}_{2}} \right)$ is given by :
$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$
So, by using the diameter points P(-2, 7) and Q(2, -5), we get the equation of circle as:
$\begin{align}
& \left( x+2 \right)\left( x-2 \right)+\left( y-7 \right)\left( y+5 \right)=0 \\
& \Rightarrow {{x}^{2}}-4+{{y}^{2}}-2y-35=0 \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}-2y-39=0 \\
\end{align}$
So, we get the locus of ${{E}_{1}}$ by the expression as ${{x}^{2}}+{{y}^{2}}-2y-39=0$.
Now, ${{E}_{2}}$ be the mid points of the points given in the question (1, 1) and the above equation centre (0, 1) and we get the locus of ${{E}_{2}}$ as:
$\begin{align}
& \left( x-1 \right)\left( x-0 \right)+\left( y-1 \right)\left( y-1 \right)=0 \\
& \Rightarrow {{x}^{2}}-x+{{y}^{2}}-2y+1=0 \\
& \Rightarrow {{x}^{2}}+{{y}^{2}}-x-2y+1=0 \\
\end{align}$
Hence, options (a) and (d) are correct.
Note: Now, to solve these types of questions we need to know some of the basics of the identities beforehand to solve the expression easily. So, the required expressions are as follows:
$\begin{align}
& {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
& \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} \\
\end{align}$
Also we must know the angle sum property of the triangle which states that sum of all the angle of a triangle is ${{180}^{\circ }}$ and straight line makes a total angle of ${{180}^{\circ }}.
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