Question

Let S be the circle in the xy-plane defined by the equation ${{x}^{2}}+{{y}^{2}}=4$. Let ${{E}_{1}}{{E}_{2}}$and ${{F}_{1}}{{F}_{2}}$ be the chord of S passing through the point ${{P}_{o}}$(1, 1) and parallel to the x-axis and the y-axis, respectively. Let ${{G}_{1}}{{G}_{2}}$ be the chord of the S passing though ${{P}_{o}}$and having slope -1. Let the tangents to S at ${{E}_{1}}$and ${{E}_{2}}$ meet at ${{E}_{3}}$, the tangents of S at ${{G}_{1}}$and ${{G}_{2}}$meet at ${{G}_{3}}$. Then, the points ${{E}_{3}}$, ${{F}_{3}}$and ${{G}_{3}}$lie on the curve.
(a) x+y=4
(b)${{\left( x-4 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=16$
(c) (x-4)(y-4)=4
(d) xy=4

Answer 1

Hint: First, before proceeding for this, we must draw the diagram of the circle with equation ${{x}^{2}}+{{y}^{2}}=4$ and all the conditions. Then, by using the conditions of the chord given we can find the point values of the chord. Then, by using the equation of tangents, we get all the required points.

Complete step by step answer:
In this question, we are supposed to find the points ${{E}_{3}}$, ${{F}_{3}}$and ${{G}_{3}}$lie on the curve when S be the circle in the xy-plane defined by the equation ${{x}^{2}}+{{y}^{2}}=4$and ${{E}_{1}}{{E}_{2}}$and ${{F}_{1}}{{F}_{2}}$ be the chord of S passing through the point ${{P}_{o}}$(1, 1) and parallel to the x-axis and the y-axis, respectively and also ${{G}_{1}}{{G}_{2}}$ be the chord of the S passing though ${{P}_{o}}$and having slope -1 and then the tangents to S at ${{E}_{1}}$and ${{E}_{2}}$ meet at ${{E}_{3}}$, the tangents of S at ${{G}_{1}}$and ${{G}_{2}}$meet at ${{G}_{3}}$.
So, before proceeding for this, we must draw the diagram of the circle with equation ${{x}^{2}}+{{y}^{2}}=4$ and all the conditions as stated above as:

So, we are given in the question that chord ${{E}_{1}}{{E}_{2}}$and ${{F}_{1}}{{F}_{2}}$passes through the point ${{P}_{o}}$(1, 1) which gives y=1 and x=1 for the chords ${{E}_{1}}{{E}_{2}}$and ${{F}_{1}}{{F}_{2}}$ respectively.
Now, from the above condition, we get the two equation for the chord ${{E}_{1}}{{E}_{2}}$as ${{x}^{2}}+{{y}^{2}}=4$and (y=1) which in turn gives value of x as:
$\begin{align}
  & {{x}^{2}}+{{1}^{2}}=4 \\
 & \Rightarrow {{x}^{2}}+1=4 \\
 & \Rightarrow {{x}^{2}}=4-1 \\
 & \Rightarrow {{x}^{2}}=3 \\
 & \Rightarrow x=\pm \sqrt{3} \\
\end{align}$
So, we get the points ${{E}_{1}}$as $\left( -\sqrt{3},1 \right)$ and ${{E}_{2}}$ as $\left( \sqrt{3},1 \right)$.
Similarly, from the above condition, we get the two equation for the chord ${{F}_{1}}{{F}_{2}}$as ${{x}^{2}}+{{y}^{2}}=4$and (x=1) which in turn gives value of x as:
$\begin{align}
  & {{1}^{2}}+{{y}^{2}}=4 \\
 & \Rightarrow 1+{{y}^{2}}=4 \\
 & \Rightarrow {{y}^{2}}=4-1 \\
 & \Rightarrow {{y}^{2}}=3 \\
 & \Rightarrow y=\pm \sqrt{3} \\
\end{align}$
So, we get the points ${{F}_{1}}$as $\left( 1,\sqrt{3} \right)$ and ${{F}_{2}}$ as $\left( 1,-\sqrt{3} \right)$.
Now, from the question, we have to use the slope form of the straight line in which slope is given as -1 in the question for ${{G}_{1}}{{G}_{2}}$ and passing through point ${{P}_{o}}$(1, 1) as:
$\begin{align}
  & \left( y-1 \right)=-1\left( x-1 \right) \\
 & \Rightarrow y-1=-x+1 \\
 & \Rightarrow x+y=2 \\
\end{align}$
So, as chords ${{F}_{1}}{{F}_{2}}$and ${{E}_{1}}{{E}_{2}}$, we get the two equations for ${{G}_{1}}{{G}_{2}}$as ${{x}^{2}}+{{y}^{2}}=4$and x+y=2 which in turn gives as:
$\begin{align}
  & {{x}^{2}}+{{\left( 2-x \right)}^{2}}=4 \\
 & \Rightarrow {{x}^{2}}+4+{{x}^{2}}-4x=4 \\
 & \Rightarrow 2{{x}^{2}}-4x=0 \\
 & \Rightarrow 2x\left( x-2 \right)=0 \\
\end{align}$
So, we get two values of x for chord ${{G}_{1}}{{G}_{2}}$as:
$\begin{align}
  & 2x=0 \\
 & \Rightarrow x=0 \\
\end{align}$ or $\begin{align}
  & x-2=0 \\
 & \Rightarrow x=2 \\
\end{align}$
Now, by substituting the value of x as 0 in equation of chord ${{G}_{1}}{{G}_{2}}$, we get corresponding value of y as:
$\begin{align}
  & 0+y=2 \\
 & \Rightarrow y=2 \\
\end{align}$
Similarly, by substituting the value of x as 2 in equation of chord ${{G}_{1}}{{G}_{2}}$, we get corresponding value of y as:
$\begin{align}
  & 2+y=2 \\
 & \Rightarrow y=0 \\
\end{align}$
So, we get the points ${{G}_{1}}$as $\left( 0,2 \right)$ and ${{G}_{2}}$ as $\left( 2,0 \right)$.
Now, by using the equation of tangents for ${{E}_{1}}$and ${{E}_{2}}$, we get:
$\begin{align}
  & -\sqrt{3}x+y=4 \\
 & \sqrt{3}x+y=4 \\
\end{align}$
So, from the above equations, we get the point of intersection ${{E}_{3}}$as (0, 4).
Similarly, by using the equation of tangents for ${{F}_{1}}$and ${{F}_{2}}$, we get:
$\begin{align}
  & x+\sqrt{3}y=4 \\
 & x-\sqrt{3}y=4 \\
\end{align}$
So, from the above equations, we get the point of intersection ${{F}_{3}}$as (4, 0).
Now, by using the equation of tangents for ${{G}_{1}}$and ${{G}_{2}}$, we get:
$\begin{align}
  & 0x+2y=4 \\
 & 2x+0y=4 \\
\end{align}$
So, from the above equations, we get the point of intersection ${{G}_{3}}$as (2, 2).
So, we can clearly see from the options that points (0, 4) , (4, 0) and (2, 2) lies on the line x+y=4.
Hence, option (a) is correct.

Note:
Now, to solve these types of questions we need to know some of the basics of the equation of the straight line in slope form and equation of tangent. So, both equation is given by:
The equation of line in slope form is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ where m is slope and $\left( {{x}_{1}},{{y}_{1}} \right)$is the point where line passes.
Equation of tangent of line is $x{{x}_{1}}+y{{y}_{1}}={{a}^{2}}$ where $\left( {{x}_{1}},{{y}_{1}} \right)$is the point where the tangent passes.