Question

The pair of straight lines joining the origin to the points of intersection of the line $y=2\sqrt{2}x+c$ and the circle ${{x}^{2}}+{{y}^{2}}=2$ are at right angles, if
1) ${{c}^{2}}-4=0$
2) ${{c}^{2}}-8=0$
3) ${{c}^{2}}-9=0$
4) ${{c}^{2}}-10=0$

Answer 1

Hint: In this problem we need to find the condition for the constant $c$ such that the given statement will be valid. Given equation of the line is $y=2\sqrt{2}x+c$ and the equation of the circle is ${{x}^{2}}+{{y}^{2}}=2$. We will first calculate the homogenous equation of the given line and circle by using mathematical operations. After having the homogeneous equation we will compare it with the standard equation of pair of straight lines which is given by $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$. We know that if the straight lines in a pair of straight lines are perpendicular to each other, then the sum of $a$ and $b$ will be zero. We will use this condition and simplify the equation to get the required solution.

Complete step-by-step solution:
The equation of the line is $y=2\sqrt{2}x+c$.
The equation of the circle is ${{x}^{2}}+{{y}^{2}}=2$.
Consider the equation of the line and it can be written as
$\begin{align}
  & y=2\sqrt{2}x+c \\
 & \Rightarrow y-2\sqrt{2}x=c \\
 & \Rightarrow \dfrac{y-2\sqrt{2}x}{c}=1......\left( \text{i} \right) \\
\end{align}$
Consider the equation of the circle and it can be modified as
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}=2 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-2=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-2{{\left( 1 \right)}^{2}}=0 \\
\end{align}$
From equation $\left( \text{i} \right)$ substitute the value of $1$ in the above equation, then we will have
${{x}^{2}}+{{y}^{2}}-2{{\left( \dfrac{y-2\sqrt{2}x}{c} \right)}^{2}}=0$
Applying the formulas ${{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{a}^{n}}}$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ in the above equation, then we will get
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}-2\times \dfrac{{{\left( y-2\sqrt{2}x \right)}^{2}}}{{{c}^{2}}}=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-\dfrac{2\left( {{y}^{2}}+{{\left( 2\sqrt{2}x \right)}^{2}}-2\left( 2\sqrt{2}x \right)\left( y \right) \right)}{{{c}^{2}}}=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-\dfrac{2\left( {{y}^{2}}+8{{x}^{2}}-4\sqrt{2}xy \right)}{{{c}^{2}}}=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-\dfrac{2{{y}^{2}}+16{{x}^{2}}-8\sqrt{2}xy}{{{c}^{2}}}=0 \\
\end{align}$
Simplifying the above equation by using LCM, then we will have
$\begin{align}
  & \dfrac{{{c}^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)-\left( 2{{y}^{2}}+16{{x}^{2}}-8\sqrt{2}xy \right)}{{{c}^{2}}}=0 \\
 & \Rightarrow {{c}^{2}}{{x}^{2}}+{{c}^{2}}{{y}^{2}}-2{{y}^{2}}-16{{x}^{2}}+8\sqrt{2}xy=0\left( {{c}^{2}} \right) \\
 & \Rightarrow {{c}^{2}}{{x}^{2}}-16{{x}^{2}}+8\sqrt{2}xy+{{c}^{2}}{{y}^{2}}-2{{y}^{2}}=0 \\
\end{align}$
Take appropriate terms as common from the above equation, then we will get
$\left( {{c}^{2}}-16 \right){{x}^{2}}+8\sqrt{2}xy+\left( {{c}^{2}}-2 \right){{y}^{2}}=0$
Comparing the above equation with standard pair of equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$, then we will get
$a={{c}^{2}}-16$ and $b={{c}^{2}}-2$
If the lines joining origin and point of intersections of the given line and circle are perpendicular to each other, then the sum of $a$ and $b$ will be zero. So,
$a+b=0$
Substituting the values $a={{c}^{2}}-16$ and $b={{c}^{2}}-2$ in the above equation, then we will have
$\begin{align}
  & {{c}^{2}}-16+{{c}^{2}}-2=0 \\
 & \Rightarrow 2{{c}^{2}}-18=0 \\
\end{align}$
Dividing the above equation with $2$ on both sides of the above equation, then we will get
$\therefore {{c}^{2}}-9=0$
Hence option 3 is the correct answer.

Note: In this problem they have mentioned that the given pair of straight lines are passing through the origin. So we have assumed the homogeneous equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$. If they have mentioned that the pair of straight lines doesn’t pass through the origin, then we will consider the nonhomogeneous equation which is given by $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ .