Question

A pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^2+y^2=4$with $x+y=a$. The set containing the value of $a$is

• A. $\left\{-2,2\right\}$
• B. $\left\{-3,3\right\}$
• C. $\left\{-4,4\right\}$
• D. $\left\{-5,5\right\}$

Answer 1

The correct option is A

$\left\{-2,2\right\}$

Explanation for correct option:

Step1. Given the equation of curve and line.

Given curve is $x^2+y^2=4\dots \left(i\right)$

Given line is $\begin{array}{rcl}x+y& =& a\end{array}$

$\Rightarrow \frac{x+y}{a}=1\begin{array}{rcl}& & \dots \left(ii\right)\end{array}$

The line will intersect the circle at $2$points.

We make $\left(i\right)$a homogeneous equation by using $\left(ii\right)$.

$$\begin{gathered} \Rightarrow x^2+y^2-4{\left(1\right)}^2=0 \\ \Rightarrow x^2+y^2-4{\left(\frac{x+y}{a}\right)}^2=0 \\ \Rightarrow a^2{x}^2+a^2{y}^2-4\left(x^2+y^2+2xy\right)=0 \\ \Rightarrow x^2\left(a^2-4\right)+y^2\left(a^2-4\right)-8xy=0 \end{gathered}$$

Since this is a perpendicular pair of straight lines,

Step2. To find the set of values of $a$:

Coefficient of $x^2+$coefficient of $y^2=0$

$$\begin{gathered} \Rightarrow (a^2–4)+(a^2–4)=0 \\ \Rightarrow 2a^2–8=0 \\ \Rightarrow a^2=4 \\ \Rightarrow a=-2,2 \end{gathered}$$

So, $a=\left\{-2,2\right\}$

Hence, the correct option is A