Question

A pair of perpendicular lines passing 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

Here we are asked to find the value of $a$ using the given data. Since it is given that the perpendicular pair of lines passes through the intersection of the two given curves, we first find the intersection of those curves, and by using the property of the pair of the perpendicular line we will find the value of $a$.

Complete step-by-step solution:
It is given that a pair of perpendicular lines are passing through the origin and the intersection of the curves ${x^2} + {y^2} = 4$and$x + y = a$. We aim to find the value of$a$.
Consider the first given equation ${x^2} + {y^2} = 4$ we can see that this is an equation of a circle with a center at the origin and radius $2$.
Now let us consider the second equation $x + y = a$ let us modify this equation for our convenience.
$x + y = a$ $\Rightarrow \dfrac{{x + y}}{a} = 1$…………..$(1)$
Now let us modify the first given equation to a homogeneous equation using the equation $(1)$
{x^2} + {y^2} = 4$$$$ \Rightarrow {x^2} + {y^2} - 4 = 0
$\Rightarrow {x^2} + {y^2} - 4{(1)^2} = 0$
Now let us substitute the equation $(1)$in the above equation.
$\Rightarrow {x^2} + {y^2} - 4{\left( {\dfrac{{x + y}}{a}} \right)^2} = 0$
$\Rightarrow {x^2} + {y^2} - 4\left( {\dfrac{{{{(x + y)}^2}}}{{{a^2}}}} \right) = 0$
$\Rightarrow {a^2}{x^2} + {a^2}{y^2} - 4{(x + y)^2} = 0$
$\Rightarrow {a^2}{x^2} + {a^2}{y^2} - 4({x^2} + {y^2} + 2xy) = 0$
$\Rightarrow \left( {{a^2} - 4} \right){x^2} + \left( {{a^2} - 4} \right){y^2} - 8xy = 0$
From the given data we know that this is a pair of perpendicular lines the sum of the coefficients of ${x^2}$and ${y^2}$will be equal to zero.
Thus, we get${a^2} - 4 + {a^2} - 4 = 0$. On simplifying this we get
$\Rightarrow 2{a^2} - 8 = 0$
$\Rightarrow 2{a^2} = 8$
$\Rightarrow {a^2} = \dfrac{8}{2}$
$\Rightarrow {a^2} = 4$
$\Rightarrow a = \pm 2$
\Rightarrow a = \left\\{ { - 2,2} \right\\}
Thus, we have found the value of$a$. Now let us see the options for a correct answer.
Option (a) \left\\{ { - 2.2} \right\\}is the correct option since we got the same value in our calculation above.
Option (b) \left\\{ { - 3,3} \right\\}is an incorrect answer as we got \left\\{ { - 2.2} \right\\}as the values of$a$.
Option (c) \left\\{ { - 4,4} \right\\}is an incorrect answer as we got \left\\{ { - 2.2} \right\\}as the values of$a$.
Option (c) \left\\{ { - 5,5} \right\\}is an incorrect answer as we got \left\\{ { - 2.2} \right\\}as the values of$a$.
Hence, option (a) \left\\{ { - 2.2} \right\\}is the correct answer.

Note: We know that the circle of the equation whose center at $(0,0)$ with radius $a$will be ${x^2} + {y^2} = {a^2}$.
Using this we have found that the first given equation is the equation of a circle. Then it is given that a pair of perpendicular lines pass through the intersection of the given curves so, we modified one equation and substituted it in the other.