Question

A pair of perpendicular straight lines passing through the origin also passes through the points of intersection of the curve $x^2+y^2=4$ with the line $x+y=a$, then value(s) of $a$ can be

• A. $2$
• B. $3$
• C. $-2$
• D. $-3$

Answer 1

Answer: (A), (C)

$-2$

Required 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 $x+y=a$ can be obtained by homogenisation of the curve with respect to line

$x^2+y^2=4\cdots \left(1\right)\frac{x+y}{a}=1\cdots \left(2\right)$

Homogenising $\left(1\right)$ with respect to $\left(2\right)$
$\Rightarrow x^2+y^2=4\cdot {\left(\frac{x+y}{a}\right)}^2\Rightarrow (1-\frac{4}{a^2})x^2+(1-\frac{4}{a^2})y^2-\frac{8xy}{a^2}=0$

$\because \text{They are perpendicular lines}\Rightarrow 1-\frac{4}{a^2}+1-\frac{4}{a^2}=0\Rightarrow a^2=4\therefore a=\pm 2$