A pair of perpendicular lines passes through the origin and also through the points of intersection of the…
A pair of perpendicular lines passes through the origin and also through the points of intersection of the curve $x^2+y^2=4$ with $x+y=a$, where $a>0$. Then $a$ is equal to
2
3
4
5
Solution
The intersection point of the curve $x^2+y^2=4$ with $x+y=a$, where $(a>0)$
$x^2+(a-x)^2=4$
$x^2+a^2+x^2-2 a x=4$
$2 x^2-2 a x+\left(a^2-4\right)=0$
$x^2-a x+\left(\frac{a^2}{2}-2\right)=0$
$x=\frac{+a \pm \sqrt{a^2-4\left(\frac{a^2}{2}-2\right)}}{2}$
$x=\frac{a \pm \sqrt{a^2-2 a^2+8}}{2}=\frac{a \pm \sqrt{8-a^2}}{2}$
Since, here the point of intersection should be real number, iff
$8-a^2 \geq 0$
ie, $\quad a^2 \leq 8, a \leq 2 \sqrt{2}$
$a \leq 2.82$
Hence, the value is $a=2$ according to option.