Question

A chord $ax+y=1$ subtends ${90}^{\circ }$ at the centre of the circle $x^2+y^2=\frac{3}{2}.$ Then the value of $a$ is

• A. $\pm \frac{1}{\sqrt{2}}$
• B. $\pm \frac{1}{\sqrt{3}}$
• C. $\pm \sqrt{2}$
• D. $\pm \sqrt{3}$

Answer 1

Answer: (B)

equation of chord: $ax+y=1$

$\Rightarrow (ax+y{)}^2=1^2$

Now, given circle is $x^2+y^2=\frac{3}{2}$

$\Rightarrow x^2+y^2=\frac{3}{2}(1{)}^2$
By homogenization,
$x^2+y^2=\frac{3}{2}(ax+y{)}^2$ [From (1)]
$\Rightarrow 2x^2+2y^2=3(ax+y{)}^2$
$\Rightarrow (3a^2-2)x^2+y^2+6axy=0$
This pair of straight lines subtends ${90}^{\circ }$ at the origin.
$\Rightarrow \text{Coefficient of}{x}^2+\text{Coefficient of}{y}^2=0$

$\Rightarrow 3a^2-2+1=0$

$\Rightarrow 3a^2-1=0$

$\Rightarrow 3a^2=1$

$\Rightarrow a^2=\frac{1}{3}$
$\Rightarrow a=\pm \frac{1}{\sqrt{3}}$

Hence, Option B is correct.