Question

The equation of circle passing through the origin and cutting off equal intercepts of $2$ units on the lines $\sqrt{3}{y}^2-\sqrt{3}{x}^2-2xy=0$ is/are

• A. $(x+\sqrt{3})(x-1)+(y-\sqrt{3})(y-1)=0$
• B. $(x+1)(x+\sqrt{3})+(y+\sqrt{3})(y-1)=0$
• C. $(x-\sqrt{3})(x+1)+(y+1)(y+\sqrt{3})=0$
• D. $(x-\sqrt{3})(x-1)+(y+1)(y-\sqrt{3})=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(x+\sqrt{3})(x-1)+(y-\sqrt{3})(y-1)=0$
B $(x+1)(x+\sqrt{3})+(y+\sqrt{3})(y-1)=0$
C $(x-\sqrt{3})(x+1)+(y+1)(y+\sqrt{3})=0$
D $(x-\sqrt{3})(x-1)+(y+1)(y-\sqrt{3})=0$



Given : $\sqrt{3}{y}^2-\sqrt{3}{x}^2-2xy=0$
$\Rightarrow (\sqrt{3}y+x)(y-\sqrt{3}x)=0$
$\Rightarrow y-\sqrt{3}x=0$
$\Rightarrow y=\sqrt{3}x\cdots \left(1\right)$
and $\sqrt{3}y+x=0$
$\Rightarrow y=-\frac{x}{\sqrt{3}}\cdots \left(2\right)$
Slope of line $\left(1\right),$ $\tan {\theta }_1=\sqrt{3}$
$\Rightarrow {\theta }_1={60}^{\circ }$
$\therefore A\equiv (2\cos {60}^{\circ },2\sin {60}^{\circ })$
or $A\equiv (1,\sqrt{3})$

Slope of line $\left(2\right),$ $\tan {\theta }_2=-\frac{1}{\sqrt{3}}$
$\Rightarrow {\theta }_2={150}^{\circ }$
$\therefore B\equiv (2\cos {150}^{\circ },2\sin {150}^{\circ })$
or $B\equiv (-\sqrt{3},1)$
Similarly, $C\equiv (2\cos {240}^{\circ },2\sin {240}^{\circ })$ or $C\equiv (-1,-\sqrt{3})$
and $D\equiv \left(2\cos \right(-30{)}^{\circ },2\sin (-30{)}^{\circ })$ or $D\equiv (\sqrt{3},-1)$

Since both these lines are perpendicular to each other so $AB,BC,CD$ and $AD$ will be diameter of the required circles.
Therefore using diametric form for equation of circles,
with $AB$ as diameter : $(x+\sqrt{3})(x-1)+(y-\sqrt{3})(y-1)=0$
with $BC$ as diameter : $(x+1)(x+\sqrt{3})+(y+\sqrt{3})(y-1)=0$
with $CD$ as diameter : $(x-\sqrt{3})(x+1)+(y+1)(y+\sqrt{3})=0$
with $AD$ as diameter : $(x-\sqrt{3})(x-1)+(y+1)(y-\sqrt{3})=0$