Question

The equation of the circle which is touched by $y=x$, has its centre on the positive direction of the $x$-axis and cuts off a chord of length $2$ units along the line $\sqrt{3}y-x=0$, is

• A. $x^2+y^2-4x+2=0$
• B. $x^2+y^2-3y+2=0$
• C. $x^2+y^2-4x-3y+2=0$
• D. $x^2+y^2=2$

Answer 1

The correct option is A $x^2+y^2-4x+2=0$

Since the required circle has its centre on positive $x$-axis, so, let the coordinates of the centre be $C(a,0)$, where $a>0$.

The circle touches $y=x$.
Therefore, Radius $=$ length of the $\perp$ from $(a,0)$ on the line $x-y=0$,
$\therefore r=\frac{a}{\sqrt{2}}$.
Also the circle cuts off a chord of length $2$ units along the line $x-\sqrt{3}y=0$

length of the perpendicular from $(a,0)$ on the line $x-\sqrt{3}y=0,$
$d=\frac{a}{2}$
Hence, length of the chord $=2\sqrt{r^2-d^2}$
$2=2\sqrt{\frac{a^2}{2}-\frac{a^2}{4}}\Rightarrow a=2(\because a>0)$
Thus, centre of the circle is at $(2,0)$ and radius $=\frac{a}{\sqrt{2}}=\sqrt{2}$
Hence, equation of required circle is
$(x-2{)}^2+(y-0{)}^2=(\sqrt{2}{)}^2$
$\Rightarrow x^2+y^2-4x+2=0$