Q38 / 69 · JEE Advanced 2008 (Paper 1) · Mathematics
Paragraph:
A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation
$\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.Question:
The equation of circle $C$ is
- A$(x-2 \sqrt{3})^2+(y-1)^2=1$
- B$(x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1$
- C$(x-\sqrt{3})^2+(y+1)^2=1$
- Correct answerD$(x-\sqrt{3})^2+(y-1)^2=1$
Solution
Let centre of circle $C$ be $(h, k)$.
Then, $\left|\frac{\sqrt{3} h+k-6}{\sqrt{3+1}}\right|=1$
$$
\Rightarrow \quad \sqrt{3} h+k-6=2,-2
$$
$$
\Rightarrow \quad \sqrt{3} h+k=4
$$
(Rejecting 2 because origin and centre of $C$ are on the same side of $P Q$ ) The point $(\sqrt{3}, 1)$ satisfies Eq. (i).
$\therefore$ Equation of circle $C$ is $(x-\sqrt{3})^2+(y-1)^2=1$.
Hence, the correct answer is option: 'D'.