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 $P Q$. The equation of circle $C$ is

(D) $(x-\sqrt{3})^{2}+(y-1)^{2}=1$

(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) $(x-\sqrt{3})^{2}+(y+1)^{2}=1$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Circle

A circle

C

of radius 1 is inscribed in an equilateral triangle

P QR

. The points of contact of

C

with the sides

P Q, QR, R P

are

D, E, F

, respectively. The line

P Q

is given by the equation

3 x + y - 6 = 0

and the point

D

(\frac{3 3}{2}, \frac{3}{2})

. Further, it is given that the origin and the centre of

C

are on the same side of the line

P Q

.

The equation of circle

C

Answer: (D)

The line

P Q

has equation

3 x + y - 6 = 0

, which can be written as

y = - 3 x + 6

.

The center

C

of the inscribed circle is perpendicular distance 1 from

P Q

(the radius). The perpendicular from

D = (\frac{3 3}{2}, \frac{3}{2})

to the center has direction vector proportional to the normal of

P Q

, which is

(3, 1)

.

The unit normal vector is

\frac{1}{2} (3, 1)

.

Since the origin

(0, 0)

and center are on the same side of

P Q

Moving from

D

perpendicular to

P Q

by distance 1 in the direction toward the origin:

C = D - 1 \cdot \frac{( 3 , 1 )}{2} = (\frac{3 3}{2}, \frac{3}{2}) - (\frac{3}{2}, \frac{1}{2}) = (3, 1)

Therefore, the equation of circle

C

is:

(x - 3)^{2} + (y - 1)^{2} = 1

Answer: (D)

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option