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JEE Advance 2008
paper 1
Question
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 PQ .
The equation of circle C is
Select the correct option:
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
$(x-\sqrt{3})^2+(y+1)^2=1$
D
$(x-\sqrt{3})^2+(y-1)^2=1$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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