QJEEAdv08Ma1303: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Ma1303 Solution: The equation of circle C is

Question

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The equation of circle C is

Accepted Answer

Let centre of circle C be (h, k) then,

$| \frac{\sqrt{3} h + k - 6}{\sqrt{3 + 1}} | = 1$

$\sqrt{3} h + k - 6 = 2, - 2$

$\sqrt{3} h + k = 4$ (rejecting 2 because origin and centre of point $(\sqrt{3}, 1)$ , satisfies eq.)

eq. of circle (C) is

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

Clearly point E and F satisfy the equation in given option D.

As ${(x - 2 \sqrt{3})}^{2} + (y - 1)^{2} = 1$ not possible.

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