If $O = (0, 0), A = (1, 0)$ and $B = (\frac{1}{2}, \frac{\sqrt{3}}{2})$ then centre of the circle for which the lines OA, OB and AB are tangents is?

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

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

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

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

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Solution

The correct option is A $(\frac{1}{2}, \frac{1}{2 \sqrt{3}})$
$\begin{matrix} I = (\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}) a = \sqrt{\frac{1}{4}, \frac{3}{4}} = 1 b = 1, c = 1 & a = b = c = 1 i s e q u i l a t e r a l S o, i t s c e n t r e i s (\frac{1 + \frac{1}{2} + 0}{3}, \frac{0 + \frac{\sqrt{3}}{2} + 0}{3}) \Rightarrow (\frac{1}{2}, \frac{1}{2 \sqrt{3}}) \end{matrix}$

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