Question

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?

• A. $(\frac{1}{2},\frac{1}{2\sqrt{3}})$
• B. $(\frac{1}{2},\frac{1}{\sqrt{3}})$
• C. $(\frac{1}{\sqrt{3}},\frac{1}{2})$
• D. $(\frac{1}{2\sqrt{3}},\frac{1}{2})$

Answer 1

The correct option is A $(\frac{1}{2},\frac{1}{2\sqrt{3}})$

$\begin{array}{cc}I=\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)& \\ a=\sqrt{\frac{1}{4},\frac{3}{4}}=1& \\ b=1,c=1& a=b=c=1isequilateral\\ So,itscentreis& \\ \left(\frac{1+\frac{1}{2}+0}{3},\frac{0+\frac{\sqrt{3}}{2}+0}{3}\right)& \\ \Rightarrow \left(\frac{1}{2},\frac{1}{2\sqrt{3}}\right)& \end{array}$