If $P(0,0), Q(1,0)$ and $R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ are three given points, then the centre of the circle for which the lines $P Q, Q R$ and $R P$ are the tangents is

• A

  $\left(\frac{1}{2}, \frac{1}{4}\right)$
• B

  $\left(\frac{1}{2}, \frac{\sqrt{3}}{4}\right)$
• Correct answer
  C

  $\left(\frac{1}{2}, \frac{1}{2 \sqrt{3}}\right)$
• D

  $\left(\frac{1}{2}, \frac{-1}{\sqrt{3}}\right)$

Solution

$\begin{aligned} &\text { } I=\left(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}\right)\\ &a=\sqrt{\frac{1}{4}+\frac{3}{4}}=1, b=1, c=1\\ &\text { Centre of the circle. }\\ &=\left(\frac{1 \times 0+1 \times 1+1 \times \frac{1}{2}}{1+1+1}, \frac{1 \times 0+1 \times 0+1 \times \sqrt{3} / 2}{1+1+1}\right)\\ &=\left(\frac{0+1+\frac{1}{2}}{3} \frac{0+0+\frac{\sqrt{3}}{2}}{3}\right)\\ &=\left(\frac{3}{\frac{2}{3}}, \frac{\frac{\sqrt{3}}{2}}{3}\right) \end{aligned}$ $=\left(\frac{3}{2} \times \frac{1}{3}, \frac{\sqrt{3}}{2} \times \frac{1}{3}\right)$ $=\left(\frac{1}{2}, \frac{1}{2 \sqrt{3}}\right)$

Hence, the correct answer is option: 'C'.