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