If an equilateral triangle of side $a$ is inscribed in the circle $x^{2} + y^{2} - 6 x - 4 y + 5 = 0,$ then the value of $a^{2}$ is

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Solution

Given : $x^{2} + y^{2} - 6 x - 4 y + 5 = 0$
Center $= (3, 2)$ ,
Radius $= \sqrt{9 + 4 - 5} = \sqrt{8} = 2 \sqrt{2}$

As $△ P Q R$ is equilateral triangle, so $C$ is orthocentre, we get

$P M = P C + C M \Rightarrow a sin 60^{\circ} = r + \frac{r}{2} (∵ \frac{P M}{P Q} = sin 60^{\circ}) \Rightarrow a = \frac{2}{\sqrt{3}} \times \frac{3 r}{2} \Rightarrow a = \sqrt{3} r = 2 \sqrt{6} ∴ a^{2} = 24$

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