Q. Area of the equilateral triangle inscribed in the circle $ {{x}^{2}}+{{y}^{2}}-7x+9y+5=0 $ is
KEAMKEAM 2009
Solution:
Given, $ {{x}^{2}}-{{y}^{2}}-7x+9y+5=0 $
$ \therefore $ $ R=\sqrt{{{\left( \frac{-7}{2} \right)}^{2}}+{{\left( \frac{9}{2} \right)}^{2}}-5} $
$=\sqrt{\frac{49}{4}+\frac{81}{4}-5}=\frac{\sqrt{110}}{2} $
In $ \Delta OAB, $ $ \cos 30{}^\circ =\frac{AB}{R} $
$ \Rightarrow $ $ \frac{\sqrt{3}}{2}=\frac{AB}{\sqrt{110/2}} $
$ \Rightarrow $ $ \frac{\sqrt{330}}{4}=AB $
$ \therefore $ Length $ AC=\frac{\sqrt{330}}{2} $
$ \therefore $ Area of equilateral $ \Delta =\frac{\sqrt{3}}{4}{{(a)}^{2}} $
$=\frac{\sqrt{3}}{4}\times \frac{330}{4}=\frac{165\sqrt{3}}{8} $ sq unit
