question_answer

Area of the equilateral triangle inscribed in the circle \[{{x}^{2}}+{{y}^{2}}-7x+9y+5=0\] is

A) \[\frac{155}{8}\sqrt{3}\] square units

B) \[\frac{165}{8}\sqrt{3}\] square units

C) \[\frac{175}{8}\sqrt{3}\] square units

D) \[\frac{165}{8}\sqrt{3}\] square units

Correct Answer: D

Solution :

[d] Given circle : \[{{x}^{2}}+{{y}^{2}}-7x+9y+5=0\]

\[\therefore \] Centre \[=\left( \frac{7}{2},\frac{-9}{2} \right)\]

Radius \[=\sqrt{\frac{49}{4}+\frac{81}{4}-5}=\frac{\sqrt{110}}{2}\]

Since \[\Delta ABC\] is an equilateral

\[\therefore \angle MAL=30{}^\circ ,\angle MLA=90{}^\circ \]

Also \[MA=\frac{\sqrt{110}}{2}\]

\[\therefore AL=MA\cos 30{}^\circ =\frac{\sqrt{110}}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{330}}{4}\]

\[\therefore \] Side of \[\Delta =2.AL=\frac{\sqrt{330}}{2}\]

Area of equilateral \[\Delta =\frac{\sqrt{3}}{4}{{a}^{2}}=\frac{\sqrt{3}}{4}\times \frac{330}{4}\]

\[=\frac{165}{8}\sqrt{3}\] sq. units