question_answer

If the area of an equilateral triangle inscribed in the circle \[{{x}^{2}}+{{y}^{2}}+10x+12y+c=0\] is \[27\sqrt{3}\] sq units then c is equal to [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

A) 20                                

B) 25    

C) -25       

D)                  13

Correct Answer: B

Solution :

\[C\left( -5,\,\,-6 \right)\] \[r=\sqrt{25+36-c}\]   \[\cos 30{}^\circ \,\,=\,\,\frac{x}{r}\] \[\frac{\sqrt{3}}{2}\,=\,\frac{x}{r}\] \[\Rightarrow \,\,x\,\,=\,\,\frac{\sqrt{3}r}{2}\,\] side of triangle \[=\,\,\sqrt{3}r\] \[Area\,\,=\,\,\frac{\sqrt{3}}{4}{{(\sqrt{3}r)}^{2}}\] \[\frac{3\sqrt{3}}{4}{{r}^{2}}\,=\,27\sqrt{3}\] \[\Rightarrow \,\,\,{{r}^{2}}\,\,=\,\,36\] \[\Rightarrow \,\,\,25+36-c=36\] \[\Rightarrow \,\,\,c=25\]