question_answer

If the incentre of an equilateral triangle is (1, 1) and the equation of its one side is \[3x+4y+3=0,\] then the equation of the circumcircle of this triangle is : [JEE Main Online Paper (Held On 11 April 2015)]  

A) \[{{x}^{2}}+{{y}^{2}}-2x-2y-2=0\]

B) \[{{x}^{2}}+{{y}^{2}}-2x-2y-14=0\]

C) \[{{x}^{2}}+{{y}^{2}}-2x-2y+2=0\]

D) \[{{x}^{2}}+{{y}^{2}}-2x-2y-7=0\]

Correct Answer: B

Solution :

  In an equilateral triangle in centre & circumcenter all same & R = 2r Now, \[r=\frac{\left| 3+4+3 \right|}{\sqrt{9+16}}\] =2 \[\Rightarrow \] \[R=4\] \[\therefore \]Equation of circumcircle is \[{{\left( x-1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=16\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-2x-2y-14=0\]