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 :

 

  • Option 1)

    x^{2}+y^{2}-2x-2y-2=0

  • Option 2)

    x^{2}+y^{2}-2x-2y-14=0

  • Option 3)

    x^{2}+y^{2}-2x-2y+2=0

  • Option 4)

    x^{2}+y^{2}-2x-2y-7=0

 

Answers (1)

As learnt in concept

Perpendicular distance of a point from a line -

\rho =\frac{\left | ax_{1}+by_{1}+c\right |}{\sqrt{a^{2}+b^{2}}}

 

 

- wherein

\rho  is the distance from the line ax+by+c=0 .

 

 

Equation of a circle -

\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

- wherein

Circle with centre \left ( h,k \right ) and radius r.

 

 

Distance of  O from BC

is \frac{\left | 3\left ( 1 \right )+4\left ( 1 \right )+3 \right |}{5}

= 2 units

Also \: \frac{AO}{DO}=\frac{2}{1}=\frac{AO}{2}

=> AO= 4 \: units

Hence circumcircle has r = 4 and centre (1,1)

\left ( x-1 \right )^{2}+\left ( y-1 \right )^{2}=4^{2}

x^{2}+y^{2}-2x-2y-14=0


Option 1)

x^{2}+y^{2}-2x-2y-2=0

Incorrect option    

Option 2)

x^{2}+y^{2}-2x-2y-14=0

Correct option

Option 3)

x^{2}+y^{2}-2x-2y+2=0

Incorrect option    

Option 4)

x^{2}+y^{2}-2x-2y-7=0

Incorrect option    

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