Question

If the incentre of an equilateral triangle is $\left( 1,1 \right)$ and the equation of its one side is $3x+4y+3=0$ then the equation of the circumcircle of this triangle is
(A)${{x}^{2}}+{{y}^{2}}-2x-2y-2=0$
(B)${{x}^{2}}+{{y}^{2}}-2x-2y+2=0$
(C)${{x}^{2}}+{{y}^{2}}-2x-2y-7=0$
(D)${{x}^{2}}+{{y}^{2}}-2x-2y-14=0$

Answer 1

Hint: For answering this question we will use the fact stating that the incentre and circumcenter of an equilateral triangle coincides with each other and also use this relation between inradius $r$ and circumradius $R$ is given by $R=2r$. From these values we will derive the equation of the circumcircle having centre $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius $r$ which is given as ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$ .

Complete step by step answer:
From the basic concept we know that the incentre and circumcenter of an equilateral triangle coincides with each other.
From the question we have the incentre as $\left( 1,1 \right)$ so the circumcentre will also be $\left( 1,1 \right)$.
The relation between inradius $r$ and circumradius $R$ is given by $R=2r$.
Here from the question we have the equation of one side of the triangle that is $3x+4y+3=0$.
The inradius can be stated as the distance between the incentre and any one side of the triangle.
As we know that the distance between a point $\left( {{x}_{1}},{{y}_{1}} \right)$ and a straight line $ax+by+c=0$ is given by $\dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ .
So now the inradius can be given as the distance between the incentre $\left( 1,1 \right)$ and the side of the triangle $3x+4y+3=0$ which is given as $\dfrac{3\left( 1 \right)+4\left( 1 \right)+3}{\sqrt{{{3}^{2}}+{{4}^{2}}}}$ .
After simplifying it we will have $\dfrac{3+4+3}{\sqrt{9+16}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2$ .
Hence the inradius is $r=2$. So the circumradius will be $R=2\left( 2 \right)=4$ .
As we know that the equation of a circle having centre $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius $r$ will be given as ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$ .
So the equation of the circumcircle having the circumcentre $\left( 1,1 \right)$ and radius $4$ will be given as ${{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{4}^{2}}$ .
By expanding the equation we will have ${{x}^{2}}+1-2x+{{y}^{2}}+1-2y=16$ .
By simplifying this we will have ${{x}^{2}}-2x+{{y}^{2}}-2y-14=0$ .
Hence, we can conclude that for an equilateral triangle having the incentre $\left( 1,1 \right)$ and the equation of its one side is $3x+4y+3=0$ then the equation of the circumcircle of this triangle will be ${{x}^{2}}-2x+{{y}^{2}}-2y-14=0$.

So, the correct answer is “Option D”.

Note: While answering questions of this type we should be careful and remember that for an equilateral triangle, the incentre and circumcenter coincides with each other and also remember the relation between inradius $r$ and circumradius $R$ is given by $R=2r$. If we forget the relationship then we will be unable to solve the question and get into a mess.