Question

The equation of circumcircle of an equilateral triangle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and one vertex of the triangle is $\left( {1,1} \right)$. The equation of incircle of the triangle is
A) $4\left( {{x^2} + {y^2}} \right) = {g^2} + {f^2}$
B) $4\left( {{x^2} + {y^2}} \right) + 8gx + 8fy = \left( {1 - g} \right)\left( {1 + 3g} \right) + \left( {1 - f} \right)\left( {1 + 3f} \right)$
C) $4\left( {{x^2} + {y^2}} \right) + 8gx + 8fy = {g^2} + {f^2}$
D) $4\left( {{x^2} + {y^2}} \right) + 8gx + 8fy = \left( {2 - g} \right)\left( {1 + 3g} \right) + \left( {2 - f} \right)\left( {2 + 3f} \right)$

Answer 1

Hint: We can find the radius of the circumcircle by converting the equation into the form ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$. As the circumcircle passes through all the vertices of, we can obtain an equation independent of x and y and eliminate c from the radius. For an equilateral triangle, the radius of the incircle is half the radius of the circumcircle and the centre for both will be the same. So, we can find the equation of the incircle using the radius and the centre.

Complete step by step solution:
We have the equation of the circumcircle as ${x^2} + {y^2} + 2gx + 2fy + c = 0$. We can write this equation as follows,
We can add ${g^2} + {f^2}$on both sides
$ \Rightarrow {x^2} + {y^2} + 2gx + 2fy + {g^2} + {f^2} = {g^2} + {f^2} - c$
Rearranging the equation, we get,
\[ \Rightarrow \left( {{x^2} + 2gx + {g^2}} \right) + \left( {{y^2} + 2fy + {f^2}} \right) = {g^2} + {f^2} - c\]
Using the algebraic identity,${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$we get,
\[ \Rightarrow {\left( {x + g} \right)^2} + {\left( {y + f} \right)^2} = {g^2} + {f^2} - c\]
Comparing this with the standard equation of the circle${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$with radius r and center $\left( {h,k} \right)$, we get,
center is $\left( { - g, - f} \right)$ and $r = \sqrt {{g^2} + {f^2} - c} $ … (1)
We know that a circumcircle is a circle that passes through all the vertices of the triangle. So the point $\left( {1,1} \right)$satisfies the equation of the circle.
\[ \Rightarrow {1^2} + {1^2} + 2g\left( 1 \right) + 2f\left( 1 \right) + c = 0\]
\[ \Rightarrow c = - 2 - 2g - 2f\]… (2)
We can substitute equation (2) in (1),
$ \Rightarrow r = \sqrt {{g^2} + {f^2} - \left( { - 2 - 2g - 2f} \right)} $
On opening the bracket we get,
$ \Rightarrow r = \sqrt {{g^2} + {f^2} + 2 + 2g + 2f} $ … (2)
For an equilateral triangle, the circumcenter and incircle have a common centre and radius of incircle is half the radius of circumcenter. Let R be the radius of incircle.
$ \Rightarrow R = \dfrac{r}{2}$ .. (3)
And center $\left( { - g, - f} \right)$

Therefore, equation of the incircle is given by,
\[{\left( {x + g} \right)^2} + {\left( {y + f} \right)^2} = {R^2}\]
Substituting (3), we get,
\[{\left( {x + g} \right)^2} + {\left( {y + f} \right)^2} = {\left( {\dfrac{r}{2}} \right)^2}\]
On substituting (2), we get,
\[{\left( {x + g} \right)^2} + {\left( {y + f} \right)^2} = {\left( {\dfrac{{\sqrt {{g^2} + {f^2} + 2 + 2g + 2f} }}{2}} \right)^2}\]
Expanding the squares, we get ,
\[ \Rightarrow \left( {{x^2} + 2gx + {g^2}} \right) + \left( {{y^2} + 2fy + {f^2}} \right) = \dfrac{1}{4}\left( {{g^2} + {f^2} + 2 + 2g + 2f} \right)\]
Multiplying both sides with 4, we get,
\[ \Rightarrow \]\[4{x^2} + 8gx + 4{g^2} + 4{y^2} + 8fy + 4{f^2} = {g^2} + {f^2} + 2 + 2g + 2f\]
On rearranging, we get,
\[ \Rightarrow \]\[4{x^2} + 4{y^2} + 8gx + 8fy = {g^2} - 4{g^2} + {f^2} - 4{f^2} + 2 + 2g + 2f\]
On simplification, we get,
\[ \Rightarrow \]\[4{x^2} + 4{y^2} + 8gx + 8fy = - 3{g^2} - 3{f^2} + 2 + 2g + 2f\]
\[ \Rightarrow \]\[4{x^2} + 4{y^2} + 8gx + 8fy = \left( { - 3{g^2} + 2g + 1} \right) + \left( { - 3{f^2} + 2f + 1} \right)\]
We can factorize the 2 terms of RHS
\[ \Rightarrow 4{x^2} + 4{y^2} + 8gx + 8fy = \left( { - 3{g^2} + 3g - g + 1} \right) + \left( { - 3{f^2} + 3f - f + 1} \right)\]
On Taking terms common we get,
\[ \Rightarrow 4{x^2} + 4{y^2} + 8gx + 8fy = \left( { - 3g\left( {g - 1} \right) - \left( {g - 1} \right)} \right) + \left( { - 3f\left( {f - 1} \right) - \left( {f - 1} \right)} \right)\]
On further simplification we get,
\[ \Rightarrow 4{x^2} + 4{y^2} + 8gx + 8fy = \left( {\left( {g - 1} \right)\left( { - 3g - 1} \right)} \right) + \left( {\left( {f - 1} \right)\left( { - 3f - 1} \right)} \right)\]
\[ \Rightarrow 4\left( {{x^2} + {y^2}} \right) + 8gx + 8fy = \left( {\left( {1 - g} \right)\left( {1 + 3g} \right)} \right) + \left( {\left( {1 - f} \right)\left( {1 + 3f} \right)} \right)\]
Therefore, the equation of the incircle is $4\left( {{x^2} + {y^2}} \right) + 8gx + 8fy = \left( {1 - g} \right)\left( {1 + 3g} \right) + \left( {1 - f} \right)\left( {1 + 3f} \right)$.

So, the correct answer is option B.

Note: A circumcircle of a triangle is the circle that passes through all the 3 vertices of the triangle. Center of the circumcircle is called the circumcircle. It is the point of intersection of the perpendicular bisectors of the sides. The incircle of the triangle is the largest circle that can be drawn inside a triangle. The sides of the triangles are tangential to the incircle. The centre of incircle is called incenter and it is the point of intersection of the angle bisectors of the triangle. For an equilateral triangle, the perpendicular bisectors are also the angle bisector of the opposite angle. So, they have common incentre and circumcenter. For an equilateral triangle, the radius of the incircle is the half of the radius of the circumcircle. This can be proved using trigonometry.