CircleHardBloom L3

Question

The equation of circumcircle of an equilateral triangle is $x^{2} + y^{2} + 2gx + 2fy + c = 0$ and one vertex of triangle is $(1,1)$. The equation of incircle of triangle is:

Options

A.$4\left( x^{2} + y^{2} \right) = g^{2} + f^{2}$
B.$4\left( x^{2} + y^{2} \right) + 8gx + 8fy = (1 - g)(1 + 3g) + (1 - f)(1 + 3f)$
C.$4\left( x^{2} + y^{2} \right) + 8gx + 8fy = g^{2} + f^{2}$
D.$4\left( x^{2} + y^{2} \right) + 8gx + 8fy = (2 - g)(1 + 3g) + (2 - f)(2 + 3f)$

Solution

[Key Concept: For an equilateral triangle the circumcentre is the centroid; relate $g,f,c$ to the side and centre.]

$r = Rcos60^{\circ} = \frac{R}{2}$

Equation of incircle is

$$\begin{matrix} & x^{2} + y^{2} + 2gx + 2fy + g^{2} + f^{2} = \frac{g^{2} + f^{2} - c}{4} = \frac{(g + 1)^{2}}{4} + \frac{(f + 1)^{2}}{4} \\ & \ \\ R^{2} = & \ (g + 1)^{2} + (f + 1)^{2} = g^{2} + f^{2} + 2g + 2f + 2 = g^{2} + f^{2} - c \end{matrix}$$

∴ Equation of incircle becomes

$$\begin{matrix} & 4\left( x^{2} + y^{2} \right) + 8gx + 8fy \\ = & 2(g + f) - 3\left( g^{2} + f^{2} \right) + 2 \\ = & \left( 2g - 3g^{2} + 1 \right) + \left( 2f - 3f^{2} + 1 \right) \\ = & \ (1 - g)(1 + 3g) + (1 - f)(1 + 3f) \end{matrix}$$

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