First slide

Definition of a circle

Question

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

Moderate

A

$4 (x^{2} + y^{2}) = g^{2} + f^{2}$
B

$4 (x^{2} + y^{2}) + 8 g x + 8 f y = g^{2} ∣ + f^{2}$
C

$4 (x^{2} + y^{2}) + 8 g x + 8 f y = (1 - g) (1 + 3 g) + (1 - f) (1 + 3 f)$
D

none of these

Solution

In an equilateral triangle the circumcentre and incentre coincide. So, the coordinates of the incentre are $(- g, - f)$ . Also, in an equilateral triangle, circumradius is twice the in-radius.
$∴$ In- radius $= \frac{1}{2}$ (Circum- radius) $= \frac{1}{2}$ $\sqrt{8^{2} + f^{2} - c}$
But, the circumcircle passes through $(1, 1)$ . Therefore
$1 + 1 + 2 g + 2 f + c = 0 \Rightarrow c = - 2 - 2 g - 2 f$
$∴ In-radius = \frac{1}{2} \sqrt{g^{2} + f^{2} + 2 + 2 g + 2 f} = \frac{1}{2} \sqrt{(g + 1)^{2} + (f + 1)^{2}}$
Hence, the equation of the in-circle is
$(x + g)^{2} + (y + f)^{2} = \frac{1}{4} [(g + 1)^{2} + (f + 1)^{2}]$
$\begin{aligned} \Rightarrow 4 (x^{2} + y^{2}) + 8 g x + 8 f y = 1 + 2 g - 3 g^{2} + 1 + 2 f - 3 f^{2} \\ \Rightarrow 4 (x^{2} + y^{2}) + 8 g x + 8 f y = (1 - g) (1 + 3 g) + (1 - f) (1 + 3 f) \end{aligned}$

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