Find the inclinations of the axes so that the following equations may represent circles, and in each case find the radius and centre;
$x^{2} - x y + y^{2} - 2 g x - 2 f y = 0$ .

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Solution

When the axes are inclined at an angle $ω$ , the general equation of a circle with center $(h, k)$ and radius $r$ can be written as
$x^{2} + y^{2} + 2 x y cos ω - 2 (h + k cos ω) x - 2 (k + h cos ω) y + h^{2} + k^{2} + 2 h k cos ω - r^{2} = 0$
When compared this equation with the equation of circle as given, we have
$2 cos ω = - 1$ or $cos ω = \frac{- 1}{2}$
$∴ ω = 120^{o}$
Also, $h + k cos ω = h - \frac{k}{2} = g . . . (1)$
$k - \frac{h}{2} = f . . . (2)$ and
$h^{2} + k^{2} - h k = r^{2}$
Multiplying equation $(1)$ by $2$ and adding that to equation $(2)$ , we get
$\frac{3 h}{2} = 2 g + f$ or $h = \frac{2 (2 g + f)}{3}$
$\Rightarrow k = \frac{h}{2} + f = \frac{2 g + f}{3} + f = \frac{2 (g + 2 f)}{3}$
Also, $r^{2} = \frac{4 (4 g^{2} + f^{2} + 4 f g) + 4 (g^{2} + 4 f^{2} + 4 f g) - 4 (2 g^{2} + 2 f^{2} + 5 f g)}{9} = \frac{4 (3 g^{2} + 3 f^{2} + 3 f g)}{9} = \frac{4 (f^{2} + g^{2} + f g)}{3}$
Center $= (\frac{2 (f + 2 g)}{3}, \frac{2 (g + 2 f)}{3})$
Radius $= \sqrt{\frac{4 (f^{2} + g^{2} + f g)}{3}}$

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