Question

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

Answer 1

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