Question

Find the inclinations of the axes so that the following equations may represent circles, and in each case find the radius and centre;
$x^2+\sqrt{3}xy+y^2-4x-6y+5=0$.

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
$2\cos \omega =\sqrt{3}$ or $\cos \omega =\frac{\sqrt{3}}{2}$
$\therefore \omega ={30}^o$
Also, $h+k\cos \omega =h+\frac{\sqrt{3}k}{2}=2$ or $2h+\sqrt{3}k=4...\left(1\right)$
$k+\frac{\sqrt{3}h}{2}=3$ or $2k+\sqrt{3}h=6...\left(2\right)$ and
Multiplying equation $\left(1\right)$ by $2$ and equation $\left(2\right)$ by $\sqrt{3}$, and subtracting the two, we get
$4h-3h=8-6\sqrt{3}$ or $h=8-6\sqrt{3}$
$\Rightarrow 2k=6-8\sqrt{3}+18=24-8\sqrt{3}$ or $k=12-4\sqrt{3}$
Also, $h^2+k^2+\sqrt{3}hk-r^2=5$
$\Rightarrow 64+108-96\sqrt{3}+144+48-96\sqrt{3}+\sqrt{3}(96-32\sqrt{3}-72\sqrt{3}+72)-r^2=5$
$\Rightarrow 364-192\sqrt{3}+168\sqrt{3}-312-r^2=5$
$\Rightarrow 52-24\sqrt{3}-r^2=5$
$\therefore r^2=47-24\sqrt{3}$
Center $=(8-6\sqrt{3},12-4\sqrt{3})$
Radius $=\sqrt{47-24\sqrt{3}}$