Question

The axes being inclined at an angle $\omega$, find the centre and radius of the circle $x^2+2xy\cos \omega -2gx-2fy=0$.

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
$h+k\cos \omega =g...\left(1\right)$
$k+h\cos \omega =f...\left(2\right)$
Multiplying equation $\left(1\right)$ by $\cos \omega$ and subtracting that from equation $\left(2\right)$, we get
$k(1-{\cos }^2\omega )=f-g\cos \omega$
$\Rightarrow k=\frac{f-g\cos \omega }{{\sin }^2\omega }$
$\Rightarrow h=g-k\cos \omega =\frac{g-g{\cos }^2\omega -f\cos \omega +g{\cos }^2\omega }{{\sin }^2\omega }$
$\Rightarrow h=\frac{g-f\cos \omega }{{\sin }^2\omega }$
Also, $h^2+k^2+2hk\cos \omega -r^2=0$
$\Rightarrow r^2=\frac{1}{{\sin }^4\omega }\times (g^2+f^2{\cos }^2\omega -2fg\cos \omega +f^2+g^2{\cos }^2\omega -2fg\cos \omega +2\cos \omega (gf+gf{\cos }^2\omega -f^2\cos \omega -$

$g^2\cos \omega \left)\right)$
$\Rightarrow r^2=\frac{1}{{\sin }^4\omega }\times (g^2{\sin }^2\omega +f^2{\sin }^2\omega -2fg\cos \omega {\sin }^2\omega )$
$\Rightarrow r^2=\frac{g^2+f^2-2fg\cos \omega }{{\sin }^2\omega }$
Center $=(\frac{g-f\cos \omega )}{{\sin }^2\omega },\frac{f-g\cos \omega )}{{\sin }^2\omega })$
Radius $=\sqrt{\frac{g^2+f^2-2fg\cos \omega }{{\sin }^2\omega }}$