Question

Prove that the equation to a circle whose radius is $a$ and which touches the axes of coordinates, which are inclined at an angle $\omega$, is
$x^2+2xy\cos \omega +y^2-2a(x+y)\cot \frac{\omega }{2}+a^2{\cot }^2\frac{\omega }{2}=0$.

Answer 1

If the circle touches both the axis, the center may be taken as $(h,h)$. If a us the radius
$a=h\sin \omega$ or $h=a\csc \omega$
$\therefore$Center$(a\csc \omega ,a\csc \omega )$ and radius 'a'
Putting these values in the general equation , we get
$x^2+y^2+2xy\cos \omega -2x(a\csc \omega +a\csc \omega \cos \omega )-2y(a\csc \omega +a\csc \omega \cos \omega )+a^2{\csc }^2\omega +$

$a^2{\csc }^2\omega +2a^2{\csc }^2\omega \cos \omega -a^2=0$
or $x^2+y^2+2xy\cos \omega -2xa\csc \omega (1+\cos \omega )-2ya\csc \omega (1+\cos \omega )+2a^2{\csc }^2\omega (1+\cos \omega )-a^2=\mathrm{0.....1}$
$\csc \omega (1+\cos \omega )=\frac{2{\csc }^2\omega /2}{2\sin \omega /2\cos \omega /2}=\cot \frac{\omega }{2}$
and $2a^2{\csc }^2\omega (1+\cos \omega )-a^2$
$=a^2[2\frac{2{\csc }^2\omega /2}{4{\csc }^2\omega /2{\sin }^2\omega /2}-1]$
$=a^2[{\csc }^2\frac{\omega }{2}-1]$
$=a^2{\cot }^2\frac{\omega }{2}$
Putting these values in $1$, we get
$x^2+y^2+2xy\cos \omega -2a(x+y)\cot \frac{\omega }{2}+a^2{\cot }^2\frac{\omega }{2}=0$