Question

A circle touches the axis of $x$ and cuts off a constant length $2l$ from the axis of $y$; prove that the equation of the locus of its centre is $y^2-x^2=l^{2}cose{c}^2\omega$, the axes being inclined at an angle $\omega$.

Answer 1

Let $C$ be the center of the circle$\equiv (h,k)$

radius be a, then equation of the circle will be

${(x-h)}^2+{(y-k)}^2+2(x-h)(y-k)\cos \omega =a^2.....1$

By $△CMN;-$

$CN=CM\sin \omega$

$a=k\sin \omega .....2$

Equation of y axis is $x=\mathrm{0....3}$

Solving $1$ and $3$

$h^2+y^2+k^2-2yk-(2hy-2hk)\cos \omega =a^2$

or $y^2-2y(k+h\cos \omega )+h^2+k^2+2hk\cos \omega =a^2......4$

If A and B be the points of intersection having co ordinates $(O,y_1)\&(O,y_2)$

$AB=y_2-y_1=2l$

${(y_2+y_1)}^2-4y_1{y}_2={\left(2l\right)}^2.....5$

$y_1+y_2=-2(k+h\cos \omega )$

$y_1{y}_2=h^2+k^2+2hk\cos \omega -a^2$

Putting in $5$ we get

$4{(k+h\cos \omega )}^2-4(h^2+k^2+2hk\cos \omega -a^2)=4l^2$

or $k^2{\sin }^2\omega -h^2(1-{\cos }^2\omega )=l^2$

or $(k^2-h^2)=\frac{l^2}{{\sin }^2\omega }$

Therefore Locus is
$y^2-x^2=\frac{l^2}{{\sin }^2\omega }=l^2{\csc }^2\omega$