# Show that the point (x, y) given

Question:

Show that the point $(x, y)$ given by $x=\frac{2 a t}{1+t^{2}}$ and $y=\frac{a\left(1-t^{2}\right)}{1+t^{2}}$ lies on a circle for all real values of $t$ such that $-1 \leq t \leq 1$ where $a$ is any given real numbers.

Solution:

Given

$x=\frac{2 a t}{1+t^{2}}, y=\frac{a\left(1-t^{2}\right)}{1+t^{2}}$

Squaring both the equations,

$\mathrm{x}^{2}=\left[\frac{2 \mathrm{at}}{1+\mathrm{t}^{2}}\right]^{2} \& \mathrm{y}^{2}=\left[\frac{\mathrm{a}\left(1-\mathrm{t}^{2}\right)}{1+\mathrm{t}^{2}}\right]^{2}$

$x^{2}+y^{2}=\frac{4 a^{2} t^{2}}{\left(1+t^{2}\right)^{2}}+\frac{a^{2}\left(1-t^{2}\right)^{2}}{\left(1+t^{2}\right)^{2}}$

Taking LCM and simplifying we get

$=\frac{a^{2}\left(4 t^{2}+\left(1-t^{2}\right)^{2}\right)}{\left(1+t^{2}\right)^{2}}$

$=\frac{\left.\mathrm{a}^{2}\left(4 \mathrm{t}^{2}+1+\mathrm{t}^{4}-2 \mathrm{t}^{2}\right)\right)}{\left(1+\mathrm{t}^{2}\right)^{2}}$

$=\frac{a^{2}\left(2 t^{2}+1+t^{4}\right)}{\left(1+t^{2}\right)^{2}}$

$=\frac{a^{2}\left(1+t^{2}\right)^{2}}{\left(1+t^{2}\right)^{2}}$

$=a^{2}$

Therefore $x^{2}+y^{2}=a^{2}$

The equation of a circle having centre $(h, k)$, having radius as $r$ units, is

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Centre $=(0,0)$ Radius $=$ a units Hence proved.