The centre and radius of a circle $x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right), y=\frac{8 a t}{1+t^{2}}$, are…
The centre and radius of a circle $x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right), y=\frac{8 a t}{1+t^{2}}$, are respectively
$(0,0)$ and $3 a$ units
$(0,0)$ and $4 a$ units
$(0,0)$ and $2 a$ units
$(0,0)$ and $a$ units
Solution
Given equation of circle in parametric form is
$x=4 a\left(\frac{1-t^{2}}{1+t^{2}}\right) \text { and } y=4 a\left(\frac{2 t}{1+t^{2}}\right)$
$x=4 a \cos 2 \theta$ and $y=4 a \sin 2 \theta$, where $t=\tan \theta$
Comparing with $x=r \cos \theta, y=r \sin \theta$, we get $r=4 a$, centre is $(0,0)$