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

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A)
Toolbox:
• equation of a circle passing through the origin is $x^2+y^2=a^2$ where $a$ is the radius.
The given points $(x,y)$ are $\big(\large\frac{2at}{1+t^2}\big),\big(\frac{a(1-t^2)}{1+t^2}\big)$ respectively.
It is also given $-1 \leq t \leq 1$
Put $t=1$ then
$x=\large\frac{2\times a\times 1}{1+1}$$=a y=\large\frac{a(1-1)}{1+t^2}$$=0$
Hence the points are $(a,0)$
Clearly $a$ is the radius of the circle.
Substituting for $x=a$ and $y=0$ in the equation of the circle $x^2+y^2=a^2$
We get $a^2+0=a^2$
This clearly indicates that the points $\large\frac{2at}{1+t^2}$ and $\large\frac{a(1-t^2)}{1+t^2}$ lies on the circle.