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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 $-1 \leq t \geq 1$, where $a$ is any given real number.
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Verified Answer
$$
\begin{aligned}
&\operatorname{Here}\left(\frac{x}{a}\right)^2+\left(\frac{y}{a}\right)^2=\left(\frac{2 t}{1+y^2}\right)^2+\left(\frac{1-t^2}{1+t^2}\right)^2 \\
&=\frac{4 t^2+\left(1-t^2\right)^2}{\left(1+t^2\right)^2}=\frac{4 t^2+1+t^4-2 t^2}{\left(1+t^2\right)^2} \\
&=\frac{\left(1+t^2\right)^2}{\left(1+t^2\right)^2}=1 \Rightarrow\left(\frac{x}{a}\right)^2+\left(\frac{y}{a}\right)^2=1 \\
&\Rightarrow \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2 \\
&\therefore(\mathrm{x}, \mathrm{y}) \text { lies on circle }
\end{aligned}
$$
\begin{aligned}
&\operatorname{Here}\left(\frac{x}{a}\right)^2+\left(\frac{y}{a}\right)^2=\left(\frac{2 t}{1+y^2}\right)^2+\left(\frac{1-t^2}{1+t^2}\right)^2 \\
&=\frac{4 t^2+\left(1-t^2\right)^2}{\left(1+t^2\right)^2}=\frac{4 t^2+1+t^4-2 t^2}{\left(1+t^2\right)^2} \\
&=\frac{\left(1+t^2\right)^2}{\left(1+t^2\right)^2}=1 \Rightarrow\left(\frac{x}{a}\right)^2+\left(\frac{y}{a}\right)^2=1 \\
&\Rightarrow \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2 \\
&\therefore(\mathrm{x}, \mathrm{y}) \text { lies on circle }
\end{aligned}
$$
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