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  • Show that the point (x, y) given by x = 2at/1+t^{2} and y=1(1-t^{2})/1+t^{2} lies on a circle for all real values of t such that –1 ≤ t ≤1 where a is any given real numbers.

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

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We have variable point as x=\frac{2at}{1+t^{2}} and y=\frac{a\left ( 1-t^{2} \right )}{1+t^{2}}

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

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