Show that the point `(x ,y)` given by `x=(2a t)/(1+t^2)a n dy=((1-t^2)/(1+t^2))` lies on a circle for all real values of `t` such that `-1lt=tlt=1,` where a is any given real number.

Explanation

We have



`x=frac{2 a t}{1+t^{2}} ldots ldots ldots . text { (i) } `


`y=a(frac{1-t^{2}}{1+t^{2}}) ldots ldots . text { (ii) }`



Squaring and adding (i) and (ii)



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


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