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

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Solution

Squaring and adding $x = \frac{2 a t}{1 + t^{2}}$ and $y = a (\frac{1 - t^{2}}{1 + t^{2}})$ , we get
$x^{2} + y^{2} = {(\frac{2 a t}{1 + t^{2}})}^{2} + a^{2} {(\frac{1 - t^{2}}{1 + t^{2}})}^{2} \Rightarrow x^{2} + y^{2} = \frac{4 a^{2} t^{2} + a^{2} - 2 a^{2} t^{2} + a^{2} t^{4}}{{(1 + t^{2})}^{2}} \Rightarrow x^{2} + y^{2} = \frac{a^{2} + 2 a^{2} t^{2} + a^{2} t^{4}}{{(1 + t^{2})}^{2}} \Rightarrow x^{2} + y^{2} = a^{2} \frac{{(1 + t^{2})}^{2}}{{(1 + t^{2})}^{2}} \Rightarrow x^{2} + y^{2} = a^{2}$
Since, the above equation represents the equation of a circle, hence points (x, y) lies on the circle.

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Show that the point (x, y) given by x=2at1+t2 and y=a1-t21+t2 lies on a circle for all real values of t such that -1≤t≤1, where a is any given real number. [NCERT EXEMPLAR]

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