Question

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

 

Answer 1

Squaring and adding \[x = \frac{2at}{1 + t^2}\] and

\[y = a\left( \frac{1 - t^2}{1 + t^2} \right)\] , we get

\[x^2 + y^2 = \left( \frac{2at}{1 + t^2} \right)^2 + a^2 \left( \frac{1 - t^2}{1 + t^2} \right)^2 \]
\[ \Rightarrow x^2 + y^2 = \frac{4 a^2 t^2 + a^2 - 2 a^2 t^2 + a^2 t^4}{\left( 1 + t^2 \right)^2}\]
\[ \Rightarrow x^2 + y^2 = \frac{a^2 + 2 a^2 t^2 + a^2 t^4}{\left( 1 + t^2 \right)^2}\]
\[ \Rightarrow x^2 + y^2 = a^2 \frac{\left( 1 + t^2 \right)^2}{\left( 1 + t^2 \right)^2}\]
\[ \Rightarrow x^2 + y^2 = a^2 \]
Since, the above equation represents the equation of a circle, hence points (x, y) lies on the circle.