Show that the points (x, y), given by x = 2at/1+t^2 and y = a(1−t^2)/ 1+t^2 lies on a circle for all real values

Question

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

Answer 1

Given, 
\(x= \frac{2at}{1+t^2}\) 
\(y=a \frac{(1-t^2)}{1+t_2}\) 

We have,

⇒ x² + y² = a²

Thus, all points (x,y) lies on a circle with centre (0, 0) and radius ‘a’ units