Question

Show that the point (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 of t such that –1 < t < 1 where a is any given real numbers.

Answer 1

Given x = `(2at)/(1 + t^2)` and y = `(a(1 - t^2))/(1 + t^2)`

⇒ x² + y² = `((2at)/(1 + t^2))^2 + ((a(1 - t^2))/(1 + t^2))^2`

= `(4a^2t^2)/((1 + t^2)^2) + (a^2(1 - t^2)^2)/((1 + t^2)^2`

= `(4a^2t^2 + a^2(1 + t^4 - 2t^2))/(1 + t^2)^2`

= `(4a^2t^2 + a^2 + a^2t^4 - 2a^2t^2)/(1 + t^2)^2`

= `(a^2 + a^2t^4 + 2a^2t^2)/(1 + t^2)^2`

= `(a^2(1 + t^4 + 2t^2))/(1 + t^2)^2`

= `(a^2(1 + t^2)^2)/(1 + t^2)^2`

= a²

∴ x² + y² = a² which is the equation of a circle.

Hence, the given points lie on a circle.