Question

Area enclosed by the curves $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$

• A. $3\pi$
• B. $2\pi$
• C. $\frac{3\pi }{2}$
• D. $\pi$

Answer 1

Answer: (D)

$\pi$

Given, $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$
Squaring and adding both we get,
$x^2+y^2=\frac{(1-t^2{)}^2+4t^2}{(1+t^2{)}^2}=\frac{(1+t^2{)}^2}{(1+t^2{)}^2}=1$
Clearly given curve is a circle with radius $1$ unit.
Hence required area is $=\pi (1{)}^2=\pi$ sq. units