Question

The locus of the point of intersection of lines $x=\frac{1-t^2}{1+t^2},y=\frac{2t}{1+t^2}$, where $t$ is a parameter is a circle whose radius is

• A. $2$
• B. $1$
• C. $4$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $1$

$x=\frac{1-t^2}{1+t^2}$

$y=\frac{2t}{1+t^2}$

$\frac{dx}{dt}=\frac{-4t}{(1+t^2{)}^2}$

$\frac{dy}{dt}=\frac{2(1-t^2)}{(1+t^2{)}^2}$

$\frac{dy}{dx}=\frac{-(1-t^2)}{2t}=-\frac{x}{y}$

$x^2+y^2=c$

$x^2+y^2=\frac{(1-t^2{)}^2+4t^2}{(1+t^2{)}^2}=1$

$x^2+y^2=1$

Therefore, radius=1.