Question

The differential equation $\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{y}$ determines a part of circle with

• A. Variable radii and fixed centre at $(0,1)$
• B. Variable radii and fixed centre at $(0,-1)$
• C. Fixed radius $1$ and variable center along the $x-$axis.
• D. Fixed radius $1$ and variable center along the $y-$axis.

Answer 1

The correct option is C Fixed radius $1$ and variable center along the $x-$axis.

$\frac{dy}{dx}=\frac{\sqrt{1-y^2}}{y}\Rightarrow \frac{-2y}{\sqrt{1-y^2}}dy+2dx=0$

$\Rightarrow 2\sqrt{1-y^2}+2x=2c\Rightarrow \sqrt{1-y^2}+x=c$

$\Rightarrow (x-c{)}^2+y^2=1$
which is a part circle of fixed radius $1$ and variable centre $(c,0)$ lying on $x-$axis.