The differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\sqrt{1-y^2}}{y}$ determines a family of…
The differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\sqrt{1-y^2}}{y}$ determines a family of circles with
variable radii and fixed centre at $(0,1)$.
variable radii and fixed centre at $(0,-1)$.
fixed radius of 1 unit and variable centre along the $\mathrm{Y}$-axis.
fixed radius of 1 unit and variable centre along the $\mathrm{X}$-axis.
Solution
$\begin{array}{ll} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\sqrt{1-y^2}}{y} \\ \therefore \quad & \int \frac{y}{\sqrt{1-y^2}} \mathrm{~d} y=\int 1 \mathrm{~d} x \\ \therefore \quad & -\sqrt{1-y^2}=x+\mathrm{c} \\ \therefore \quad & (x+\mathrm{c})^2=1-y^2\end{array}$
$\therefore \quad(x+\mathrm{c})^2+y^2=1$
$\therefore \quad$ Radius is fixed, which is 1 and the centre is $(-c, 0)$ which is a variable centre on the $\mathrm{X}$-axis.