Question
The differential equation $$\frac{{dy}}{{dx}} = \frac{{\sqrt {1 - {y^2}} }}{y}$$ determines a family of circles with-
A.
variable radii and a fixed centre at $$\left( {0,\, 1} \right)$$
B.
variable radii and a fixed centre at $$\left( {0,\, - 1} \right)$$
C.
fixed radius 1 and variable centres along the $$x$$-axis.
D.
fixed radius 1 and variable centres along the $$y$$-axis.
Answer :
fixed radius 1 and variable centres along the $$x$$-axis.
Solution :
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{\sqrt {1 - {y^2}} }}{y} \cr
& \Rightarrow \frac{{ - 2y}}{{\sqrt {1 - {y^2}} }}dy + 2dx = 0 \cr
& \Rightarrow 2\sqrt {1 - {y^2}} + 2x = 2c \cr
& \Rightarrow \sqrt {1 - {y^2}} + x = c \cr
& \Rightarrow {\left( {x - c} \right)^2} + {y^2} = 1 \cr} $$
which is a circle of fixed radius 1 and variable centre $$\left( {c,\,0} \right)$$ lying on $$x$$-axis.