Question
The differential equation of the family of circles with fixed radius $$5$$ units and centre on the line $$y = 2$$ is-
A.
$$\left( {x - 2} \right)y{'^2} = 25 - {\left( {y - 2} \right)^2}$$
B.
$$\left( {y - 2} \right)y{'^2} = 25 - {\left( {y - 2} \right)^2}$$
C.
$${\left( {y - 2} \right)^2}y{'^2} = 25 - {\left( {y - 2} \right)^2}$$
D.
$${\left( {x - 2} \right)^2}y{'^2} = 25 - {\left( {y - 2} \right)^2}$$
Answer :
$${\left( {y - 2} \right)^2}y{'^2} = 25 - {\left( {y - 2} \right)^2}$$
Solution :
Let the centre of the circle be $$\left( {h,\,2} \right)$$
$$\therefore $$ Equation of circle is
$${\left( {x - h} \right)^2} + {\left( {y - 2} \right)^2} = 25.....(1)$$
Differentiating with respect to $$x,$$ we get
$$\eqalign{
& 2\left( {x - h} \right) + 2\left( {y - 2} \right)\frac{{dy}}{{dx}} = 0 \cr
& \Rightarrow x - h = - \left( {y - 2} \right)\frac{{dy}}{{dx}} \cr} $$
Substituting in equation (1), we get
$$\eqalign{
& {\left( {y - 2} \right)^2}{\left( {\frac{{dy}}{{dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25 \cr
& \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2} \cr} $$