Question

Differential equation of all circles of radius r is of
$(a)$ Third order second degree
$(b)$ Second order second degree
$(c)$ Second order third degree
$(d)$ None of these

Answer 1

Hint: In this question use the general equation of circle with center at h, k and radius r which is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$. Use product rule of derivative along with chain rule of derivative to differentiate this general equation. The differentiation should be done twice as radius if given and there are only two variables h, k that we are dealing with.

Complete step-by-step answer:
As we know that the general equation of the circle is
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ Where (h, k) and r is the center and radius of the circle respectively.
Now differentiate this equation w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}{\left( {x - h} \right)^2} + \dfrac{d}{{dx}}{\left( {y - k} \right)^2} = \dfrac{d}{{dx}}{r^2}$
Now as we know that the differentiation of constant terms is zero.
$ \Rightarrow 2\left( {x - h} \right)\dfrac{d}{{dx}}\left( {x - h} \right) + 2\left( {y - k} \right)\dfrac{d}{{dx}}\left( {y - k} \right) = 0$
$ \Rightarrow 2\left( {x - h} \right) + 2\left( {y - k} \right)\dfrac{{dy}}{{dx}} = 0$
Now divide by 2 throughout we have,
$ \Rightarrow \left( {x - h} \right) + \left( {y - k} \right)\dfrac{{dy}}{{dx}} = 0$…………………………. (1)
Now again differentiate equation (1) we have,
$ \Rightarrow \dfrac{d}{{dx}}\left( {x - h} \right) + \dfrac{d}{{dx}}\left( {\left( {y - k} \right)\dfrac{{dy}}{{dx}}} \right) = 0$
Now apply product rule of differentiation $\dfrac{d}{{dx}}ab = a\dfrac{d}{{dx}}b + b\dfrac{d}{{dx}}a$ we have
$ \Rightarrow \dfrac{{dx}}{{dx}} - 0 + \dfrac{{dy}}{{dx}}\left( {\dfrac{d}{{dx}}\left( {y - k} \right)} \right) + \left( {y - k} \right)\left( {\dfrac{d}{{dx}}\dfrac{{dy}}{{dx}}} \right) = 0$
$ \Rightarrow 1 + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \left( {y - k} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = 0$
So this is the required differential equation of the circle.
Now as we know order of differential equation is the highest order of derivative and the degree of the differential equation is the highest power of the highest order of derivative.
So in the above equation the highest order of derivative is 2 and the highest power of the highest order of derivative is 1.
Thus the differential equation of all the circles has second order and single degrees.
Hence option (d) none of these is correct.

Note: A differential equation is an equation involving derivatives of a function or functions. Whenever we face such types of problems the key concept is that the general equation is differentiated only the number of times till we are able to get rid of the constants of the equation. It is advised to remember the general equation involving circles as it helps to save a lot of time.