Spotting the error in finding order and degree of differential equation whose solution is $(x-h)^2 + (y-k)^2 = a^2$

Question

Find order and degree of differential equation whose solution is: $(x-h)^2 + (y-k)^2 = a^2$ (where $a$ is a constant)

Attempt:

$x- h + yy' - ky' = 0 $

$\implies 1+ yy'' + (y')^2 - ky'' = 0 $

So, order $= 2$

degree $= 1$

But answer is: order $ =2$ , degree $=2$

Answer 1

Hint:

You need to eliminate three constants, and for this you will need to differentiate three times. Hence order $3$.

$$a^2=(x-h)^2+(y-k)^2=a^2\to0=(x-h)+(y-k)y',$$

$$h=x+(y-k)y'\to0=1+y'^2+(y-k)y'',$$

$$k=\frac{1+y'^2}{y''}+y\to0=\left(\frac{y'^2+1}{y''}\right)'+1.$$