Let X=x-a, Y=y-b, Z=z-c, then X²+Y²+Z²=r².
Differentiating wrt to another variable t, 2XdX/dt + 2YdY/dt + 2ZdZ/dt = 0, so
XdX/dt + YdY/dt + ZdZ/dt = 0. Note that dX/dt can be replaced by dx/dt, etc.
Differentiating again:
(dX/dt)²+Xd²Xdt²+(dY/dt)²+Yd²Y/dt²+(dZ/dt)²+Zd²Z/dt²=0. Note that d²X/dt² can be replaced by d²x/dt², etc.
The original equation is that of a sphere centre (a,b,c), radius r. By transformation of axes so that the new origin is the centre of the sphere, a, b and c can be eliminated, and DEs created as above.
If dt=dx then we have:
X+Yy'+Zz'=0 (first degree)
1+Yy"+(y')²+Zz"+(z')²=0 (second degree)
The prime means d/dx. Double prime is d²/dx².
By further differentiation we arrive at three DEs containing three variables X, Y and Z and differentials involving x, y and z. From such a system of equations, X, Y and Z can be found in terms of differentials only. The algebra becomes very complicated. Once found, the variables can be replaced in X²+Y²+Z²=r². This will produce a rather complicated DE with no reference at all to a, b or c.
The third degree DE is:
Yy'''+3y'y"+Zz'''+3z'z"=0.
The second and third degrees DEs form a system of simultaneous equations containing only Y and Z, so both can found. These can then be substituted into the first degree DE to find X.
Finally, substitute X, Y and Z into X²+Y²+Z²=r² to arrive at a rather complicated equation involving x and derivatives wrt x, without any references to a, b or c.
To illustrate and simplify the process, I offer the following.
Yy"+Zz"=-A
Yy'''+Zz'''=-3B
where A=1+(y')²+(z')² and B=y'y"+z'z".
So, Y=(3Bz"-Az''')/(y"z'''-y'''z") and Z=(3By"-Ay''')/(y'''z"-y"z''').
And X=-(Yy'+Zz')=(3By"z'-3By'z"+Ay'z'''-Ay'''z')/(y"z'''-y'''z") and X²+Y²+Z²=r².