Difference between a fixed and variable radius in differential equations

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Homework Statement


Find a differential equation whose solution is a family of circles with centers in the xy-plane and of variable radii. Hint: Write the equation of the family as x^2+y^2-2ax-2by+2c=0


Homework Equations


The previous questions asks to find a differential equation whose solution is a family of circles with centers at (h,k) and of fixed radius. Using the equation (x-h)^2+(y-k)^2=r^2 gave me the correct answer.


The Attempt at a Solution


So, I'm not actually asking how to do the problem. What is the meaning/difference of "fixed radius" and "variable radius"? The equations for the two problems seem to be the same.
 
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From the hint given, I would just look for a differential equation which has this family as solution. Otherwise, I have no idea what "variable radius" could be.
 
Using [/itex](x- h)^2+ (y- k)^2= r^2[/itex] should give you the correct equation for either problem. In the previous problem, r was fixed and can appear in the differential equation. For this new problem, r is variable and you want a differential equation that does not contain h, k, or r.

x^2- 2hx+ h^2+ y^2- 2ky+ k^2= r^2[/itex]<br /> <br /> Differentiating with respect to x, 2x- 2h+ (2y- 2k)(dy/dx)= 0. Differentiating again will get rid of h: 2- 2(dy/dx)^2+ (2y- 2k)(d^2y/dx^2)= 0<br /> Now, you need to eliminate k from that equation.
 
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Interesting wording, but it make sense now.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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