Difference between a fixed and variable radius in differential equations

[df606]

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.

 

[grey_earl]

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.

 

[HallsofIvy]

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.

[tex]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.[/tex]

 

[grey_earl]

Interesting wording, but it make sense now.