Obtain the differential equation of the family of plane curves

[Portal.Leaf]

Homework Statement

Obtain the differential equation of the family of plane curves described:
Circles tangent to the x-axis.

Homework Equations

(x-h)^2 + (y-k)^2 = r^2

The Attempt at a Solution

I tried to answer this question using the same way I did on a problem very similar to this (Circles with fixed radius r and tangent to the x-axis), but now I'm getting a different answer.
The answer provided by the book for the problem above is [1+(y')^2]^3 = [yy''+1+(y')^2]^2. I have no idea how it's done.

I want to ask the difference between the ways of how to solve these two problems:
(1) circles tangent to the x-axis.
(2) circles with fixed radius tangent to the axis.

I can solve question (2) because of the hint that h=r, but doing the same with question (1) doesn't seem to work and it's making me crazy already. Please give me some clue on how to solve this one. Thanks a bunch!

 

[LCKurtz]

Homework Statement

Obtain the differential equation of the family of plane curves described:
Circles tangent to the x-axis.

Homework Equations

(x-h)^2 + (y-k)^2 = r^2

The Attempt at a Solution

I tried to answer this question using the same way I did on a problem very similar to this (Circles with fixed radius r and tangent to the x-axis), but now I'm getting a different answer.
The answer provided by the book for the problem above is [1+(y')^2]^3 = [yy''+1+(y')^2]^2. I have no idea how it's done.

I want to ask the difference between the ways of how to solve these two problems:
(1) circles tangent to the x-axis.
(2) circles with fixed radius tangent to the axis.

I can solve question (2) because of the hint that h=r, but doing the same with question (1) doesn't seem to work and it's making me crazy already. Please give me some clue on how to solve this one. Thanks a bunch!

I haven't worked it all the way out for 1, but I will get you started. For circles tangent to the x axis, you don't know the point of tangency, call it $(a,0)$ or the radius $r$. The general equation for that family of circles would be $(x-a)^2 + (y-r)^2 = r^2$. You can differentiate that twice, implicitly with respect to $x$:$$
2(x-a) + 2(y-r)y' = 0$$ $$
2 +2y' + 2(y-r)y'' = 0$$Now your problem becomes using your equations to get rid of the $r$ and $a$, or maybe easier, get rid of $(x-a)$ and $y-r$ by expressing them in terms of $y$ and its derivatives.

 

[Portal.Leaf]

Thanks! I'll be working on it now. :)

UPDATE: I followed your instructions and had a lot of substitutions. Finally, I got it right! Thank you so much!

 

[Kenyow]

I haven't worked it all the way out for 1, but I will get you started. For circles tangent to the x axis, you don't know the point of tangency, call it $(a,0)$ or the radius $r$. The general equation for that family of circles would be $(x-a)^2 + (y-r)^2 = r^2$. You can differentiate that twice, implicitly with respect to $x$:$$
2(x-a) + 2(y-r)y' = 0$$ $$
2 +2y' + 2(y-r)y'' = 0$$Now your problem becomes using your equations to get rid of the $r$ and $a$, or maybe easier, get rid of $(x-a)$ and $y-r$ by expressing them in terms of $y$ and its derivatives.

is the second derivative of the equation's right?

 

[LCKurtz]

is the second derivative of the equation's right?

No. Good catch. Apparently the OP got the idea anyway though.