Implicit differentiation - geometric form of solution?

Written in differentials, this equation takes the form,

(x dy - y dx)^2 = (dy)^2 + (dx)^2,

or, in polar coordinates, where x = r cos φ and y = r sin φ,

r^4 (d φ)^2 = r^2 (d φ)^2 + (d r)^2,

Eq.(1): r √(r^2-1) d φ = d r.

The first solution is r = 1. If r>1, we can integrate this equation to get φ(r).

The easier way is to notice that the equation of a straight line in the polar coordinates is

x = r cos φ = h.

Here, the x-axis is directed normally to the line, and h is the distance from the coordinate origin to the line.

Differentiating this relation gives

cos φ d r - r sin φ d φ = 0.

Substituting here cos φ = h/r and sin φ = √ (1 - h^2/r^2), we get

Eq.(2): r √(r^2-h^2) d φ = h d r.

Equation (2) coincides with equation (1) if h=1.

Thus, the curves representing solutions of the equation are:

(A) the unit circle,

(B) straight lines tangent to the unit circle.

For |x| > 1:

Write y = Y(x)*sqrt(x² - 1)

==> dy/dx = {dY/dx + x/(x^2-1)}*sqrt(x² - 1)

-->

dY/dx = +/- sqrt(Y^2+1)/(x^2-1).

The variables can be separated and terms integrated to:

log(Y + sqrt(1+Y^2)) = +/-(1/2)*log(C*|x-1|/|x+1|),

where C is any positive constant.

This can be solved for Y, and consequently,

y = Ax -/+ sqrt(A^2+1), ---- (*)

where A is any real constant.

So for |x| > 1, we have a linear form for y.

For |x| < 1:

write y = Z(x) * sqrt(1-x^2).

Similar working reveals that:

dZ/dx = +/- sqrt(Z^2-1)/(1-x^2).

As for Y, we can find a solution for Z which leads to a linear form for y:

--->

y = Bx +/- sqrt(B^2+1), --- (†)

where B is any real constant.

As an extra however, Z = +/-1 is a solution,

i.e. y = +/-(1-x^2)^(1/2). --- (‡)

Re. geometric interpretation of solutions:

(A) Clearly, (‡) represents the circumference of the unit disc.

(B) The other solutions have the general form:

y = ax +/- sqrt(a^2+1), for any real number a, with restrictions on x,

so these are parts of straight lines in the x-y axis.

If t := x^2 + y^2 (square of distance from line to origin at point(x,y)), then:

t = (a^2+1)*(x^2+1) +/- 2*a*sqrt(a^2+1)*x,

and a bit of calculus shows that

min. value of t is achieved at (x,y) = -/+(a/sqrt(a^2+1),-1/sqrt(a^2+1)),

with t =1.

This confirms Zo Maar's observation that these lines are tangents to the unit circle (since t > 1, except at the single minimum point).