Find the curve $y=y(x)$ which satisfies this condition This is the problem that I'm trying to solve:

Let $y=y(x)$ be a curve that passes through the point $(1,1)$. We also know
  that y(x) satisfies the following property: for each $P(x,y) \in
 y$, the tangent at $P$ is at a distance from $(0,0)$ equal to
  the distance from $(0,y)$ to $(0,0)$. Find $y=y(x).$

I made a sketch that (hopefully) will help understand the problem a tad better:
$\hskip 2.2in$
By the definition of that property, and without doing any maths, I understood $y(x)$ needed to be a circumference. But let us assume we don't know this for the moment.
This is how I proceeded:


*

*I found the general formula for the tangent to $y(x)$ at $A(x_0, y_0)$: $T(x,y)=0$, where $T(x,y)=y-y_0-y^\prime(x_0)(x-x_0)$ $$\Rightarrow \boxed{y-y_0-y^\prime(x_0)(x-x_0)=0}$$

*I found the direction vector of the perpendicular line to the tangent at $A(x_0, y_0)$: $\vec{\nabla}T = \left(-y^\prime(x_0),\,1\right)$

*Therefore, the equation of the perpendicular to the tangent to $y(x)$ (sorry for the tongue-twister) that crosses the origin $(0,0)$ will be: $R(x,y)=0$, where: $R(x,y)=(x-0)\partial_x T+(y-0)\partial_y T=-xy^\prime(x_0)+y$ $$\Rightarrow \boxed{-xy^\prime(x_0)+y=0}$$

*I found the intersection point between the tangent at $A(x_0,\,y_0)$ and the perpendicular line to this tangent that crosses the origin: $$T(x,y)=R(x,y)=0 \Rightarrow y^\prime(x_0)=\dfrac{y_0}{x_0}$$
which doesn't really give me an intersection, but a condition that $y^\prime(x_0)$ needs to meet.


However, the point $(x_0,\,y_0)$ was arbitrarily taken. So the above condition can be written, more generally, as: $$y^\prime(x)=\dfrac{y(x)}{x}$$
I thought that solving for this differential equation would yield the equation of $y(x)$, but the solution that I got is far from the real solution: $$\dfrac{dx}{x}=\dfrac{dy}{y} \Rightarrow \boxed{\log(y)=log(x)+C}$$ Since $y(x)$ passes through $(1,1)$ according to the problem description, then $C=0$ and we are left with the equation of a line: $y=x$.
I tried to solve this problem in the most general way possible. The solution is obviously wrong. I'd like to know where I went wrong in my calculations.
Thanks
 A: Does your condition from step (4) describe the point A on the circle? I don't think it does. I think it describes the point where the tangent line intersects the perpendicular line to the origin, which is not on the curve y(x) as your diagram shows. I don't think you've made a computational error at all. You just changed the meaning of (x0, y0) halfway through the problem.
Just to confirm, you're right that the solution curve is a circle. Here's a purely geometric proof of that.
Let the point O be the origin (0,0). Let the point of tangency be P. Let the point (0,y) be Q1. Let the point where the line of tangency approaches O most closely be Q2;
Then length of OQ2 = length of OQ1 = y by supposition. Angle OQ1P is right because that's how the coordinate system works. Angle OQ2P is right because OQ2 is the perpendicular distance from O to line PQ2. Segment OP is shared.
Thus Triangles OQ1P and OQ2P are congruent, and the angle OPQ2 is equal in measure to angle OPQ1.
Now construct the perpendicular to the tangent line at P and extend it to meet the y-axis at Z.
The measure of angle ZOP = the complement of angle between OP and the x-axis = the complement of angle OPQ1 because PQ1 is parallel to the x-axis.
The measure of angle ZPO = the complement of angle OPQ2 = the complement of angle OPQ1 as well.
Thus triangle ZOP is isosceles and ZO = ZP. Furthermore, because ZP is perpendicular to the tangent line PQ2 by construction, PQ2 is tangent to the circle centered Z with radius ZP.
The information that (1,1) is on the circle ZP fixes the point Z at (0,1).
So you are right that it should be a circumference.
