Slope of tangent in $(x,y)$ on a circle $K=\{(x,y)\in\mathbb{R}^2|x^2+y^2=r^2\}$ with initial conditions I came across an exercise with a sample solution that I unfortunately don't fully understand given that it's shortened.
Let $(x,y)$ be a point on a circle $K=\{(x,y)\in\mathbb{R}^2|x^2+y^2=r^2\}$ with $y\neq0$. Show that the slope of the tangent on $K$ in $(x,y)$ is $-\frac{x}{y}$ and find two solutions for the differential equation $y'=-\frac{x}{y}$ that adhere to the initial conditions $y(1)=1$ or $y(-1)=2$.
The sample solution is as follows:
$$x^2+y^2=r^2$$
$$2x+2y\frac{dy}{dx}=0$$
$$\Rightarrow\frac{dy}{dx}=-\frac{x}{y}$$
$$\int y\;dy=\int -x\;dx$$
$$y^2=c-x^2$$
$$c=2\;or\;c=5$$
$$\Rightarrow x^2+y^2=2\;or\;x^2+y^2=5$$
First of all I'm interested in line 2.
From what I could find and understand, we come to this point through implicit differentiation of line 1. I assume the $r^2$ disappeared because it is treated as a constant but why exactly is that the case?
Furthermore, why is the constant of integration omitted on one side of the equation and how exactly do we find the solutions $2$ and $5$ for it?
Thank you very much in advance!
 A: *

*$r$ is indeed constant (otherwise $K$ wouldn't be a circle)

*It doesn't matter if separate constants of integration are included on both sides. (Two arbitrary constants can simply be combined: $c_1 + c_2$ is still just a constant).

*$c=2$ and $c=5$ come from the initial values: $y(1) = 1$ and $y(-1) = 2$, respectively. In the first case $x=1$, $y=1$ which gives $c=2$. Similarly for the second case.
A: *

*$r$ is indeed a constant. Think about it this way - the distance from the origin, which would be $x^2 + y^2$ remains constant. Geometrically, that's a circle. In another sense, if $r$ weren't a constant, the implication is that it depends on $x$. A circle's radius shouldn't depend on anything.

*The constant of integration is omitted on one side because the sample solution "skipped a step." Really, both integrals should end up with a constant $c$. However, since both constants are arbitrary, we can just lump them into one constant and put it on one side of the equation.

*You can solve for $c$ by plugging in the initial values for $x$ and $y$ into your solution equation.
A: I purpose another solution using the parametric representation of the circle. Consider the mapping $c :[0,2\pi) \to \mathbb{R}^2$ given by:
$$c(t)=(r\cos t, r\sin t)$$
It is obvious that $c([0,2\pi))$ is your circle, notice that the interval is open in $2\pi$ so that we don't pass through the point $(1,0)$ twice. Now, $c$ is the parametric representation of your circle, and so the derivative at a point is the tangent vector at the point. We have:
$$c'(t)=(-r\sin t,r \cos t)$$
Now fix $t_0$ so you fix a point $(x_0,y_0) = c(t_0)$ on the circle. Now the tangent line at $t_0$ is the line with direction vector $c'(t_0)$ and with starting point $c(t_0)$. The slope of that line is just the component of the tangent vector on the $y$ direction divided by the component of that vector in the $x$ direction, so we have:
$$m = \frac{\phantom{-}r\cos t_0}{- r\sin t_0 } = -\cot t_0$$
However, as you probably can see from the initial construction of $c(t)$, the value of $t$ is related to the point $(x,y)$ on the circle by $t = \arctan(y/x)$ so that:
$$m = - \frac{x_0}{y_0}$$
And this is the slope of the tangent to the circle at the point $(x_0,y_0)$.
