The circle quadratic equation [br][center][color=#980000][i]x[/i][sup]2[/sup][/color] + [color=#0000ff][i]y[/i][sup]2[/sup][/color] = [color=#38761d]r[sup]2[/sup][/color] = const [1][br][/center]means that the sum of two variable areas [color=#980000][i]x[/i][sup]2[/sup][/color] and [color=#0000ff][i]y[/i][sup]2[/sup][/color] equals the constant area [color=#38761d]r[sup]2[/sup][/color]. Here we will show what these areas look like.[br][br]Taking differential of that quadratic equation, we have the differential equation [br][center][i]x[/i]⋅d[i]x[/i] + [i]y[/i]⋅d[i]y[/i] = 0 [2].[/center][br]This differential equation gives us a tool to [b]project the differentials of the quarter arc [color=#38761d]OY[sub]0[/sub]X[sub]0[/sub][/color][/b] onto x-axis and y-axis, where each [b]image is a [/b][b]differential sector[/b] swept by [i][b]r[/b][/i] along a differential arc δ as well as along a differential angle d[i]φ[/i] = δ/r, as shown by Projection lines (in the applet above). That means the [color=#38761d]circular area[/color] [color=#38761d]∫d[i]x⋅[/i]d[i]y[/i][/color] along [color=#38761d]OY[sub]0[/sub]X[sub]0[/sub][/color] equals to the [i]linear area[/i] ½r⋅δ of each differential sector which corresponds to the constant [color=#38761d]r[sup]2[/sup][/color] via ½r⋅δ = ½[color=#38761d]r[sup]2[/sup][/color](δ/r) = ½[color=#38761d]r[sup]2[/sup][/color]d[i]φ[/i]. [br][br]The quarter arc [color=#38761d]OY[sub]0[/sub]X[sub]0[/sub][/color] is split by [i][b]r[/b][/i] into two arcs [color=#0000ff]OY[sub]0[/sub]P[/color] and [color=#980000]OPX[sub]0[/sub][/color] whose images on axes correspond to [color=#0000ff][i]y[/i][sup]2[/sup][/color] and [color=#980000][i]x[/i][sup]2[/sup][/color], respectively, with areas ½[color=#0000ff][i]y[/i][sup]2[/sup][/color]d[i]φ[/i] and ½[color=#980000][i]x[/i][sup]2[/sup][/color]d[i]φ[/i].

The eq.[2] means the differential tangent vector d[i][b]r[/b][/i] = (d[i]x[/i], d[i]y[/i]) is orthogonal to the radius vector [i][b]r[/b][/i] = ([i]x[/i], [i]y[/i]), because their scalar product is zero. By choosing an arbitrarily small value δ = |d[i][b]r[/b][/i]| to set the length of all differential tangent vectors, eq.[2] also means that the radius rectangle OXPY = [[i]x[/i], [i]y[/i]] can be [b]rotated 90° and scaled down a factor δ/r[/b] to become the differential tangent rectangle [-d[i]x[/i], d[i]y[/i]]:[br][center][i][/i][i]x[/i](-d[i]x[/i]) = [i]y[/i]⋅d[i]y [/i] [2'],[br][math]\frac{x}{y}=\frac{dy}{-dx}[/math] [3],[br][i][/i][math]\frac{dy}{x}=\frac{-dx}{y}=\frac{d\vec{r}}{r}=\frac{\delta}{r}=d\varphi[/math] [4]. [/center][br]Then we can translate the differential tangent rectangle [-d[i]x[/i], d[i]y[/i]] to x-axis and to y-axis; Applying eq.[4] we have:[br][br][center][math]\left(-dx\right)dy=\left(-dx\right)x\frac{\delta}{r}=\frac{\delta}{r}y\cdot dy[/math] [5];[/center][list][*]On x-axis: The area of the differential quarter arc OY[sub]0[/sub]X[sub]0[/sub] [br][math]\int_{Y_0}^{X_0}\left(-dx\right)dy=\int_r^0\left(-dx\right)x\frac{\delta}{r}=\frac{\delta}{r}\int_0^rx\cdot dx=\frac{\delta}{r}\cdot\frac{r^2}{2}=\frac{r^2}{2}d\varphi[/math] [br]is the area of the differential sector OY[sub]0[/sub]Y[sub]1[/sub].[/*][*]On y-axis: The area of the differential quarter arc OY[sub]0[/sub]X[sub]0[/sub] [br][math]\int_{Y_0}^{X_0}\left(-dx\right)\cdot dy=\int_0^r\frac{\delta}{r}y\cdot dy=\frac{\delta}{r}\int_0^ry\cdot dy=\frac{\delta}{r}\cdot\frac{r^2}{2}=\frac{r^2}{2}d\varphi[/math] [br]is the area of the differential sector OX[sub]0[/sub]X[sub]1.[/sub][/*][/list][br]

The [color=#38761d]circular area[/color] [color=#38761d]∫d[i]x⋅[/i]d[i]y[/i][/color] along the arc [color=#38761d]OY[sub]0[/sub]X[sub]0[/sub][/color] is split by [b][i]r[/i][/b] (P) into two arcs [color=#0000ff]OY[sub]0[/sub]P[/color] and [color=#980000]OPX[sub]0[/sub][/color], [br]The images of P on x-axis and y-axis also split differential sectors into (differential) top triangles and (differential) bottom trapezoids, which correspond to the two arcs [color=#0000ff]OY[sub]0[/sub]P[/color] and [color=#980000]OPX[sub]0[/sub][/color]:[br][list][*]The [color=#0000ff]y[sup]2[/sup][/color] part as shown by Projection lines > y[sup]2[/sup]:[br][math]\int_{Y_0}^P\left(-dx\right)dy=\frac{\delta}{r}\left(\int_0^yy\cdot dy=\int_x^rx\cdot dx\right)=\frac{\delta}{r}\left(\frac{y^2}{2}=\frac{r^2-x^2}{2}\right)[/math][br]The area along arc [color=#0000ff]OY[sub]0[/sub]P[/color] = top triangle on y-axis [color=#0000ff]OYY' (y[sup]2[/sup])[/color] = bottom trapezoid on x-axis XY[sub]0[/sub]Y[sub]1[/sub]X' ([color=#38761d]r[sup]2[/sup][/color] - [color=#980000]x[sup]2[/sup][/color]).[/*][*]The [color=#980000]x[sup]2[/sup][/color] part as shown by Projection lines > x[sup]2[/sup]:[br][math]\int_P^{X_0}\left(-dx\right)dy=\frac{\delta}{r}\left(\int_0^xx\cdot dx=\int_y^ry\cdot dy\right)=\frac{\delta}{r}\left(\frac{x^2}{2}=\frac{r^2-y^2}{2}\right)[/math][br]The area along arc [color=#980000]OPX[sub]0[/sub][/color] = top triangle on x-axis [color=#980000]OXX' (x[sup]2[/sup])[/color] = bottom trapezoid on y-axis XX[sub]0[/sub]X[sub]1[/sub]Y' ([color=#38761d]r[sup]2[/sup][/color] - [color=#0000ff]y[sup]2[/sup][/color]).[/*][/list][br][br]

The result of differential projection of the quarter arc OY[sub]0[/sub]X[sub]0[/sub] onto axes is sawtooth waveforms in the differential sectors. Considering the sectors as flat cones, whose two edges of r are identical ("wrapped around" like in the flat torus), those sawtooth waveforms become conical spirals.[br][br]When the Oxy plane is a position-momentum phase space, the flat cone of x-axis shows the wrapped-around total space of that phase space preserving the full momentum (velocity) vectors, whereas the momentum (velocity) vectors on x-axis are just collapsed (projected) image of these vectors.

The circle can be viewed a [url=https://en.wikipedia.org/wiki/Phase_portrait]phase portrait[/url] of a [url=https://en.wikipedia.org/wiki/Harmonic_oscillator]simple harmonic oscillator[/url] [center][math]F=ma=-ks[/math]; [math]\omega=\sqrt{\frac{k}{m}}[/math].[br][/center]There, x-coordinate is the position [i]s[/i] and y-coordinate is the scaled momentum [br][center][/center][center][math]x=s=r\cos\left(\omega t\right)[/math][br][math]y=\frac{p}{\sqrt{km}}=v\sqrt{\frac{m}{k}}=\frac{v}{\omega}=-r\sin\left(\omega t\right)[/math];[br][math]\frac{\delta}{r}=d\varphi=\omega dt[/math];[br][/center]The differential phase d[i]φ[/i] = δ/r = [i]ω[/i]d[i]t[/i] sweeps a differential sector A = ½[color=#38761d]r[sup]2[/sup][/color]d[i]φ[/i] corresponding to time-differential energy [br][center][math]E_t=d_tE=\frac{2}{\pi}Ed\varphi=\frac{r^2}{\pi}kd\varphi=\frac{2k}{\pi}A[/math] [br][/center]and the differential tangent rectangle [i]a[/i][-d[i]x[/i], d[i]y[/i]] corresponds to spacetime-differential energy[br][center][math]E_{st}=d_sd_tS=F\cdot ds\cdot dt=a\sqrt{km}[/math][/center]The current position [b][i]P[/i][/b] splits the time-differential energy [i]E[sub]t[/sub][/i] into two components: potential energy [color=#980000]U[/color] ~ [color=#980000][i]x[/i][sup]2[/sup][/color] and kinetic [color=#0000ff]K[/color] ~ [color=#0000ff][i]y[/i][sup]2[/sup][/color].[br][center][math]U=\frac{x^2}{\pi}kd\varphi[/math]; [math]K=\frac{y^2}{\pi}kd\varphi[/math][/center][br]When [i]ω[/i] = 1, we have simple formulae: [i]x[/i] = [i]s[/i] = [i]r[/i] cos([i]t[/i]); [i]y[/i] = [i]v[/i] = -[i]r[/i] sin([i]t[/i]).[br][br][br]