What does the $z-z_0=m(y-y_0)$ equation do in this article? and why is it written into the sphere equation?, and how this 2 equations derived? I'm reading this topic Analytic Treatment of the Perspective View of a Circle in this article. I don't understand the subject, so I decided to analyze it piece by piece.

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*$z-z_0 = m(y-y_0)$ What is the equation used for?

*And why do they put it inside the sphere equation?

$(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2$
-->
$(x - x_0)^2 + (1 + m^2)(y - y_0)^2 = r^2$
Edit:Finally got it, thanks to your contributions. I just couldn't figure out how the 2 equations at the end were derived, I would appreciate it if you could explain.
$y-y_0=\frac{z_0-vy_0}{v-m}$
$x-x_0=uy-x_0=\frac{u(z_0-my_0)-vx_0+mx_0}{v-m}$
 A: $z-z_0=m(y-y_0)$ and $x$ arbitrary describes a plane with normal direction $(0,1,m)$. Projection of the sphere in this normal direction onto the $xy$-plane will leave an elliptical shadow that is described by the ellipse equation $$(x-x_0)^2+(1+m^2)(y-y_0)^2=r^2.$$

As to the wider context of the question: The equations given describe a scene in 3D space, specifically a circle defined as intersection of a sphere and a plane.
A camera is situated with its lense or hole (for a pinhole camera) at $(0,0,0)$. The physically correct image plane would be behind the lense, but mathematically it makes no difference (except a point-reflection) to use a plane at the same distance in-front of the lense.
The forward direction of the camera is here the y-direction, the image plane is positioned orthogonal to that at $y=1$. The optical ray for a point in the image plane is thus $(x,y,z)=(su,s,sv)=(yu,y,yv)$.
Together this results in an over-determined system connecting points in the image plane to the object in the scene
\begin{align}
x&=uy\\
z&=vy\\
z-z_0&=m(y-y_0)\\
r^2&=(x-x_0)^2+(y-y_0)^2+(z-z_0)^2
\end{align}
Use these equations to eliminate $x,y,z$ starting with $x,z$
\begin{align}
vy_0-z_0&=(m-v)(y-y_0)\\
r^2&=(u(y-y_0)+(uy_0-x_0))^2+(y-y_0)^2+(v(y-y_0)+(vy_0-z_0))^2
\end{align}
and then $y$, to get
\begin{align}
r^2(m-v)^2&=(u(vy_0-z_0)+(m-v)(uy_0-x_0))^2+(vy_0-z_0)^2+m^2(vy_0-z_0)^2
\\
&=((vx_0-uz_0)+m(uy_0-x_0))^2+(1+m^2)(vy_0-z_0)^2
\end{align}
for the equation of the projection in the image plane. Now one would need to expand and sort the terms to get the normal form for the resulting quadric.
