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The equation of plane straight line is: $$y=Kx+b$$

In spherical geometry,the spherical circle equation is:

$$ \beta=\arctan ((1+(\tan \theta )^2) K\sin \alpha + \tan \theta \sqrt {(1+(1+(\tan \theta )^2)(K \sin \alpha)^2} )$$

This is the circle circle equation on the sphere. The alpha is longitude, and the beta is latitudes. $K$ is the slope of a circle, the tangent of angle between plane of a circle and plane of equator. $\theta$ is the distance between point of circle and equator when alpha equals zero. When alpha equals zero, theta equals the value of beta.

When diameter of a sphere tends to infinity, the equation of spherical circle can be changed to:

$$ \beta=K\alpha+ \theta $$

So the plane circles are only a special case of spherical circles. The spherical circles are essentially straight lines.

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You can say that lines in the plane is a limiting case of great circles on spheres, but they are not a special case of great circles, because they are not actually great circles.

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  • $\begingroup$ The equations of my spherical circles are the equations of the great circles, and the equations of the small circles. I mean, the plane lines are the special cases of the great circles and the small circles, when the diameter of the sphere tends to infinity. $\endgroup$ – enbin zheng Apr 3 '18 at 1:30
  • $\begingroup$ @enbinzheng: Lines are not small circles either -- they're not circles at all. $\endgroup$ – Henning Makholm Apr 3 '18 at 1:34
  • $\begingroup$ But,my equation contains big circles and small circles. My equation is the general equation. $\endgroup$ – enbin zheng Apr 3 '18 at 1:41
  • $\begingroup$ @enbinzheng The general equation of what? Lines in Euclidean geometry aren’t circles. If you start at any point on a circle and travel along the curve, you eventually return to your starting point. If you start at any point on a straight line and travel in either direction along the line, when do you return to the starting point? $\endgroup$ – amd Apr 3 '18 at 1:52
  • $\begingroup$ @amd I mean that my equation is general equation of circles on a sphere. $\endgroup$ – enbin zheng Apr 3 '18 at 1:58
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Using the stereographic projection, we can identify the plane $\mathbb{R}^2$ as the subset $S^2 \setminus \{N\}$ of the sphere $S^2$, where $N$ is the north pole. This is actually a diffeomorphism of smooth manifolds, if that means anything to you. If not, don't worry about it. All that really matters here is that the stereographic projection preserves angles and circles.

Under the sterographic projection, lines in the plane are identified with great circles on $S^2$ that pass through the north pole. Other great circles on $S^2$ get identified with circles in the plane. So in this sense, yes, lines in the plane are a special case of great circles.

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  • $\begingroup$ The equations of my spherical circles are the equations of the great circles, and the equations of the small circles. I mean, the plane lines are the special cases of the great circles and the small circles, when the diameter of the sphere tends to infinity. $\endgroup$ – enbin zheng Apr 3 '18 at 1:34
  • $\begingroup$ Right. In terms of the stereographic projection, as a great circle on the sphere gets closer and closer to crossing the north pole, the corresponding circle in the plane has radius tending to infinity. The great thing about the stereographic projection is that you can actually say what happens "at infinity," because a great circle through the north pole corresponds to a line in the plane. $\endgroup$ – Joshua Ruiter Apr 3 '18 at 2:00
  • $\begingroup$ @enbinzheng: That doesn't become true no matter how many times you repeat it in response to people telling you it isn't true. $\endgroup$ – Henning Makholm Apr 3 '18 at 2:00
  • $\begingroup$ @JoshuaRuiter I mean that the small circles are also straight lines. $\endgroup$ – enbin zheng Apr 3 '18 at 2:14
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    $\begingroup$ Based on your past history of questions and extended attempts to justify yourself in comments, I'm going to assume you aren't going to listen to me, so I'm not going to continue to explain my answer. $\endgroup$ – Joshua Ruiter Apr 3 '18 at 4:04

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