Checking Whether the Points $(x, y, w)^T = (1, pm i, 0)^T$ Lie in the Circle $(x - aw)^2 + (y - bw)^2 = r^2...
Page 5 of my computer vision textbook, Multiple View Geometry in Computer Vision, says the following:
The equation for a circle in homogeneous coordinates $(x, y, w)$ is of the form
$$(x - aw)^2 + (y - bw)^2 = r^2 w^2$$
This represents the circle with centre represented in homogeneous coordinates as $(x_0, y_0, w_0)^T = (a, b, 1)^T$. It is quickly verified that the points $(x, y, w)^T = (1, pm i, 0)^T$ lie on every such circle.
The aforementioned section of the textbook is available freely here.
Now, as I understand it, I can verify that the points lie in the circle by using the Euclidean distance:
$$d(x, y, w) = sqrt{(x - x_0)^2 + (y - y_0)^2 + (w - w_0)^2}$$
However, if I plug in the provided values, then it seems that we would end up with the following two messy expressions:
$$d(1, i, 0) = sqrt{(1 - a)^2 + (i - b)^2 + 1} tag{1}$$
$$d(1, -i, 0) = sqrt{(1 - a)^2 + (-i - b)^2 + 1} tag{2}$$
It is not clear to me how/whether this tells us that the points lie in the circle?
So how does one verify that the points lie on every such circle? The author states that this is "quickly verifiable".
I would greatly appreciate it if people could please take the time to clarify this.
geometry euclidean-geometry circle projective-geometry
asked Nov 26 '18 at 18:49
The Pointer
2,60221335
|
show 1 more comment
Page 5 of my computer vision textbook, Multiple View Geometry in Computer Vision, says the following:
The equation for a circle in homogeneous coordinates $(x, y, w)$ is of the form
$$(x - aw)^2 + (y - bw)^2 = r^2 w^2$$
This represents the circle with centre represented in homogeneous coordinates as $(x_0, y_0, w_0)^T = (a, b, 1)^T$. It is quickly verified that the points $(x, y, w)^T = (1, pm i, 0)^T$ lie on every such circle.
The aforementioned section of the textbook is available freely here.
Now, as I understand it, I can verify that the points lie in the circle by using the Euclidean distance:
$$d(x, y, w) = sqrt{(x - x_0)^2 + (y - y_0)^2 + (w - w_0)^2}$$
However, if I plug in the provided values, then it seems that we would end up with the following two messy expressions:
$$d(1, i, 0) = sqrt{(1 - a)^2 + (i - b)^2 + 1} tag{1}$$
$$d(1, -i, 0) = sqrt{(1 - a)^2 + (-i - b)^2 + 1} tag{2}$$
It is not clear to me how/whether this tells us that the points lie in the circle?
So how does one verify that the points lie on every such circle? The author states that this is "quickly verifiable".
I would greatly appreciate it if people could please take the time to clarify this.
geometry euclidean-geometry circle projective-geometry
asked Nov 26 '18 at 18:49
The Pointer
2,60221335
When you say the you "plug in the values into the equation for the circle in homogeneous coordinates", are you plugging $x$ in for $aw$ and $y$ in for $bw$? Maybe you could give some more detail on what exactly you did at this step.
– gd1035
Nov 26 '18 at 18:56
@gd1035 I made an oversight. Will change question now.
– The Pointer
Nov 26 '18 at 18:58
What do you mean by the Euclidean distance between an imaginary point (either of the circular points) and a real one (the circle’s center)?
– amd
Nov 26 '18 at 21:34
1
Distance isn’t meaningful in projective geometry in the first place, let alone the distance between a real and imaginary point. Trying to verify that those points lie on any circle by trying to compute distances was the wrong approach.
– amd
Nov 26 '18 at 23:03
1
One of the main points of that section is that you impose an Euclidean geometry on the projective plane by choosing a pair of points to serve as the circular points. All non-degenerate conics are projectively equivalent, so you can only distinguish between ellipses, parabolas and hyperbolas by choosing a line to be the line at infinity. Similarly, distinguishing circles from ellipses requires imposing a Euclidean geometry on the plane: circles are then those ellipses that pass through the two distinguished points.
– amd
Nov 26 '18 at 23:07
|
show 1 more comment
Page 5 of my computer vision textbook, Multiple View Geometry in Computer Vision, says the following:
The equation for a circle in homogeneous coordinates $(x, y, w)$ is of the form
$$(x - aw)^2 + (y - bw)^2 = r^2 w^2$$
This represents the circle with centre represented in homogeneous coordinates as $(x_0, y_0, w_0)^T = (a, b, 1)^T$. It is quickly verified that the points $(x, y, w)^T = (1, pm i, 0)^T$ lie on every such circle.
The aforementioned section of the textbook is available freely here.
Now, as I understand it, I can verify that the points lie in the circle by using the Euclidean distance:
$$d(x, y, w) = sqrt{(x - x_0)^2 + (y - y_0)^2 + (w - w_0)^2}$$
However, if I plug in the provided values, then it seems that we would end up with the following two messy expressions:
$$d(1, i, 0) = sqrt{(1 - a)^2 + (i - b)^2 + 1} tag{1}$$
$$d(1, -i, 0) = sqrt{(1 - a)^2 + (-i - b)^2 + 1} tag{2}$$
It is not clear to me how/whether this tells us that the points lie in the circle?
So how does one verify that the points lie on every such circle? The author states that this is "quickly verifiable".
I would greatly appreciate it if people could please take the time to clarify this.
geometry euclidean-geometry circle projective-geometry
asked Nov 26 '18 at 18:49
The Pointer
2,60221335
Page 5 of my computer vision textbook, Multiple View Geometry in Computer Vision, says the following:
The equation for a circle in homogeneous coordinates $(x, y, w)$ is of the form
$$(x - aw)^2 + (y - bw)^2 = r^2 w^2$$
This represents the circle with centre represented in homogeneous coordinates as $(x_0, y_0, w_0)^T = (a, b, 1)^T$. It is quickly verified that the points $(x, y, w)^T = (1, pm i, 0)^T$ lie on every such circle.
The aforementioned section of the textbook is available freely here.
Now, as I understand it, I can verify that the points lie in the circle by using the Euclidean distance:
$$d(x, y, w) = sqrt{(x - x_0)^2 + (y - y_0)^2 + (w - w_0)^2}$$
However, if I plug in the provided values, then it seems that we would end up with the following two messy expressions:
$$d(1, i, 0) = sqrt{(1 - a)^2 + (i - b)^2 + 1} tag{1}$$
$$d(1, -i, 0) = sqrt{(1 - a)^2 + (-i - b)^2 + 1} tag{2}$$
It is not clear to me how/whether this tells us that the points lie in the circle?
So how does one verify that the points lie on every such circle? The author states that this is "quickly verifiable".
I would greatly appreciate it if people could please take the time to clarify this.
geometry euclidean-geometry circle projective-geometry
geometry euclidean-geometry circle projective-geometry
asked Nov 26 '18 at 18:49
The Pointer
2,60221335
asked Nov 26 '18 at 18:49
The Pointer
2,60221335
edited Nov 26 '18 at 19:10
asked Nov 26 '18 at 18:49
The Pointer
2,60221335
asked Nov 26 '18 at 18:49
The Pointer
2,60221335
asked Nov 26 '18 at 18:49
The Pointer
2,60221335
2,60221335
When you say the you "plug in the values into the equation for the circle in homogeneous coordinates", are you plugging $x$ in for $aw$ and $y$ in for $bw$? Maybe you could give some more detail on what exactly you did at this step.
– gd1035
Nov 26 '18 at 18:56
@gd1035 I made an oversight. Will change question now.
– The Pointer
Nov 26 '18 at 18:58
What do you mean by the Euclidean distance between an imaginary point (either of the circular points) and a real one (the circle’s center)?
– amd
Nov 26 '18 at 21:34
1
Distance isn’t meaningful in projective geometry in the first place, let alone the distance between a real and imaginary point. Trying to verify that those points lie on any circle by trying to compute distances was the wrong approach.
– amd
Nov 26 '18 at 23:03
1
One of the main points of that section is that you impose an Euclidean geometry on the projective plane by choosing a pair of points to serve as the circular points. All non-degenerate conics are projectively equivalent, so you can only distinguish between ellipses, parabolas and hyperbolas by choosing a line to be the line at infinity. Similarly, distinguishing circles from ellipses requires imposing a Euclidean geometry on the plane: circles are then those ellipses that pass through the two distinguished points.
– amd
Nov 26 '18 at 23:07
|
show 1 more comment
When you say the you "plug in the values into the equation for the circle in homogeneous coordinates", are you plugging $x$ in for $aw$ and $y$ in for $bw$? Maybe you could give some more detail on what exactly you did at this step.
– gd1035
Nov 26 '18 at 18:56
@gd1035 I made an oversight. Will change question now.
– The Pointer
Nov 26 '18 at 18:58
What do you mean by the Euclidean distance between an imaginary point (either of the circular points) and a real one (the circle’s center)?
– amd
Nov 26 '18 at 21:34
1
Distance isn’t meaningful in projective geometry in the first place, let alone the distance between a real and imaginary point. Trying to verify that those points lie on any circle by trying to compute distances was the wrong approach.
– amd
Nov 26 '18 at 23:03
1
One of the main points of that section is that you impose an Euclidean geometry on the projective plane by choosing a pair of points to serve as the circular points. All non-degenerate conics are projectively equivalent, so you can only distinguish between ellipses, parabolas and hyperbolas by choosing a line to be the line at infinity. Similarly, distinguishing circles from ellipses requires imposing a Euclidean geometry on the plane: circles are then those ellipses that pass through the two distinguished points.
– amd
Nov 26 '18 at 23:07
When you say the you "plug in the values into the equation for the circle in homogeneous coordinates", are you plugging $x$ in for $aw$ and $y$ in for $bw$? Maybe you could give some more detail on what exactly you did at this step.
– gd1035
Nov 26 '18 at 18:56
When you say the you "plug in the values into the equation for the circle in homogeneous coordinates", are you plugging $x$ in for $aw$ and $y$ in for $bw$? Maybe you could give some more detail on what exactly you did at this step.
– gd1035
Nov 26 '18 at 18:56
@gd1035 I made an oversight. Will change question now.
– The Pointer
Nov 26 '18 at 18:58
@gd1035 I made an oversight. Will change question now.
– The Pointer
Nov 26 '18 at 18:58
What do you mean by the Euclidean distance between an imaginary point (either of the circular points) and a real one (the circle’s center)?
– amd
Nov 26 '18 at 21:34
What do you mean by the Euclidean distance between an imaginary point (either of the circular points) and a real one (the circle’s center)?
– amd
Nov 26 '18 at 21:34
1
1
Distance isn’t meaningful in projective geometry in the first place, let alone the distance between a real and imaginary point. Trying to verify that those points lie on any circle by trying to compute distances was the wrong approach.
– amd
Nov 26 '18 at 23:03
Distance isn’t meaningful in projective geometry in the first place, let alone the distance between a real and imaginary point. Trying to verify that those points lie on any circle by trying to compute distances was the wrong approach.
– amd
Nov 26 '18 at 23:03
1
1
One of the main points of that section is that you impose an Euclidean geometry on the projective plane by choosing a pair of points to serve as the circular points. All non-degenerate conics are projectively equivalent, so you can only distinguish between ellipses, parabolas and hyperbolas by choosing a line to be the line at infinity. Similarly, distinguishing circles from ellipses requires imposing a Euclidean geometry on the plane: circles are then those ellipses that pass through the two distinguished points.
– amd
Nov 26 '18 at 23:07
One of the main points of that section is that you impose an Euclidean geometry on the projective plane by choosing a pair of points to serve as the circular points. All non-degenerate conics are projectively equivalent, so you can only distinguish between ellipses, parabolas and hyperbolas by choosing a line to be the line at infinity. Similarly, distinguishing circles from ellipses requires imposing a Euclidean geometry on the plane: circles are then those ellipses that pass through the two distinguished points.
– amd
Nov 26 '18 at 23:07
|
show 1 more comment
1 Answer
1
active
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votes
Plug $x=1$, $y=pm i$ amd $w=0$ into the equation, to get:
$$
(1-0)^2+(pm i-0)^2 = 0,
$$
which is obviously always true.
answered Nov 26 '18 at 19:19
Aretino
22.7k21442
Ahh, I’m not sure if you saw the post before the edit, but this is the direction I was going with. But how does one interpret this? What is always true?
– The Pointer
Nov 26 '18 at 19:23
It's true that those points obey the equation, and then by definition lie on the circle represented by that equation.
– Aretino
Nov 26 '18 at 19:24
So if the points obey the equation, then they must, by definition, be part of the circle?
– The Pointer
Nov 26 '18 at 19:26
It depends on how you define a circle, but whether by definition or not they do lie on the circle.
– Aretino
Nov 26 '18 at 20:25
okay, thanks for the clarification.
– The Pointer
Nov 26 '18 at 20:27
add a comment |
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Plug $x=1$, $y=pm i$ amd $w=0$ into the equation, to get:
$$
(1-0)^2+(pm i-0)^2 = 0,
$$
which is obviously always true.
answered Nov 26 '18 at 19:19
Aretino
22.7k21442
Ahh, I’m not sure if you saw the post before the edit, but this is the direction I was going with. But how does one interpret this? What is always true?
– The Pointer
Nov 26 '18 at 19:23
It's true that those points obey the equation, and then by definition lie on the circle represented by that equation.
– Aretino
Nov 26 '18 at 19:24
So if the points obey the equation, then they must, by definition, be part of the circle?
– The Pointer
Nov 26 '18 at 19:26
It depends on how you define a circle, but whether by definition or not they do lie on the circle.
– Aretino
Nov 26 '18 at 20:25
okay, thanks for the clarification.
– The Pointer
Nov 26 '18 at 20:27
add a comment |
Plug $x=1$, $y=pm i$ amd $w=0$ into the equation, to get:
$$
(1-0)^2+(pm i-0)^2 = 0,
$$
which is obviously always true.
answered Nov 26 '18 at 19:19
Aretino
22.7k21442
Ahh, I’m not sure if you saw the post before the edit, but this is the direction I was going with. But how does one interpret this? What is always true?
– The Pointer
Nov 26 '18 at 19:23
It's true that those points obey the equation, and then by definition lie on the circle represented by that equation.
– Aretino
Nov 26 '18 at 19:24
So if the points obey the equation, then they must, by definition, be part of the circle?
– The Pointer
Nov 26 '18 at 19:26
It depends on how you define a circle, but whether by definition or not they do lie on the circle.
– Aretino
Nov 26 '18 at 20:25
okay, thanks for the clarification.
– The Pointer
Nov 26 '18 at 20:27
add a comment |
Plug $x=1$, $y=pm i$ amd $w=0$ into the equation, to get:
$$
(1-0)^2+(pm i-0)^2 = 0,
$$
which is obviously always true.
answered Nov 26 '18 at 19:19
Aretino
22.7k21442
Plug $x=1$, $y=pm i$ amd $w=0$ into the equation, to get:
$$
(1-0)^2+(pm i-0)^2 = 0,
$$
which is obviously always true.
answered Nov 26 '18 at 19:19
Aretino
22.7k21442
answered Nov 26 '18 at 19:19
Aretino
22.7k21442
answered Nov 26 '18 at 19:19
Aretino
22.7k21442
answered Nov 26 '18 at 19:19
Aretino
22.7k21442
22.7k21442
Ahh, I’m not sure if you saw the post before the edit, but this is the direction I was going with. But how does one interpret this? What is always true?
– The Pointer
Nov 26 '18 at 19:23
It's true that those points obey the equation, and then by definition lie on the circle represented by that equation.
– Aretino
Nov 26 '18 at 19:24
So if the points obey the equation, then they must, by definition, be part of the circle?
– The Pointer
Nov 26 '18 at 19:26
It depends on how you define a circle, but whether by definition or not they do lie on the circle.
– Aretino
Nov 26 '18 at 20:25
okay, thanks for the clarification.
– The Pointer
Nov 26 '18 at 20:27
add a comment |
Ahh, I’m not sure if you saw the post before the edit, but this is the direction I was going with. But how does one interpret this? What is always true?
– The Pointer
Nov 26 '18 at 19:23
It's true that those points obey the equation, and then by definition lie on the circle represented by that equation.
– Aretino
Nov 26 '18 at 19:24
So if the points obey the equation, then they must, by definition, be part of the circle?
– The Pointer
Nov 26 '18 at 19:26
It depends on how you define a circle, but whether by definition or not they do lie on the circle.
– Aretino
Nov 26 '18 at 20:25
okay, thanks for the clarification.
– The Pointer
Nov 26 '18 at 20:27
Ahh, I’m not sure if you saw the post before the edit, but this is the direction I was going with. But how does one interpret this? What is always true?
– The Pointer
Nov 26 '18 at 19:23
Ahh, I’m not sure if you saw the post before the edit, but this is the direction I was going with. But how does one interpret this? What is always true?
– The Pointer
Nov 26 '18 at 19:23
It's true that those points obey the equation, and then by definition lie on the circle represented by that equation.
– Aretino
Nov 26 '18 at 19:24
It's true that those points obey the equation, and then by definition lie on the circle represented by that equation.
– Aretino
Nov 26 '18 at 19:24
So if the points obey the equation, then they must, by definition, be part of the circle?
– The Pointer
Nov 26 '18 at 19:26
So if the points obey the equation, then they must, by definition, be part of the circle?
– The Pointer
Nov 26 '18 at 19:26
It depends on how you define a circle, but whether by definition or not they do lie on the circle.
– Aretino
Nov 26 '18 at 20:25
It depends on how you define a circle, but whether by definition or not they do lie on the circle.
– Aretino
Nov 26 '18 at 20:25
okay, thanks for the clarification.
– The Pointer
Nov 26 '18 at 20:27
okay, thanks for the clarification.
– The Pointer
Nov 26 '18 at 20:27
add a comment |
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When you say the you "plug in the values into the equation for the circle in homogeneous coordinates", are you plugging $x$ in for $aw$ and $y$ in for $bw$? Maybe you could give some more detail on what exactly you did at this step.
– gd1035
Nov 26 '18 at 18:56
@gd1035 I made an oversight. Will change question now.
– The Pointer
Nov 26 '18 at 18:58
What do you mean by the Euclidean distance between an imaginary point (either of the circular points) and a real one (the circle’s center)?
– amd
Nov 26 '18 at 21:34
1
Distance isn’t meaningful in projective geometry in the first place, let alone the distance between a real and imaginary point. Trying to verify that those points lie on any circle by trying to compute distances was the wrong approach.
– amd
Nov 26 '18 at 23:03
1
One of the main points of that section is that you impose an Euclidean geometry on the projective plane by choosing a pair of points to serve as the circular points. All non-degenerate conics are projectively equivalent, so you can only distinguish between ellipses, parabolas and hyperbolas by choosing a line to be the line at infinity. Similarly, distinguishing circles from ellipses requires imposing a Euclidean geometry on the plane: circles are then those ellipses that pass through the two distinguished points.
– amd
Nov 26 '18 at 23:07