Circle whose radius is infinite
$begingroup$
I have the intuition that a circle whose radius is infinite is a straight line.
Nonetheless, I don’t feel that what I’ve just stated is really scientific as it has some vagueness and lacks precision.
What I would faithfully ask for is a more accurate expression of this idea ,if it were true, in addition to a proof thereof.
I would appreciate it if I were provided ,besides your own answers, with some sources to refer to.
Many thanks for considering my request.
geometry differential-geometry circle hyperbolic-geometry inversive-geometry
$endgroup$
|
show 2 more comments
$begingroup$
I have the intuition that a circle whose radius is infinite is a straight line.
Nonetheless, I don’t feel that what I’ve just stated is really scientific as it has some vagueness and lacks precision.
What I would faithfully ask for is a more accurate expression of this idea ,if it were true, in addition to a proof thereof.
I would appreciate it if I were provided ,besides your own answers, with some sources to refer to.
Many thanks for considering my request.
geometry differential-geometry circle hyperbolic-geometry inversive-geometry
$endgroup$
2
$begingroup$
There is no such thing as a circle with infinite radius.
$endgroup$
– Ben W
Jan 1 at 18:13
5
$begingroup$
There is such a thing as the limit of a circle whose radius tends to infinity and it passes through two fixed points, and that limit is the line passing through the fixed points.
$endgroup$
– Oscar Lanzi
Jan 1 at 18:17
1
$begingroup$
en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
$endgroup$
– Will Jagy
Jan 1 at 19:14
1
$begingroup$
There are many situations in which thinking of a line as a circle of infinite radius is helpful. See, for instance, Descartes' Theorem relating the curvatures of four mutually-tangent circles. The formulas "just work" when one or more of the circles is a line, and the corresponding curvature is $0$. (Note that curvature is the reciprocal of radius, so "zero curvature" matches "infinite radius".)
$endgroup$
– Blue
Jan 1 at 22:07
1
$begingroup$
@BenW: In geometry with the euclidean plane extended to the projective plane, it is common to talk about the line at infinity, which is also a circle with infinite radius. Any non-degenerate circle with centre on the line at infinity is a line in the plane. In the projective plane, any two distinct points have a unique line through them, and any two distinct lines have a unique intersection, and any two distinct intersecting circles have exactly two intersections counting multiplicity. Blue gave an example of a formula involving radius that requires a line to have infinite radius.
$endgroup$
– user21820
Jan 2 at 10:18
|
show 2 more comments
$begingroup$
I have the intuition that a circle whose radius is infinite is a straight line.
Nonetheless, I don’t feel that what I’ve just stated is really scientific as it has some vagueness and lacks precision.
What I would faithfully ask for is a more accurate expression of this idea ,if it were true, in addition to a proof thereof.
I would appreciate it if I were provided ,besides your own answers, with some sources to refer to.
Many thanks for considering my request.
geometry differential-geometry circle hyperbolic-geometry inversive-geometry
$endgroup$
I have the intuition that a circle whose radius is infinite is a straight line.
Nonetheless, I don’t feel that what I’ve just stated is really scientific as it has some vagueness and lacks precision.
What I would faithfully ask for is a more accurate expression of this idea ,if it were true, in addition to a proof thereof.
I would appreciate it if I were provided ,besides your own answers, with some sources to refer to.
Many thanks for considering my request.
geometry differential-geometry circle hyperbolic-geometry inversive-geometry
geometry differential-geometry circle hyperbolic-geometry inversive-geometry
edited Jan 1 at 18:12
asked Jan 1 at 18:08
user630906
2
$begingroup$
There is no such thing as a circle with infinite radius.
$endgroup$
– Ben W
Jan 1 at 18:13
5
$begingroup$
There is such a thing as the limit of a circle whose radius tends to infinity and it passes through two fixed points, and that limit is the line passing through the fixed points.
$endgroup$
– Oscar Lanzi
Jan 1 at 18:17
1
$begingroup$
en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
$endgroup$
– Will Jagy
Jan 1 at 19:14
1
$begingroup$
There are many situations in which thinking of a line as a circle of infinite radius is helpful. See, for instance, Descartes' Theorem relating the curvatures of four mutually-tangent circles. The formulas "just work" when one or more of the circles is a line, and the corresponding curvature is $0$. (Note that curvature is the reciprocal of radius, so "zero curvature" matches "infinite radius".)
$endgroup$
– Blue
Jan 1 at 22:07
1
$begingroup$
@BenW: In geometry with the euclidean plane extended to the projective plane, it is common to talk about the line at infinity, which is also a circle with infinite radius. Any non-degenerate circle with centre on the line at infinity is a line in the plane. In the projective plane, any two distinct points have a unique line through them, and any two distinct lines have a unique intersection, and any two distinct intersecting circles have exactly two intersections counting multiplicity. Blue gave an example of a formula involving radius that requires a line to have infinite radius.
$endgroup$
– user21820
Jan 2 at 10:18
|
show 2 more comments
2
$begingroup$
There is no such thing as a circle with infinite radius.
$endgroup$
– Ben W
Jan 1 at 18:13
5
$begingroup$
There is such a thing as the limit of a circle whose radius tends to infinity and it passes through two fixed points, and that limit is the line passing through the fixed points.
$endgroup$
– Oscar Lanzi
Jan 1 at 18:17
1
$begingroup$
en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
$endgroup$
– Will Jagy
Jan 1 at 19:14
1
$begingroup$
There are many situations in which thinking of a line as a circle of infinite radius is helpful. See, for instance, Descartes' Theorem relating the curvatures of four mutually-tangent circles. The formulas "just work" when one or more of the circles is a line, and the corresponding curvature is $0$. (Note that curvature is the reciprocal of radius, so "zero curvature" matches "infinite radius".)
$endgroup$
– Blue
Jan 1 at 22:07
1
$begingroup$
@BenW: In geometry with the euclidean plane extended to the projective plane, it is common to talk about the line at infinity, which is also a circle with infinite radius. Any non-degenerate circle with centre on the line at infinity is a line in the plane. In the projective plane, any two distinct points have a unique line through them, and any two distinct lines have a unique intersection, and any two distinct intersecting circles have exactly two intersections counting multiplicity. Blue gave an example of a formula involving radius that requires a line to have infinite radius.
$endgroup$
– user21820
Jan 2 at 10:18
2
2
$begingroup$
There is no such thing as a circle with infinite radius.
$endgroup$
– Ben W
Jan 1 at 18:13
$begingroup$
There is no such thing as a circle with infinite radius.
$endgroup$
– Ben W
Jan 1 at 18:13
5
5
$begingroup$
There is such a thing as the limit of a circle whose radius tends to infinity and it passes through two fixed points, and that limit is the line passing through the fixed points.
$endgroup$
– Oscar Lanzi
Jan 1 at 18:17
$begingroup$
There is such a thing as the limit of a circle whose radius tends to infinity and it passes through two fixed points, and that limit is the line passing through the fixed points.
$endgroup$
– Oscar Lanzi
Jan 1 at 18:17
1
1
$begingroup$
en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
$endgroup$
– Will Jagy
Jan 1 at 19:14
$begingroup$
en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
$endgroup$
– Will Jagy
Jan 1 at 19:14
1
1
$begingroup$
There are many situations in which thinking of a line as a circle of infinite radius is helpful. See, for instance, Descartes' Theorem relating the curvatures of four mutually-tangent circles. The formulas "just work" when one or more of the circles is a line, and the corresponding curvature is $0$. (Note that curvature is the reciprocal of radius, so "zero curvature" matches "infinite radius".)
$endgroup$
– Blue
Jan 1 at 22:07
$begingroup$
There are many situations in which thinking of a line as a circle of infinite radius is helpful. See, for instance, Descartes' Theorem relating the curvatures of four mutually-tangent circles. The formulas "just work" when one or more of the circles is a line, and the corresponding curvature is $0$. (Note that curvature is the reciprocal of radius, so "zero curvature" matches "infinite radius".)
$endgroup$
– Blue
Jan 1 at 22:07
1
1
$begingroup$
@BenW: In geometry with the euclidean plane extended to the projective plane, it is common to talk about the line at infinity, which is also a circle with infinite radius. Any non-degenerate circle with centre on the line at infinity is a line in the plane. In the projective plane, any two distinct points have a unique line through them, and any two distinct lines have a unique intersection, and any two distinct intersecting circles have exactly two intersections counting multiplicity. Blue gave an example of a formula involving radius that requires a line to have infinite radius.
$endgroup$
– user21820
Jan 2 at 10:18
$begingroup$
@BenW: In geometry with the euclidean plane extended to the projective plane, it is common to talk about the line at infinity, which is also a circle with infinite radius. Any non-degenerate circle with centre on the line at infinity is a line in the plane. In the projective plane, any two distinct points have a unique line through them, and any two distinct lines have a unique intersection, and any two distinct intersecting circles have exactly two intersections counting multiplicity. Blue gave an example of a formula involving radius that requires a line to have infinite radius.
$endgroup$
– user21820
Jan 2 at 10:18
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
A "circle", by definition, has finite radius. One can say that the limit of a sequence of circles, all passing through two given points, is the straight line passing through those two points.
answered Jan 1 at 18:17
user247327user247327
11.2k1515
$endgroup$
add a comment |
$begingroup$
The answer you are looking for kind of depends, as your question is not truely well-defined.
The answer of @user247327, assuming additionally 2 given points and describing a limiting process, which focusses on those provided points, might serve as one possibility.
But there are answers too, which do focus onto the circle center instead.
Eg. one such answer you'd get, when considering circles around the origin with increasing radii within the real projective plane. Then too you'd get the "line of infinity" as the according limit.
A different answer would be obtained, when considering the one-point compactification of $mathbb{C}=mathbb{R}^2$. There circles around the origin with ever increasing radii finally result in that additional single point of infinity only. That one then is nothing but the opposite point to the origin on the Riemann sphere.
(For according references just google after the provided terms.)
--- rk
answered Jan 1 at 20:13
Dr. Richard KlitzingDr. Richard Klitzing
1,75016
$endgroup$
$begingroup$
I would add yet another possibility: when considering the family of circles that pass through two fixed points in the plane, it can be useful to consider the line through those points as a member of the family. In the projective plane, its “center” is the point at infinity of the points’ bisector. (Although, to even talk about circles on the projective plane, one must impose a Euclidean geometry on it.)
$endgroup$
– amd
Jan 1 at 20:33
add a comment |
$begingroup$
Suppose you take a straight line $L$ and fix a point $P in L$. Consider a variable point $C in L$, and consider the circle with center $C$ and radius $CP$. This circle crosses the line at right angles at the point $P$. As $C$ goes to infinity in one direction along the line, what can happen? Is there some limiting shape to this circle as $C$ approaches an infinite end of the line $L$?
If you were doing this in the Euclidean plane, then the answer is given by @user247327, the limiting shape is the straight line through $P$ perpendicular to $L$.
However, I notice that you have the tag hyperbolic-geometry on your question. In the hyperbolic plane, the answer is different: the limit is a horocycle, a curve of constant and nonzero curvature which does not close up to form a circle, but the two endpoints of that horocycle instead converge (in the compactification of the hyperbolic plane with its circle at infinity) to the same ideal endpoint of $L$ that $C$ is approaching.
answered Jan 4 at 3:16
Lee MosherLee Mosher
49.9k33686
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A "circle", by definition, has finite radius. One can say that the limit of a sequence of circles, all passing through two given points, is the straight line passing through those two points.
answered Jan 1 at 18:17
user247327user247327
11.2k1515
$endgroup$
add a comment |
$begingroup$
A "circle", by definition, has finite radius. One can say that the limit of a sequence of circles, all passing through two given points, is the straight line passing through those two points.
answered Jan 1 at 18:17
user247327user247327
11.2k1515
$endgroup$
add a comment |
$begingroup$
A "circle", by definition, has finite radius. One can say that the limit of a sequence of circles, all passing through two given points, is the straight line passing through those two points.
answered Jan 1 at 18:17
user247327user247327
11.2k1515
$endgroup$
A "circle", by definition, has finite radius. One can say that the limit of a sequence of circles, all passing through two given points, is the straight line passing through those two points.
answered Jan 1 at 18:17
user247327user247327
11.2k1515
answered Jan 1 at 18:17
user247327user247327
11.2k1515
answered Jan 1 at 18:17
user247327user247327
11.2k1515
answered Jan 1 at 18:17
user247327user247327
11.2k1515
11.2k1515
add a comment |
add a comment |
$begingroup$
The answer you are looking for kind of depends, as your question is not truely well-defined.
The answer of @user247327, assuming additionally 2 given points and describing a limiting process, which focusses on those provided points, might serve as one possibility.
But there are answers too, which do focus onto the circle center instead.
Eg. one such answer you'd get, when considering circles around the origin with increasing radii within the real projective plane. Then too you'd get the "line of infinity" as the according limit.
A different answer would be obtained, when considering the one-point compactification of $mathbb{C}=mathbb{R}^2$. There circles around the origin with ever increasing radii finally result in that additional single point of infinity only. That one then is nothing but the opposite point to the origin on the Riemann sphere.
(For according references just google after the provided terms.)
--- rk
answered Jan 1 at 20:13
Dr. Richard KlitzingDr. Richard Klitzing
1,75016
$endgroup$
$begingroup$
I would add yet another possibility: when considering the family of circles that pass through two fixed points in the plane, it can be useful to consider the line through those points as a member of the family. In the projective plane, its “center” is the point at infinity of the points’ bisector. (Although, to even talk about circles on the projective plane, one must impose a Euclidean geometry on it.)
$endgroup$
– amd
Jan 1 at 20:33
add a comment |
$begingroup$
The answer you are looking for kind of depends, as your question is not truely well-defined.
The answer of @user247327, assuming additionally 2 given points and describing a limiting process, which focusses on those provided points, might serve as one possibility.
But there are answers too, which do focus onto the circle center instead.
Eg. one such answer you'd get, when considering circles around the origin with increasing radii within the real projective plane. Then too you'd get the "line of infinity" as the according limit.
A different answer would be obtained, when considering the one-point compactification of $mathbb{C}=mathbb{R}^2$. There circles around the origin with ever increasing radii finally result in that additional single point of infinity only. That one then is nothing but the opposite point to the origin on the Riemann sphere.
(For according references just google after the provided terms.)
--- rk
answered Jan 1 at 20:13
Dr. Richard KlitzingDr. Richard Klitzing
1,75016
$endgroup$
$begingroup$
I would add yet another possibility: when considering the family of circles that pass through two fixed points in the plane, it can be useful to consider the line through those points as a member of the family. In the projective plane, its “center” is the point at infinity of the points’ bisector. (Although, to even talk about circles on the projective plane, one must impose a Euclidean geometry on it.)
$endgroup$
– amd
Jan 1 at 20:33
add a comment |
$begingroup$
The answer you are looking for kind of depends, as your question is not truely well-defined.
The answer of @user247327, assuming additionally 2 given points and describing a limiting process, which focusses on those provided points, might serve as one possibility.
But there are answers too, which do focus onto the circle center instead.
Eg. one such answer you'd get, when considering circles around the origin with increasing radii within the real projective plane. Then too you'd get the "line of infinity" as the according limit.
A different answer would be obtained, when considering the one-point compactification of $mathbb{C}=mathbb{R}^2$. There circles around the origin with ever increasing radii finally result in that additional single point of infinity only. That one then is nothing but the opposite point to the origin on the Riemann sphere.
(For according references just google after the provided terms.)
--- rk
answered Jan 1 at 20:13
Dr. Richard KlitzingDr. Richard Klitzing
1,75016
$endgroup$
The answer you are looking for kind of depends, as your question is not truely well-defined.
The answer of @user247327, assuming additionally 2 given points and describing a limiting process, which focusses on those provided points, might serve as one possibility.
But there are answers too, which do focus onto the circle center instead.
Eg. one such answer you'd get, when considering circles around the origin with increasing radii within the real projective plane. Then too you'd get the "line of infinity" as the according limit.
A different answer would be obtained, when considering the one-point compactification of $mathbb{C}=mathbb{R}^2$. There circles around the origin with ever increasing radii finally result in that additional single point of infinity only. That one then is nothing but the opposite point to the origin on the Riemann sphere.
(For according references just google after the provided terms.)
--- rk
answered Jan 1 at 20:13
Dr. Richard KlitzingDr. Richard Klitzing
1,75016
answered Jan 1 at 20:13
Dr. Richard KlitzingDr. Richard Klitzing
1,75016
answered Jan 1 at 20:13
Dr. Richard KlitzingDr. Richard Klitzing
1,75016
answered Jan 1 at 20:13
Dr. Richard KlitzingDr. Richard Klitzing
1,75016
1,75016
$begingroup$
I would add yet another possibility: when considering the family of circles that pass through two fixed points in the plane, it can be useful to consider the line through those points as a member of the family. In the projective plane, its “center” is the point at infinity of the points’ bisector. (Although, to even talk about circles on the projective plane, one must impose a Euclidean geometry on it.)
$endgroup$
– amd
Jan 1 at 20:33
add a comment |
$begingroup$
I would add yet another possibility: when considering the family of circles that pass through two fixed points in the plane, it can be useful to consider the line through those points as a member of the family. In the projective plane, its “center” is the point at infinity of the points’ bisector. (Although, to even talk about circles on the projective plane, one must impose a Euclidean geometry on it.)
$endgroup$
– amd
Jan 1 at 20:33
$begingroup$
I would add yet another possibility: when considering the family of circles that pass through two fixed points in the plane, it can be useful to consider the line through those points as a member of the family. In the projective plane, its “center” is the point at infinity of the points’ bisector. (Although, to even talk about circles on the projective plane, one must impose a Euclidean geometry on it.)
$endgroup$
– amd
Jan 1 at 20:33
$begingroup$
I would add yet another possibility: when considering the family of circles that pass through two fixed points in the plane, it can be useful to consider the line through those points as a member of the family. In the projective plane, its “center” is the point at infinity of the points’ bisector. (Although, to even talk about circles on the projective plane, one must impose a Euclidean geometry on it.)
$endgroup$
– amd
Jan 1 at 20:33
add a comment |
$begingroup$
Suppose you take a straight line $L$ and fix a point $P in L$. Consider a variable point $C in L$, and consider the circle with center $C$ and radius $CP$. This circle crosses the line at right angles at the point $P$. As $C$ goes to infinity in one direction along the line, what can happen? Is there some limiting shape to this circle as $C$ approaches an infinite end of the line $L$?
If you were doing this in the Euclidean plane, then the answer is given by @user247327, the limiting shape is the straight line through $P$ perpendicular to $L$.
However, I notice that you have the tag hyperbolic-geometry on your question. In the hyperbolic plane, the answer is different: the limit is a horocycle, a curve of constant and nonzero curvature which does not close up to form a circle, but the two endpoints of that horocycle instead converge (in the compactification of the hyperbolic plane with its circle at infinity) to the same ideal endpoint of $L$ that $C$ is approaching.
answered Jan 4 at 3:16
Lee MosherLee Mosher
49.9k33686
$endgroup$
add a comment |
$begingroup$
Suppose you take a straight line $L$ and fix a point $P in L$. Consider a variable point $C in L$, and consider the circle with center $C$ and radius $CP$. This circle crosses the line at right angles at the point $P$. As $C$ goes to infinity in one direction along the line, what can happen? Is there some limiting shape to this circle as $C$ approaches an infinite end of the line $L$?
If you were doing this in the Euclidean plane, then the answer is given by @user247327, the limiting shape is the straight line through $P$ perpendicular to $L$.
However, I notice that you have the tag hyperbolic-geometry on your question. In the hyperbolic plane, the answer is different: the limit is a horocycle, a curve of constant and nonzero curvature which does not close up to form a circle, but the two endpoints of that horocycle instead converge (in the compactification of the hyperbolic plane with its circle at infinity) to the same ideal endpoint of $L$ that $C$ is approaching.
answered Jan 4 at 3:16
Lee MosherLee Mosher
49.9k33686
$endgroup$
add a comment |
$begingroup$
Suppose you take a straight line $L$ and fix a point $P in L$. Consider a variable point $C in L$, and consider the circle with center $C$ and radius $CP$. This circle crosses the line at right angles at the point $P$. As $C$ goes to infinity in one direction along the line, what can happen? Is there some limiting shape to this circle as $C$ approaches an infinite end of the line $L$?
If you were doing this in the Euclidean plane, then the answer is given by @user247327, the limiting shape is the straight line through $P$ perpendicular to $L$.
However, I notice that you have the tag hyperbolic-geometry on your question. In the hyperbolic plane, the answer is different: the limit is a horocycle, a curve of constant and nonzero curvature which does not close up to form a circle, but the two endpoints of that horocycle instead converge (in the compactification of the hyperbolic plane with its circle at infinity) to the same ideal endpoint of $L$ that $C$ is approaching.
answered Jan 4 at 3:16
Lee MosherLee Mosher
49.9k33686
$endgroup$
Suppose you take a straight line $L$ and fix a point $P in L$. Consider a variable point $C in L$, and consider the circle with center $C$ and radius $CP$. This circle crosses the line at right angles at the point $P$. As $C$ goes to infinity in one direction along the line, what can happen? Is there some limiting shape to this circle as $C$ approaches an infinite end of the line $L$?
If you were doing this in the Euclidean plane, then the answer is given by @user247327, the limiting shape is the straight line through $P$ perpendicular to $L$.
However, I notice that you have the tag hyperbolic-geometry on your question. In the hyperbolic plane, the answer is different: the limit is a horocycle, a curve of constant and nonzero curvature which does not close up to form a circle, but the two endpoints of that horocycle instead converge (in the compactification of the hyperbolic plane with its circle at infinity) to the same ideal endpoint of $L$ that $C$ is approaching.
answered Jan 4 at 3:16
Lee MosherLee Mosher
49.9k33686
edited Jan 4 at 3:22
answered Jan 4 at 3:16
Lee MosherLee Mosher
49.9k33686
answered Jan 4 at 3:16
Lee MosherLee Mosher
49.9k33686
answered Jan 4 at 3:16
Lee MosherLee Mosher
49.9k33686
49.9k33686
add a comment |
add a comment |
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There is no such thing as a circle with infinite radius.
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– Ben W
Jan 1 at 18:13
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There is such a thing as the limit of a circle whose radius tends to infinity and it passes through two fixed points, and that limit is the line passing through the fixed points.
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– Oscar Lanzi
Jan 1 at 18:17
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en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
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– Will Jagy
Jan 1 at 19:14
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There are many situations in which thinking of a line as a circle of infinite radius is helpful. See, for instance, Descartes' Theorem relating the curvatures of four mutually-tangent circles. The formulas "just work" when one or more of the circles is a line, and the corresponding curvature is $0$. (Note that curvature is the reciprocal of radius, so "zero curvature" matches "infinite radius".)
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– Blue
Jan 1 at 22:07
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@BenW: In geometry with the euclidean plane extended to the projective plane, it is common to talk about the line at infinity, which is also a circle with infinite radius. Any non-degenerate circle with centre on the line at infinity is a line in the plane. In the projective plane, any two distinct points have a unique line through them, and any two distinct lines have a unique intersection, and any two distinct intersecting circles have exactly two intersections counting multiplicity. Blue gave an example of a formula involving radius that requires a line to have infinite radius.
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– user21820
Jan 2 at 10:18