Circle whose radius is infinite












0














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







share|cite|improve this question




















  • 2




    There is no such thing as a circle with infinite radius.
    – Ben W
    2 days ago






  • 5




    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.
    – Oscar Lanzi
    2 days ago






  • 1




    en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
    – Will Jagy
    2 days ago






  • 1




    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".)
    – Blue
    2 days ago








  • 1




    @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.
    – user21820
    2 days ago
















0














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







share|cite|improve this question




















  • 2




    There is no such thing as a circle with infinite radius.
    – Ben W
    2 days ago






  • 5




    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.
    – Oscar Lanzi
    2 days ago






  • 1




    en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
    – Will Jagy
    2 days ago






  • 1




    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".)
    – Blue
    2 days ago








  • 1




    @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.
    – user21820
    2 days ago














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked 2 days ago







user630906















  • 2




    There is no such thing as a circle with infinite radius.
    – Ben W
    2 days ago






  • 5




    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.
    – Oscar Lanzi
    2 days ago






  • 1




    en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
    – Will Jagy
    2 days ago






  • 1




    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".)
    – Blue
    2 days ago








  • 1




    @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.
    – user21820
    2 days ago














  • 2




    There is no such thing as a circle with infinite radius.
    – Ben W
    2 days ago






  • 5




    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.
    – Oscar Lanzi
    2 days ago






  • 1




    en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
    – Will Jagy
    2 days ago






  • 1




    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".)
    – Blue
    2 days ago








  • 1




    @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.
    – user21820
    2 days ago








2




2




There is no such thing as a circle with infinite radius.
– Ben W
2 days ago




There is no such thing as a circle with infinite radius.
– Ben W
2 days ago




5




5




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.
– Oscar Lanzi
2 days ago




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.
– Oscar Lanzi
2 days ago




1




1




en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
– Will Jagy
2 days ago




en.wikipedia.org/wiki/Horocycle and then en.wikipedia.org/wiki/Hypercycle_(geometry)
– Will Jagy
2 days ago




1




1




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".)
– Blue
2 days ago






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".)
– Blue
2 days ago






1




1




@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.
– user21820
2 days ago




@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.
– user21820
2 days ago










3 Answers
3






active

oldest

votes


















5














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.






share|cite|improve this answer





























    1














    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






    share|cite|improve this answer





















    • 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.)
      – amd
      2 days ago





















    0














    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.






    share|cite|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      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.






      share|cite|improve this answer


























        5














        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.






        share|cite|improve this answer
























          5












          5








          5






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          user247327

          10.4k1515




          10.4k1515























              1














              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






              share|cite|improve this answer





















              • 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.)
                – amd
                2 days ago


















              1














              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






              share|cite|improve this answer





















              • 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.)
                – amd
                2 days ago
















              1












              1








              1






              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






              share|cite|improve this answer












              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







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 days ago









              Dr. Richard Klitzing

              1,3516




              1,3516












              • 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.)
                – amd
                2 days ago




















              • 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.)
                – amd
                2 days ago


















              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.)
              – amd
              2 days ago






              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.)
              – amd
              2 days ago













              0














              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.






              share|cite|improve this answer




























                0














                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.






                share|cite|improve this answer


























                  0












                  0








                  0






                  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.






                  share|cite|improve this answer














                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 9 hours ago

























                  answered 9 hours ago









                  Lee Mosher

                  48.2k33681




                  48.2k33681






























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                      The Binding of Isaac: Rebirth/Afterbirth

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                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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                      What does “Dominus providebit” mean?

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                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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