Mixing
Inscribing tangential circle in non-tangential circles
$begingroup$
Given the positions ($p_1$, $p_2$, $p_3$) and radii ($r_1$, $r_2$, $r_3$) of three circles that are not pairwise tangential, how do you calculate the location and radius of the circle that is pairwise tangential with each of the three original and contains none of them?
The circle that contains all three is a duplicate of this question. This answer to that question shows a geometric way of calculating $p_5$ (and it looks like $p_4$, too), but does not include the formulae needed to calculate the hyperbolae and solve the equations.
Edit: Given three discrete circles, there are 8 circles that can be created which are each pairwise tangent to the discrete circles:
- 1 that contains none of the circles
- 3 that contain one circle
- 3 that contain two circles
- 1 that contains all three circles
It is the circle that contains none of the original three that I am interested in (shown in blue above and below).
(The original version of this question included the desire to find $p_5$/$r_5$ as well; I'm not actually interested in that circle, as this edit reflects.)
geometry euclidean-geometry
asked Jan 8 at 18:34
PhrogzPhrogz
85411115
$endgroup$
|
show 7 more comments
$begingroup$
Given the positions ($p_1$, $p_2$, $p_3$) and radii ($r_1$, $r_2$, $r_3$) of three circles that are not pairwise tangential, how do you calculate the location and radius of the circle that is pairwise tangential with each of the three original and contains none of them?
The circle that contains all three is a duplicate of this question. This answer to that question shows a geometric way of calculating $p_5$ (and it looks like $p_4$, too), but does not include the formulae needed to calculate the hyperbolae and solve the equations.
Edit: Given three discrete circles, there are 8 circles that can be created which are each pairwise tangent to the discrete circles:
- 1 that contains none of the circles
- 3 that contain one circle
- 3 that contain two circles
- 1 that contains all three circles
It is the circle that contains none of the original three that I am interested in (shown in blue above and below).
(The original version of this question included the desire to find $p_5$/$r_5$ as well; I'm not actually interested in that circle, as this edit reflects.)
geometry euclidean-geometry
asked Jan 8 at 18:34
PhrogzPhrogz
85411115
$endgroup$
$begingroup$
As the answer in the link you provides alludes to, calculating P5 is gonna be a mess.
$endgroup$
– T. Fo
Jan 8 at 18:40
$begingroup$
@T.Ford Thanks. I'd still like to get it for completeness, but FWIW $p_4$ and $r_4$ are what I really care about. (On the off chance that they are simpler to calculate.)
$endgroup$
– Phrogz
Jan 8 at 18:43
$begingroup$
Also, in your question when you say the 'smallest (p4, r4)' isnt the circle you are looking for unique? Like more generally I feel as though there should be a result for finding a (unique?) inscribed circle between three non-intersecting circles
$endgroup$
– T. Fo
Jan 8 at 18:47
$begingroup$
All of the equations for the analytic analogue of that geometric construction are easily found, and, I dare say you already know most of them. Surely you can construct an equation of a circle with given radius and center and that of a line through two points. Solving that system (or some simple vector arithmetic) gets you $G$ and $H$. You might not know off the top of your head how to construct a hyperbola from its foci and a vertex, but you can find that with a search, too. Intersecting the hyperbolas is tedious—it involves solving cubics or quartics—but conceptually simple.
$endgroup$
– amd
Jan 8 at 19:32
2
$begingroup$
Tangent: I'm curious when this problem even has a solution. For example, if we label your three balls $B_1$, $B_2$, and $B_3$, and they all of the same radius and their centers are colinear then there will be no solution as you describe. But to what degree can we shift one of the circles off the line and induce a solution. i.e. for what configurations of the balls is this even solvable?
$endgroup$
– T. Fo
Jan 8 at 21:34
|
show 7 more comments
$begingroup$
Given the positions ($p_1$, $p_2$, $p_3$) and radii ($r_1$, $r_2$, $r_3$) of three circles that are not pairwise tangential, how do you calculate the location and radius of the circle that is pairwise tangential with each of the three original and contains none of them?
The circle that contains all three is a duplicate of this question. This answer to that question shows a geometric way of calculating $p_5$ (and it looks like $p_4$, too), but does not include the formulae needed to calculate the hyperbolae and solve the equations.
Edit: Given three discrete circles, there are 8 circles that can be created which are each pairwise tangent to the discrete circles:
- 1 that contains none of the circles
- 3 that contain one circle
- 3 that contain two circles
- 1 that contains all three circles
It is the circle that contains none of the original three that I am interested in (shown in blue above and below).
(The original version of this question included the desire to find $p_5$/$r_5$ as well; I'm not actually interested in that circle, as this edit reflects.)
geometry euclidean-geometry
asked Jan 8 at 18:34
PhrogzPhrogz
85411115
$endgroup$
Given the positions ($p_1$, $p_2$, $p_3$) and radii ($r_1$, $r_2$, $r_3$) of three circles that are not pairwise tangential, how do you calculate the location and radius of the circle that is pairwise tangential with each of the three original and contains none of them?
The circle that contains all three is a duplicate of this question. This answer to that question shows a geometric way of calculating $p_5$ (and it looks like $p_4$, too), but does not include the formulae needed to calculate the hyperbolae and solve the equations.
Edit: Given three discrete circles, there are 8 circles that can be created which are each pairwise tangent to the discrete circles:
- 1 that contains none of the circles
- 3 that contain one circle
- 3 that contain two circles
- 1 that contains all three circles
It is the circle that contains none of the original three that I am interested in (shown in blue above and below).
(The original version of this question included the desire to find $p_5$/$r_5$ as well; I'm not actually interested in that circle, as this edit reflects.)
geometry euclidean-geometry
geometry euclidean-geometry
asked Jan 8 at 18:34
PhrogzPhrogz
85411115
asked Jan 8 at 18:34
PhrogzPhrogz
85411115
edited Jan 8 at 21:12
Phrogz
asked Jan 8 at 18:34
PhrogzPhrogz
85411115
asked Jan 8 at 18:34
PhrogzPhrogz
85411115
asked Jan 8 at 18:34
PhrogzPhrogz
85411115
85411115
$begingroup$
As the answer in the link you provides alludes to, calculating P5 is gonna be a mess.
$endgroup$
– T. Fo
Jan 8 at 18:40
$begingroup$
@T.Ford Thanks. I'd still like to get it for completeness, but FWIW $p_4$ and $r_4$ are what I really care about. (On the off chance that they are simpler to calculate.)
$endgroup$
– Phrogz
Jan 8 at 18:43
$begingroup$
Also, in your question when you say the 'smallest (p4, r4)' isnt the circle you are looking for unique? Like more generally I feel as though there should be a result for finding a (unique?) inscribed circle between three non-intersecting circles
$endgroup$
– T. Fo
Jan 8 at 18:47
$begingroup$
All of the equations for the analytic analogue of that geometric construction are easily found, and, I dare say you already know most of them. Surely you can construct an equation of a circle with given radius and center and that of a line through two points. Solving that system (or some simple vector arithmetic) gets you $G$ and $H$. You might not know off the top of your head how to construct a hyperbola from its foci and a vertex, but you can find that with a search, too. Intersecting the hyperbolas is tedious—it involves solving cubics or quartics—but conceptually simple.
$endgroup$
– amd
Jan 8 at 19:32
2
$begingroup$
Tangent: I'm curious when this problem even has a solution. For example, if we label your three balls $B_1$, $B_2$, and $B_3$, and they all of the same radius and their centers are colinear then there will be no solution as you describe. But to what degree can we shift one of the circles off the line and induce a solution. i.e. for what configurations of the balls is this even solvable?
$endgroup$
– T. Fo
Jan 8 at 21:34
|
show 7 more comments
$begingroup$
As the answer in the link you provides alludes to, calculating P5 is gonna be a mess.
$endgroup$
– T. Fo
Jan 8 at 18:40
$begingroup$
@T.Ford Thanks. I'd still like to get it for completeness, but FWIW $p_4$ and $r_4$ are what I really care about. (On the off chance that they are simpler to calculate.)
$endgroup$
– Phrogz
Jan 8 at 18:43
$begingroup$
Also, in your question when you say the 'smallest (p4, r4)' isnt the circle you are looking for unique? Like more generally I feel as though there should be a result for finding a (unique?) inscribed circle between three non-intersecting circles
$endgroup$
– T. Fo
Jan 8 at 18:47
$begingroup$
All of the equations for the analytic analogue of that geometric construction are easily found, and, I dare say you already know most of them. Surely you can construct an equation of a circle with given radius and center and that of a line through two points. Solving that system (or some simple vector arithmetic) gets you $G$ and $H$. You might not know off the top of your head how to construct a hyperbola from its foci and a vertex, but you can find that with a search, too. Intersecting the hyperbolas is tedious—it involves solving cubics or quartics—but conceptually simple.
$endgroup$
– amd
Jan 8 at 19:32
2
$begingroup$
Tangent: I'm curious when this problem even has a solution. For example, if we label your three balls $B_1$, $B_2$, and $B_3$, and they all of the same radius and their centers are colinear then there will be no solution as you describe. But to what degree can we shift one of the circles off the line and induce a solution. i.e. for what configurations of the balls is this even solvable?
$endgroup$
– T. Fo
Jan 8 at 21:34
$begingroup$
As the answer in the link you provides alludes to, calculating P5 is gonna be a mess.
$endgroup$
– T. Fo
Jan 8 at 18:40
$begingroup$
As the answer in the link you provides alludes to, calculating P5 is gonna be a mess.
$endgroup$
– T. Fo
Jan 8 at 18:40
$begingroup$
@T.Ford Thanks. I'd still like to get it for completeness, but FWIW $p_4$ and $r_4$ are what I really care about. (On the off chance that they are simpler to calculate.)
$endgroup$
– Phrogz
Jan 8 at 18:43
$begingroup$
@T.Ford Thanks. I'd still like to get it for completeness, but FWIW $p_4$ and $r_4$ are what I really care about. (On the off chance that they are simpler to calculate.)
$endgroup$
– Phrogz
Jan 8 at 18:43
$begingroup$
Also, in your question when you say the 'smallest (p4, r4)' isnt the circle you are looking for unique? Like more generally I feel as though there should be a result for finding a (unique?) inscribed circle between three non-intersecting circles
$endgroup$
– T. Fo
Jan 8 at 18:47
$begingroup$
Also, in your question when you say the 'smallest (p4, r4)' isnt the circle you are looking for unique? Like more generally I feel as though there should be a result for finding a (unique?) inscribed circle between three non-intersecting circles
$endgroup$
– T. Fo
Jan 8 at 18:47
$begingroup$
All of the equations for the analytic analogue of that geometric construction are easily found, and, I dare say you already know most of them. Surely you can construct an equation of a circle with given radius and center and that of a line through two points. Solving that system (or some simple vector arithmetic) gets you $G$ and $H$. You might not know off the top of your head how to construct a hyperbola from its foci and a vertex, but you can find that with a search, too. Intersecting the hyperbolas is tedious—it involves solving cubics or quartics—but conceptually simple.
$endgroup$
– amd
Jan 8 at 19:32
$begingroup$
All of the equations for the analytic analogue of that geometric construction are easily found, and, I dare say you already know most of them. Surely you can construct an equation of a circle with given radius and center and that of a line through two points. Solving that system (or some simple vector arithmetic) gets you $G$ and $H$. You might not know off the top of your head how to construct a hyperbola from its foci and a vertex, but you can find that with a search, too. Intersecting the hyperbolas is tedious—it involves solving cubics or quartics—but conceptually simple.
$endgroup$
– amd
Jan 8 at 19:32
2
2
$begingroup$
Tangent: I'm curious when this problem even has a solution. For example, if we label your three balls $B_1$, $B_2$, and $B_3$, and they all of the same radius and their centers are colinear then there will be no solution as you describe. But to what degree can we shift one of the circles off the line and induce a solution. i.e. for what configurations of the balls is this even solvable?
$endgroup$
– T. Fo
Jan 8 at 21:34
$begingroup$
Tangent: I'm curious when this problem even has a solution. For example, if we label your three balls $B_1$, $B_2$, and $B_3$, and they all of the same radius and their centers are colinear then there will be no solution as you describe. But to what degree can we shift one of the circles off the line and induce a solution. i.e. for what configurations of the balls is this even solvable?
$endgroup$
– T. Fo
Jan 8 at 21:34
|
show 7 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066556%2finscribing-tangential-circle-in-non-tangential-circles%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066556%2finscribing-tangential-circle-in-non-tangential-circles%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
As the answer in the link you provides alludes to, calculating P5 is gonna be a mess.
$endgroup$
– T. Fo
Jan 8 at 18:40
$begingroup$
@T.Ford Thanks. I'd still like to get it for completeness, but FWIW $p_4$ and $r_4$ are what I really care about. (On the off chance that they are simpler to calculate.)
$endgroup$
– Phrogz
Jan 8 at 18:43
$begingroup$
Also, in your question when you say the 'smallest (p4, r4)' isnt the circle you are looking for unique? Like more generally I feel as though there should be a result for finding a (unique?) inscribed circle between three non-intersecting circles
$endgroup$
– T. Fo
Jan 8 at 18:47
$begingroup$
All of the equations for the analytic analogue of that geometric construction are easily found, and, I dare say you already know most of them. Surely you can construct an equation of a circle with given radius and center and that of a line through two points. Solving that system (or some simple vector arithmetic) gets you $G$ and $H$. You might not know off the top of your head how to construct a hyperbola from its foci and a vertex, but you can find that with a search, too. Intersecting the hyperbolas is tedious—it involves solving cubics or quartics—but conceptually simple.
$endgroup$
– amd
Jan 8 at 19:32
2
$begingroup$
Tangent: I'm curious when this problem even has a solution. For example, if we label your three balls $B_1$, $B_2$, and $B_3$, and they all of the same radius and their centers are colinear then there will be no solution as you describe. But to what degree can we shift one of the circles off the line and induce a solution. i.e. for what configurations of the balls is this even solvable?
$endgroup$
– T. Fo
Jan 8 at 21:34