find a circle given two tangent points and angles
0
I've searched and found similar questions, but with less information given, and I was hoping that the additional information would allow for a more streamlined solution.
Given two points, and their tangent angles (not sure if that's the proper term), what's the fewest number of steps to get the center and radius of the circle?
Circle given the information indicated
geometry circle tangent-line
asked Nov 24 at 2:10
jd19457
1
add a comment |
0
I've searched and found similar questions, but with less information given, and I was hoping that the additional information would allow for a more streamlined solution.
Given two points, and their tangent angles (not sure if that's the proper term), what's the fewest number of steps to get the center and radius of the circle?
Circle given the information indicated
geometry circle tangent-line
asked Nov 24 at 2:10
jd19457
1
What exactly do you mean by fewest number of steps?
– Mike Pierce
Nov 24 at 2:14
1
Having the tangent angles and the points, you basically have two lines that are tangent to the circle. You can construct the lines that are perpendicular to each of those at the given points, and the intersection of those two perpendicular lines will be the center of the circle. And the distance from that center to either of the two starting points is the radius.
– Mike Pierce
Nov 24 at 2:15
Ok. Thanks. That's what I had figured. I was hoping there was some trick to getting it without having to manually find the intersection.
– jd19457
Nov 24 at 2:19
I have to echo @MikePierce’s comment: how are you measuring the number of steps? What is a “step,” anyway? You might, for instance, construct a pair of degenerate conics $Q_1$ and $Q_2$ from the two tangent lines and the secant through the two points and then find the value of $lambda$ that makes the cross term in $(1-lambda)Q_1+lambda Q_2$ vanish. How does that compare to the number of “steps” for the intersecting perpendiculars method?
– amd
Nov 24 at 5:10
I know this isn't a programming stackexchange, but separate computational steps. This will be in a tight loop, and I want the shortest number of computational steps. I realize some computations are more costly than others, but I just wanted to keep the question strictly math.
– jd19457
Nov 24 at 22:15
add a comment |
0
0
0
I've searched and found similar questions, but with less information given, and I was hoping that the additional information would allow for a more streamlined solution.
Given two points, and their tangent angles (not sure if that's the proper term), what's the fewest number of steps to get the center and radius of the circle?
Circle given the information indicated
geometry circle tangent-line
asked Nov 24 at 2:10
jd19457
1
I've searched and found similar questions, but with less information given, and I was hoping that the additional information would allow for a more streamlined solution.
Given two points, and their tangent angles (not sure if that's the proper term), what's the fewest number of steps to get the center and radius of the circle?
Circle given the information indicated
geometry circle tangent-line
geometry circle tangent-line
asked Nov 24 at 2:10
jd19457
1
asked Nov 24 at 2:10
jd19457
1
asked Nov 24 at 2:10
jd19457
1
asked Nov 24 at 2:10
jd19457
1
asked Nov 24 at 2:10
jd19457
1
1
What exactly do you mean by fewest number of steps?
– Mike Pierce
Nov 24 at 2:14
1
Having the tangent angles and the points, you basically have two lines that are tangent to the circle. You can construct the lines that are perpendicular to each of those at the given points, and the intersection of those two perpendicular lines will be the center of the circle. And the distance from that center to either of the two starting points is the radius.
– Mike Pierce
Nov 24 at 2:15
Ok. Thanks. That's what I had figured. I was hoping there was some trick to getting it without having to manually find the intersection.
– jd19457
Nov 24 at 2:19
I have to echo @MikePierce’s comment: how are you measuring the number of steps? What is a “step,” anyway? You might, for instance, construct a pair of degenerate conics $Q_1$ and $Q_2$ from the two tangent lines and the secant through the two points and then find the value of $lambda$ that makes the cross term in $(1-lambda)Q_1+lambda Q_2$ vanish. How does that compare to the number of “steps” for the intersecting perpendiculars method?
– amd
Nov 24 at 5:10
I know this isn't a programming stackexchange, but separate computational steps. This will be in a tight loop, and I want the shortest number of computational steps. I realize some computations are more costly than others, but I just wanted to keep the question strictly math.
– jd19457
Nov 24 at 22:15
add a comment |
What exactly do you mean by fewest number of steps?
– Mike Pierce
Nov 24 at 2:14
1
Having the tangent angles and the points, you basically have two lines that are tangent to the circle. You can construct the lines that are perpendicular to each of those at the given points, and the intersection of those two perpendicular lines will be the center of the circle. And the distance from that center to either of the two starting points is the radius.
– Mike Pierce
Nov 24 at 2:15
Ok. Thanks. That's what I had figured. I was hoping there was some trick to getting it without having to manually find the intersection.
– jd19457
Nov 24 at 2:19
I have to echo @MikePierce’s comment: how are you measuring the number of steps? What is a “step,” anyway? You might, for instance, construct a pair of degenerate conics $Q_1$ and $Q_2$ from the two tangent lines and the secant through the two points and then find the value of $lambda$ that makes the cross term in $(1-lambda)Q_1+lambda Q_2$ vanish. How does that compare to the number of “steps” for the intersecting perpendiculars method?
– amd
Nov 24 at 5:10
I know this isn't a programming stackexchange, but separate computational steps. This will be in a tight loop, and I want the shortest number of computational steps. I realize some computations are more costly than others, but I just wanted to keep the question strictly math.
– jd19457
Nov 24 at 22:15
What exactly do you mean by fewest number of steps?
– Mike Pierce
Nov 24 at 2:14
What exactly do you mean by fewest number of steps?
– Mike Pierce
Nov 24 at 2:14
1
1
Having the tangent angles and the points, you basically have two lines that are tangent to the circle. You can construct the lines that are perpendicular to each of those at the given points, and the intersection of those two perpendicular lines will be the center of the circle. And the distance from that center to either of the two starting points is the radius.
– Mike Pierce
Nov 24 at 2:15
Having the tangent angles and the points, you basically have two lines that are tangent to the circle. You can construct the lines that are perpendicular to each of those at the given points, and the intersection of those two perpendicular lines will be the center of the circle. And the distance from that center to either of the two starting points is the radius.
– Mike Pierce
Nov 24 at 2:15
Ok. Thanks. That's what I had figured. I was hoping there was some trick to getting it without having to manually find the intersection.
– jd19457
Nov 24 at 2:19
Ok. Thanks. That's what I had figured. I was hoping there was some trick to getting it without having to manually find the intersection.
– jd19457
Nov 24 at 2:19
I have to echo @MikePierce’s comment: how are you measuring the number of steps? What is a “step,” anyway? You might, for instance, construct a pair of degenerate conics $Q_1$ and $Q_2$ from the two tangent lines and the secant through the two points and then find the value of $lambda$ that makes the cross term in $(1-lambda)Q_1+lambda Q_2$ vanish. How does that compare to the number of “steps” for the intersecting perpendiculars method?
– amd
Nov 24 at 5:10
I have to echo @MikePierce’s comment: how are you measuring the number of steps? What is a “step,” anyway? You might, for instance, construct a pair of degenerate conics $Q_1$ and $Q_2$ from the two tangent lines and the secant through the two points and then find the value of $lambda$ that makes the cross term in $(1-lambda)Q_1+lambda Q_2$ vanish. How does that compare to the number of “steps” for the intersecting perpendiculars method?
– amd
Nov 24 at 5:10
I know this isn't a programming stackexchange, but separate computational steps. This will be in a tight loop, and I want the shortest number of computational steps. I realize some computations are more costly than others, but I just wanted to keep the question strictly math.
– jd19457
Nov 24 at 22:15
I know this isn't a programming stackexchange, but separate computational steps. This will be in a tight loop, and I want the shortest number of computational steps. I realize some computations are more costly than others, but I just wanted to keep the question strictly math.
– jd19457
Nov 24 at 22:15
add a comment |
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
});
}
});
draft saved
draft discarded
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%2f3011097%2ffind-a-circle-given-two-tangent-points-and-angles%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
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%2f3011097%2ffind-a-circle-given-two-tangent-points-and-angles%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
What exactly do you mean by fewest number of steps?
– Mike Pierce
Nov 24 at 2:14
1
Having the tangent angles and the points, you basically have two lines that are tangent to the circle. You can construct the lines that are perpendicular to each of those at the given points, and the intersection of those two perpendicular lines will be the center of the circle. And the distance from that center to either of the two starting points is the radius.
– Mike Pierce
Nov 24 at 2:15
Ok. Thanks. That's what I had figured. I was hoping there was some trick to getting it without having to manually find the intersection.
– jd19457
Nov 24 at 2:19
I have to echo @MikePierce’s comment: how are you measuring the number of steps? What is a “step,” anyway? You might, for instance, construct a pair of degenerate conics $Q_1$ and $Q_2$ from the two tangent lines and the secant through the two points and then find the value of $lambda$ that makes the cross term in $(1-lambda)Q_1+lambda Q_2$ vanish. How does that compare to the number of “steps” for the intersecting perpendiculars method?
– amd
Nov 24 at 5:10
I know this isn't a programming stackexchange, but separate computational steps. This will be in a tight loop, and I want the shortest number of computational steps. I realize some computations are more costly than others, but I just wanted to keep the question strictly math.
– jd19457
Nov 24 at 22:15