Point on the square with a circle inscribed inside it
$begingroup$
I giving a second try to this question. Hopefully, with better problem definition.
I have a circle inscribed inside a square and would like to know the point the radius touches when extended. In the figure A, we have calculated the angle(θ), C(center) , D and E. How do i calculate the (x,y) of A and B?
geometry trigonometry coordinate-systems
edited Dec 31 '18 at 15:45
saulspatz
17k31435
asked Dec 31 '18 at 15:40
brainfreakbrainfreak
13
$endgroup$
add a comment |
$begingroup$
I giving a second try to this question. Hopefully, with better problem definition.
I have a circle inscribed inside a square and would like to know the point the radius touches when extended. In the figure A, we have calculated the angle(θ), C(center) , D and E. How do i calculate the (x,y) of A and B?
geometry trigonometry coordinate-systems
edited Dec 31 '18 at 15:45
saulspatz
17k31435
asked Dec 31 '18 at 15:40
brainfreakbrainfreak
13
$endgroup$
add a comment |
$begingroup$
I giving a second try to this question. Hopefully, with better problem definition.
I have a circle inscribed inside a square and would like to know the point the radius touches when extended. In the figure A, we have calculated the angle(θ), C(center) , D and E. How do i calculate the (x,y) of A and B?
geometry trigonometry coordinate-systems
edited Dec 31 '18 at 15:45
saulspatz
17k31435
asked Dec 31 '18 at 15:40
brainfreakbrainfreak
13
$endgroup$
I giving a second try to this question. Hopefully, with better problem definition.
I have a circle inscribed inside a square and would like to know the point the radius touches when extended. In the figure A, we have calculated the angle(θ), C(center) , D and E. How do i calculate the (x,y) of A and B?
geometry trigonometry coordinate-systems
geometry trigonometry coordinate-systems
edited Dec 31 '18 at 15:45
saulspatz
17k31435
asked Dec 31 '18 at 15:40
brainfreakbrainfreak
13
edited Dec 31 '18 at 15:45
saulspatz
17k31435
asked Dec 31 '18 at 15:40
brainfreakbrainfreak
13
edited Dec 31 '18 at 15:45
saulspatz
17k31435
edited Dec 31 '18 at 15:45
saulspatz
17k31435
edited Dec 31 '18 at 15:45
saulspatz
17k31435
17k31435
asked Dec 31 '18 at 15:40
brainfreakbrainfreak
13
asked Dec 31 '18 at 15:40
brainfreakbrainfreak
13
asked Dec 31 '18 at 15:40
brainfreakbrainfreak
13
13
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In the case you've drawn, you already know the $x$ value, assuming the circle has center in $(C_x,C_y)$ and radius $r$, $A_x=B_x=C_x+r.$ As for the $y,$ a little trigonometry helps: $A_y=C_y+r·tan theta.$
answered Dec 31 '18 at 15:56
PatricioPatricio
3427
$endgroup$
add a comment |
$begingroup$
If you know the coordinates of the center then you add $r$ to the $x$ coordinate and you add $r tan (theta)$ to the $y$ coordinate of the center to get coordinates of $A$
Similarly you can find coordinates of $B$
answered Dec 31 '18 at 15:55
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Describe the circle as
$$vec x=vec m+begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}.$$
Now consider the ray
$$vec y=vec m+lambdabegin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}$$
with $lambda>0$.
You want to have the first coordinate for $tin(-pi/2,pi/2)$ of $vec y$ to be $m_1+r$, hence $lambda=1/cos(t)$ and the desired point is
$$vec m+frac{1}{cos(t)}begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}=begin{pmatrix} m_1+r\ m_2+rtan(t)
end{pmatrix}.$$
answered Dec 31 '18 at 16:14
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
add a comment |
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%2f3057819%2fpoint-on-the-square-with-a-circle-inscribed-inside-it%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In the case you've drawn, you already know the $x$ value, assuming the circle has center in $(C_x,C_y)$ and radius $r$, $A_x=B_x=C_x+r.$ As for the $y,$ a little trigonometry helps: $A_y=C_y+r·tan theta.$
answered Dec 31 '18 at 15:56
PatricioPatricio
3427
$endgroup$
add a comment |
$begingroup$
In the case you've drawn, you already know the $x$ value, assuming the circle has center in $(C_x,C_y)$ and radius $r$, $A_x=B_x=C_x+r.$ As for the $y,$ a little trigonometry helps: $A_y=C_y+r·tan theta.$
answered Dec 31 '18 at 15:56
PatricioPatricio
3427
$endgroup$
add a comment |
$begingroup$
In the case you've drawn, you already know the $x$ value, assuming the circle has center in $(C_x,C_y)$ and radius $r$, $A_x=B_x=C_x+r.$ As for the $y,$ a little trigonometry helps: $A_y=C_y+r·tan theta.$
answered Dec 31 '18 at 15:56
PatricioPatricio
3427
$endgroup$
In the case you've drawn, you already know the $x$ value, assuming the circle has center in $(C_x,C_y)$ and radius $r$, $A_x=B_x=C_x+r.$ As for the $y,$ a little trigonometry helps: $A_y=C_y+r·tan theta.$
answered Dec 31 '18 at 15:56
PatricioPatricio
3427
answered Dec 31 '18 at 15:56
PatricioPatricio
3427
answered Dec 31 '18 at 15:56
PatricioPatricio
3427
answered Dec 31 '18 at 15:56
PatricioPatricio
3427
3427
add a comment |
add a comment |
$begingroup$
If you know the coordinates of the center then you add $r$ to the $x$ coordinate and you add $r tan (theta)$ to the $y$ coordinate of the center to get coordinates of $A$
Similarly you can find coordinates of $B$
answered Dec 31 '18 at 15:55
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
If you know the coordinates of the center then you add $r$ to the $x$ coordinate and you add $r tan (theta)$ to the $y$ coordinate of the center to get coordinates of $A$
Similarly you can find coordinates of $B$
answered Dec 31 '18 at 15:55
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
If you know the coordinates of the center then you add $r$ to the $x$ coordinate and you add $r tan (theta)$ to the $y$ coordinate of the center to get coordinates of $A$
Similarly you can find coordinates of $B$
answered Dec 31 '18 at 15:55
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
If you know the coordinates of the center then you add $r$ to the $x$ coordinate and you add $r tan (theta)$ to the $y$ coordinate of the center to get coordinates of $A$
Similarly you can find coordinates of $B$
answered Dec 31 '18 at 15:55
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 15:55
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 15:55
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 15:55
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
$begingroup$
Describe the circle as
$$vec x=vec m+begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}.$$
Now consider the ray
$$vec y=vec m+lambdabegin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}$$
with $lambda>0$.
You want to have the first coordinate for $tin(-pi/2,pi/2)$ of $vec y$ to be $m_1+r$, hence $lambda=1/cos(t)$ and the desired point is
$$vec m+frac{1}{cos(t)}begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}=begin{pmatrix} m_1+r\ m_2+rtan(t)
end{pmatrix}.$$
answered Dec 31 '18 at 16:14
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
add a comment |
$begingroup$
Describe the circle as
$$vec x=vec m+begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}.$$
Now consider the ray
$$vec y=vec m+lambdabegin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}$$
with $lambda>0$.
You want to have the first coordinate for $tin(-pi/2,pi/2)$ of $vec y$ to be $m_1+r$, hence $lambda=1/cos(t)$ and the desired point is
$$vec m+frac{1}{cos(t)}begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}=begin{pmatrix} m_1+r\ m_2+rtan(t)
end{pmatrix}.$$
answered Dec 31 '18 at 16:14
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
add a comment |
$begingroup$
Describe the circle as
$$vec x=vec m+begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}.$$
Now consider the ray
$$vec y=vec m+lambdabegin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}$$
with $lambda>0$.
You want to have the first coordinate for $tin(-pi/2,pi/2)$ of $vec y$ to be $m_1+r$, hence $lambda=1/cos(t)$ and the desired point is
$$vec m+frac{1}{cos(t)}begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}=begin{pmatrix} m_1+r\ m_2+rtan(t)
end{pmatrix}.$$
answered Dec 31 '18 at 16:14
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
Describe the circle as
$$vec x=vec m+begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}.$$
Now consider the ray
$$vec y=vec m+lambdabegin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}$$
with $lambda>0$.
You want to have the first coordinate for $tin(-pi/2,pi/2)$ of $vec y$ to be $m_1+r$, hence $lambda=1/cos(t)$ and the desired point is
$$vec m+frac{1}{cos(t)}begin{pmatrix} rcos(t)\ rsin(t)
end{pmatrix}=begin{pmatrix} m_1+r\ m_2+rtan(t)
end{pmatrix}.$$
answered Dec 31 '18 at 16:14
Michael HoppeMichael Hoppe
11.2k31837
answered Dec 31 '18 at 16:14
Michael HoppeMichael Hoppe
11.2k31837
answered Dec 31 '18 at 16:14
Michael HoppeMichael Hoppe
11.2k31837
answered Dec 31 '18 at 16:14
Michael HoppeMichael Hoppe
11.2k31837
11.2k31837
add a comment |
add a comment |
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%2f3057819%2fpoint-on-the-square-with-a-circle-inscribed-inside-it%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
