Circles within ellipses












0












$begingroup$


What is the largest circle centered at $(x_0, y_0)$ that is totally enveloped by an ellipse with a major axis $A$ and minor axis $B$? In this problem, assume a constraint s.t. the major axis is in the $x$ direction. Alternate cases of the problem would be solved by a rotation of the major axis to the $x$ axis as a first step. Additionally, assume $(x_0, y_0)$ is within the ellipse. The ellipse is centered at $(0,0)$.



I think that the closed form solution is not guaranteed if neither $x_0$ nor $y_0$ are zero. I know a closed form solution can be found if either $x_0$ or $y_0$ is equal to zero.



Thank you for your time






calculus geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    In other words, you want the minimum distance from any point on the ellipse to $(x_0,y_0).$ Finding the minimum (or maximum) distance is also involved in math.stackexchange.com/questions/2641562/… and math.stackexchange.com/questions/1846062/…
    $endgroup$
    – David K
    Jan 4 at 13:09










  • $begingroup$
    Where is the center of the ellipse?
    $endgroup$
    – Jens
    Jan 4 at 14:53
















0












$begingroup$


What is the largest circle centered at $(x_0, y_0)$ that is totally enveloped by an ellipse with a major axis $A$ and minor axis $B$? In this problem, assume a constraint s.t. the major axis is in the $x$ direction. Alternate cases of the problem would be solved by a rotation of the major axis to the $x$ axis as a first step. Additionally, assume $(x_0, y_0)$ is within the ellipse. The ellipse is centered at $(0,0)$.



I think that the closed form solution is not guaranteed if neither $x_0$ nor $y_0$ are zero. I know a closed form solution can be found if either $x_0$ or $y_0$ is equal to zero.



Thank you for your time






calculus geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    In other words, you want the minimum distance from any point on the ellipse to $(x_0,y_0).$ Finding the minimum (or maximum) distance is also involved in math.stackexchange.com/questions/2641562/… and math.stackexchange.com/questions/1846062/…
    $endgroup$
    – David K
    Jan 4 at 13:09










  • $begingroup$
    Where is the center of the ellipse?
    $endgroup$
    – Jens
    Jan 4 at 14:53














0












0








0


1



$begingroup$


What is the largest circle centered at $(x_0, y_0)$ that is totally enveloped by an ellipse with a major axis $A$ and minor axis $B$? In this problem, assume a constraint s.t. the major axis is in the $x$ direction. Alternate cases of the problem would be solved by a rotation of the major axis to the $x$ axis as a first step. Additionally, assume $(x_0, y_0)$ is within the ellipse. The ellipse is centered at $(0,0)$.



I think that the closed form solution is not guaranteed if neither $x_0$ nor $y_0$ are zero. I know a closed form solution can be found if either $x_0$ or $y_0$ is equal to zero.



Thank you for your time






calculus geometry







share|cite|improve this question











$endgroup$




What is the largest circle centered at $(x_0, y_0)$ that is totally enveloped by an ellipse with a major axis $A$ and minor axis $B$? In this problem, assume a constraint s.t. the major axis is in the $x$ direction. Alternate cases of the problem would be solved by a rotation of the major axis to the $x$ axis as a first step. Additionally, assume $(x_0, y_0)$ is within the ellipse. The ellipse is centered at $(0,0)$.



I think that the closed form solution is not guaranteed if neither $x_0$ nor $y_0$ are zero. I know a closed form solution can be found if either $x_0$ or $y_0$ is equal to zero.



Thank you for your time






calculus geometry




calculus geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 17:43







szatkosa

















asked Jan 4 at 11:49









szatkosaszatkosa

585




585












  • $begingroup$
    In other words, you want the minimum distance from any point on the ellipse to $(x_0,y_0).$ Finding the minimum (or maximum) distance is also involved in math.stackexchange.com/questions/2641562/… and math.stackexchange.com/questions/1846062/…
    $endgroup$
    – David K
    Jan 4 at 13:09










  • $begingroup$
    Where is the center of the ellipse?
    $endgroup$
    – Jens
    Jan 4 at 14:53


















  • $begingroup$
    In other words, you want the minimum distance from any point on the ellipse to $(x_0,y_0).$ Finding the minimum (or maximum) distance is also involved in math.stackexchange.com/questions/2641562/… and math.stackexchange.com/questions/1846062/…
    $endgroup$
    – David K
    Jan 4 at 13:09










  • $begingroup$
    Where is the center of the ellipse?
    $endgroup$
    – Jens
    Jan 4 at 14:53
















$begingroup$
In other words, you want the minimum distance from any point on the ellipse to $(x_0,y_0).$ Finding the minimum (or maximum) distance is also involved in math.stackexchange.com/questions/2641562/… and math.stackexchange.com/questions/1846062/…
$endgroup$
– David K
Jan 4 at 13:09




$begingroup$
In other words, you want the minimum distance from any point on the ellipse to $(x_0,y_0).$ Finding the minimum (or maximum) distance is also involved in math.stackexchange.com/questions/2641562/… and math.stackexchange.com/questions/1846062/…
$endgroup$
– David K
Jan 4 at 13:09












$begingroup$
Where is the center of the ellipse?
$endgroup$
– Jens
Jan 4 at 14:53




$begingroup$
Where is the center of the ellipse?
$endgroup$
– Jens
Jan 4 at 14:53










1 Answer
1






active

oldest

votes


















1












$begingroup$

Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:




  1. Find the line normal to the ellipse passing through a generic point on the ellipse $(x_1, y_1)$.


The ellipse is $frac{x^2}{A^2} + frac{y^2}{B^2}=1$ (I). Taking derivatives: $frac{2x}{A^2} + frac{2 y y'}{B^2} = 0 Rightarrow y'= - frac{B^2 x}{A^2 y}$, which is the slope of the tangent line. The normal line is perpendicular to the tangent line, so its slope is $m_n=frac{A^2 y}{B^2x}$. The normal line passing through $(x_1, y_1)$ is then: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$




  1. Impose that the normal line goes through the given point $(x_0, y_0)$ and that the point $(x_1, y_1)$ satisfies the equation of the ellipse. That is two equations and two unknowns; get $(x_1, y_1)$ as a function of $(x_0, y_0)$. In general, there will be two solutions.


The point $(x_0, y_0)$ should satisfy $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$, then: $y_0-y_1 = frac{A^2 y_1}{B^2x_1} (x_0-x_1)$ (II)



Solving the system of equations (I), (II) yields at most two solutions for $(x_1,y_1)$, say $(x_{11},y_{11})$ and $(x_{12},y_{12})$.




  1. The radius of the desired circle is the smallest of the distances between $(x_0, y_0)$ and, respectively, $(x_{11},y_{11})$ and $(x_{12},y_{12})$ (call it $d$).


$d=min_{i in{1,2}} sqrt{(x_{1i}-x_0)^2 + (y_{1i}-y_0)^2} $, then the equation of the desired circumference (around the desired circle) is $(x-x_0)^2 + (y-y_0)^2 = d^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sure, thank you!, corrected
    $endgroup$
    – pendermath
    Jan 4 at 19:44










  • $begingroup$
    The relevant normal might be vertical ($x_0=0$), so it’s generally better to work with a different form of equation from slope-intercept.
    $endgroup$
    – amd
    Jan 4 at 19:44










  • $begingroup$
    Indeed, although in those cases finding the circle should be fairly easy. I implicitly assumed that $(x_0,y_0)$ is not in the axes of the ellipse. Thanks again!
    $endgroup$
    – pendermath
    Jan 4 at 19:47










  • $begingroup$
    You are right, again! I will update the solution accordingly Thx
    $endgroup$
    – pendermath
    Jan 4 at 20:01











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061568%2fcircles-within-ellipses%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:




  1. Find the line normal to the ellipse passing through a generic point on the ellipse $(x_1, y_1)$.


The ellipse is $frac{x^2}{A^2} + frac{y^2}{B^2}=1$ (I). Taking derivatives: $frac{2x}{A^2} + frac{2 y y'}{B^2} = 0 Rightarrow y'= - frac{B^2 x}{A^2 y}$, which is the slope of the tangent line. The normal line is perpendicular to the tangent line, so its slope is $m_n=frac{A^2 y}{B^2x}$. The normal line passing through $(x_1, y_1)$ is then: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$




  1. Impose that the normal line goes through the given point $(x_0, y_0)$ and that the point $(x_1, y_1)$ satisfies the equation of the ellipse. That is two equations and two unknowns; get $(x_1, y_1)$ as a function of $(x_0, y_0)$. In general, there will be two solutions.


The point $(x_0, y_0)$ should satisfy $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$, then: $y_0-y_1 = frac{A^2 y_1}{B^2x_1} (x_0-x_1)$ (II)



Solving the system of equations (I), (II) yields at most two solutions for $(x_1,y_1)$, say $(x_{11},y_{11})$ and $(x_{12},y_{12})$.




  1. The radius of the desired circle is the smallest of the distances between $(x_0, y_0)$ and, respectively, $(x_{11},y_{11})$ and $(x_{12},y_{12})$ (call it $d$).


$d=min_{i in{1,2}} sqrt{(x_{1i}-x_0)^2 + (y_{1i}-y_0)^2} $, then the equation of the desired circumference (around the desired circle) is $(x-x_0)^2 + (y-y_0)^2 = d^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sure, thank you!, corrected
    $endgroup$
    – pendermath
    Jan 4 at 19:44










  • $begingroup$
    The relevant normal might be vertical ($x_0=0$), so it’s generally better to work with a different form of equation from slope-intercept.
    $endgroup$
    – amd
    Jan 4 at 19:44










  • $begingroup$
    Indeed, although in those cases finding the circle should be fairly easy. I implicitly assumed that $(x_0,y_0)$ is not in the axes of the ellipse. Thanks again!
    $endgroup$
    – pendermath
    Jan 4 at 19:47










  • $begingroup$
    You are right, again! I will update the solution accordingly Thx
    $endgroup$
    – pendermath
    Jan 4 at 20:01
















1












$begingroup$

Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:




  1. Find the line normal to the ellipse passing through a generic point on the ellipse $(x_1, y_1)$.


The ellipse is $frac{x^2}{A^2} + frac{y^2}{B^2}=1$ (I). Taking derivatives: $frac{2x}{A^2} + frac{2 y y'}{B^2} = 0 Rightarrow y'= - frac{B^2 x}{A^2 y}$, which is the slope of the tangent line. The normal line is perpendicular to the tangent line, so its slope is $m_n=frac{A^2 y}{B^2x}$. The normal line passing through $(x_1, y_1)$ is then: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$




  1. Impose that the normal line goes through the given point $(x_0, y_0)$ and that the point $(x_1, y_1)$ satisfies the equation of the ellipse. That is two equations and two unknowns; get $(x_1, y_1)$ as a function of $(x_0, y_0)$. In general, there will be two solutions.


The point $(x_0, y_0)$ should satisfy $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$, then: $y_0-y_1 = frac{A^2 y_1}{B^2x_1} (x_0-x_1)$ (II)



Solving the system of equations (I), (II) yields at most two solutions for $(x_1,y_1)$, say $(x_{11},y_{11})$ and $(x_{12},y_{12})$.




  1. The radius of the desired circle is the smallest of the distances between $(x_0, y_0)$ and, respectively, $(x_{11},y_{11})$ and $(x_{12},y_{12})$ (call it $d$).


$d=min_{i in{1,2}} sqrt{(x_{1i}-x_0)^2 + (y_{1i}-y_0)^2} $, then the equation of the desired circumference (around the desired circle) is $(x-x_0)^2 + (y-y_0)^2 = d^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sure, thank you!, corrected
    $endgroup$
    – pendermath
    Jan 4 at 19:44










  • $begingroup$
    The relevant normal might be vertical ($x_0=0$), so it’s generally better to work with a different form of equation from slope-intercept.
    $endgroup$
    – amd
    Jan 4 at 19:44










  • $begingroup$
    Indeed, although in those cases finding the circle should be fairly easy. I implicitly assumed that $(x_0,y_0)$ is not in the axes of the ellipse. Thanks again!
    $endgroup$
    – pendermath
    Jan 4 at 19:47










  • $begingroup$
    You are right, again! I will update the solution accordingly Thx
    $endgroup$
    – pendermath
    Jan 4 at 20:01














1












1








1





$begingroup$

Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:




  1. Find the line normal to the ellipse passing through a generic point on the ellipse $(x_1, y_1)$.


The ellipse is $frac{x^2}{A^2} + frac{y^2}{B^2}=1$ (I). Taking derivatives: $frac{2x}{A^2} + frac{2 y y'}{B^2} = 0 Rightarrow y'= - frac{B^2 x}{A^2 y}$, which is the slope of the tangent line. The normal line is perpendicular to the tangent line, so its slope is $m_n=frac{A^2 y}{B^2x}$. The normal line passing through $(x_1, y_1)$ is then: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$




  1. Impose that the normal line goes through the given point $(x_0, y_0)$ and that the point $(x_1, y_1)$ satisfies the equation of the ellipse. That is two equations and two unknowns; get $(x_1, y_1)$ as a function of $(x_0, y_0)$. In general, there will be two solutions.


The point $(x_0, y_0)$ should satisfy $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$, then: $y_0-y_1 = frac{A^2 y_1}{B^2x_1} (x_0-x_1)$ (II)



Solving the system of equations (I), (II) yields at most two solutions for $(x_1,y_1)$, say $(x_{11},y_{11})$ and $(x_{12},y_{12})$.




  1. The radius of the desired circle is the smallest of the distances between $(x_0, y_0)$ and, respectively, $(x_{11},y_{11})$ and $(x_{12},y_{12})$ (call it $d$).


$d=min_{i in{1,2}} sqrt{(x_{1i}-x_0)^2 + (y_{1i}-y_0)^2} $, then the equation of the desired circumference (around the desired circle) is $(x-x_0)^2 + (y-y_0)^2 = d^2$.






share|cite|improve this answer











$endgroup$



Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:




  1. Find the line normal to the ellipse passing through a generic point on the ellipse $(x_1, y_1)$.


The ellipse is $frac{x^2}{A^2} + frac{y^2}{B^2}=1$ (I). Taking derivatives: $frac{2x}{A^2} + frac{2 y y'}{B^2} = 0 Rightarrow y'= - frac{B^2 x}{A^2 y}$, which is the slope of the tangent line. The normal line is perpendicular to the tangent line, so its slope is $m_n=frac{A^2 y}{B^2x}$. The normal line passing through $(x_1, y_1)$ is then: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$




  1. Impose that the normal line goes through the given point $(x_0, y_0)$ and that the point $(x_1, y_1)$ satisfies the equation of the ellipse. That is two equations and two unknowns; get $(x_1, y_1)$ as a function of $(x_0, y_0)$. In general, there will be two solutions.


The point $(x_0, y_0)$ should satisfy $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$, then: $y_0-y_1 = frac{A^2 y_1}{B^2x_1} (x_0-x_1)$ (II)



Solving the system of equations (I), (II) yields at most two solutions for $(x_1,y_1)$, say $(x_{11},y_{11})$ and $(x_{12},y_{12})$.




  1. The radius of the desired circle is the smallest of the distances between $(x_0, y_0)$ and, respectively, $(x_{11},y_{11})$ and $(x_{12},y_{12})$ (call it $d$).


$d=min_{i in{1,2}} sqrt{(x_{1i}-x_0)^2 + (y_{1i}-y_0)^2} $, then the equation of the desired circumference (around the desired circle) is $(x-x_0)^2 + (y-y_0)^2 = d^2$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 20:18

























answered Jan 4 at 17:50









pendermathpendermath

58612




58612












  • $begingroup$
    Sure, thank you!, corrected
    $endgroup$
    – pendermath
    Jan 4 at 19:44










  • $begingroup$
    The relevant normal might be vertical ($x_0=0$), so it’s generally better to work with a different form of equation from slope-intercept.
    $endgroup$
    – amd
    Jan 4 at 19:44










  • $begingroup$
    Indeed, although in those cases finding the circle should be fairly easy. I implicitly assumed that $(x_0,y_0)$ is not in the axes of the ellipse. Thanks again!
    $endgroup$
    – pendermath
    Jan 4 at 19:47










  • $begingroup$
    You are right, again! I will update the solution accordingly Thx
    $endgroup$
    – pendermath
    Jan 4 at 20:01


















  • $begingroup$
    Sure, thank you!, corrected
    $endgroup$
    – pendermath
    Jan 4 at 19:44










  • $begingroup$
    The relevant normal might be vertical ($x_0=0$), so it’s generally better to work with a different form of equation from slope-intercept.
    $endgroup$
    – amd
    Jan 4 at 19:44










  • $begingroup$
    Indeed, although in those cases finding the circle should be fairly easy. I implicitly assumed that $(x_0,y_0)$ is not in the axes of the ellipse. Thanks again!
    $endgroup$
    – pendermath
    Jan 4 at 19:47










  • $begingroup$
    You are right, again! I will update the solution accordingly Thx
    $endgroup$
    – pendermath
    Jan 4 at 20:01
















$begingroup$
Sure, thank you!, corrected
$endgroup$
– pendermath
Jan 4 at 19:44




$begingroup$
Sure, thank you!, corrected
$endgroup$
– pendermath
Jan 4 at 19:44












$begingroup$
The relevant normal might be vertical ($x_0=0$), so it’s generally better to work with a different form of equation from slope-intercept.
$endgroup$
– amd
Jan 4 at 19:44




$begingroup$
The relevant normal might be vertical ($x_0=0$), so it’s generally better to work with a different form of equation from slope-intercept.
$endgroup$
– amd
Jan 4 at 19:44












$begingroup$
Indeed, although in those cases finding the circle should be fairly easy. I implicitly assumed that $(x_0,y_0)$ is not in the axes of the ellipse. Thanks again!
$endgroup$
– pendermath
Jan 4 at 19:47




$begingroup$
Indeed, although in those cases finding the circle should be fairly easy. I implicitly assumed that $(x_0,y_0)$ is not in the axes of the ellipse. Thanks again!
$endgroup$
– pendermath
Jan 4 at 19:47












$begingroup$
You are right, again! I will update the solution accordingly Thx
$endgroup$
– pendermath
Jan 4 at 20:01




$begingroup$
You are right, again! I will update the solution accordingly Thx
$endgroup$
– pendermath
Jan 4 at 20:01


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061568%2fcircles-within-ellipses%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more