Circles within ellipses
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
asked 16 hours ago
szatkosa
564
add a comment |
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
asked 16 hours ago
szatkosa
564
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/…
– David K
14 hours ago
Where is the center of the ellipse?
– Jens
12 hours ago
add a comment |
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
asked 16 hours ago
szatkosa
564
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
asked 16 hours ago
szatkosa
564
asked 16 hours ago
szatkosa
564
edited 10 hours ago
asked 16 hours ago
szatkosa
564
asked 16 hours ago
szatkosa
564
asked 16 hours ago
szatkosa
564
564
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/…
– David K
14 hours ago
Where is the center of the ellipse?
– Jens
12 hours ago
add a comment |
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/…
– David K
14 hours ago
Where is the center of the ellipse?
– Jens
12 hours ago
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/…
– David K
14 hours ago
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/…
– David K
14 hours ago
Where is the center of the ellipse?
– Jens
12 hours ago
Where is the center of the ellipse?
– Jens
12 hours ago
add a comment |
1 Answer
1
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oldest
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Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:
- 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: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$
- 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 $(x_{12},y_{12})$.
- 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 $(x-x_0)^2 + (y-y_0)^2 = d^2$.
answered 10 hours ago
pendermath
743
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sure, thank you!, corrected
– pendermath
8 hours ago
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.
– amd
8 hours ago
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!
– pendermath
8 hours ago
You are right, again! I will update the solution accordingly Thx
– pendermath
7 hours ago
add a comment |
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Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:
- 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: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$
- 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 $(x_{12},y_{12})$.
- 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 $(x-x_0)^2 + (y-y_0)^2 = d^2$.
answered 10 hours ago
pendermath
743
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sure, thank you!, corrected
– pendermath
8 hours ago
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.
– amd
8 hours ago
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!
– pendermath
8 hours ago
You are right, again! I will update the solution accordingly Thx
– pendermath
7 hours ago
add a comment |
Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:
- 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: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$
- 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 $(x_{12},y_{12})$.
- 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 $(x-x_0)^2 + (y-y_0)^2 = d^2$.
answered 10 hours ago
pendermath
743
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sure, thank you!, corrected
– pendermath
8 hours ago
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.
– amd
8 hours ago
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!
– pendermath
8 hours ago
You are right, again! I will update the solution accordingly Thx
– pendermath
7 hours ago
add a comment |
Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:
- 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: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$
- 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 $(x_{12},y_{12})$.
- 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 $(x-x_0)^2 + (y-y_0)^2 = d^2$.
answered 10 hours ago
pendermath
743
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Assuming the ellipse center is at $(0,0)$, here are some steps that can be followed to come up with the solution:
- 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: $y-y_1 = frac{A^2 y_1}{B^2x_1} (x-x_1)$
- 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 $(x_{12},y_{12})$.
- 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 $(x-x_0)^2 + (y-y_0)^2 = d^2$.
answered 10 hours ago
pendermath
743
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
answered 10 hours ago
pendermath
743
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 10 hours ago
pendermath
743
answered 10 hours ago
pendermath
743
743
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sure, thank you!, corrected
– pendermath
8 hours ago
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.
– amd
8 hours ago
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!
– pendermath
8 hours ago
You are right, again! I will update the solution accordingly Thx
– pendermath
7 hours ago
add a comment |
Sure, thank you!, corrected
– pendermath
8 hours ago
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.
– amd
8 hours ago
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!
– pendermath
8 hours ago
You are right, again! I will update the solution accordingly Thx
– pendermath
7 hours ago
Sure, thank you!, corrected
– pendermath
8 hours ago
Sure, thank you!, corrected
– pendermath
8 hours ago
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.
– amd
8 hours ago
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.
– amd
8 hours ago
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!
– pendermath
8 hours ago
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!
– pendermath
8 hours ago
You are right, again! I will update the solution accordingly Thx
– pendermath
7 hours ago
You are right, again! I will update the solution accordingly Thx
– pendermath
7 hours ago
add a comment |
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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/…
– David K
14 hours ago
Where is the center of the ellipse?
– Jens
12 hours ago