Mixing
Circles within ellipses
$begingroup$
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
asked Jan 4 at 11:49
szatkosaszatkosa
585
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 11:49
szatkosaszatkosa
585
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 4 at 11:49
szatkosaszatkosa
585
$endgroup$
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
asked Jan 4 at 11:49
szatkosaszatkosa
585
asked Jan 4 at 11:49
szatkosaszatkosa
585
edited Jan 4 at 17:43
szatkosa
asked Jan 4 at 11:49
szatkosaszatkosa
585
asked Jan 4 at 11:49
szatkosaszatkosa
585
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 4 at 17:50
pendermathpendermath
52910
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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 Jan 4 at 17:50
pendermathpendermath
52910
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 Jan 4 at 17:50
pendermathpendermath
52910
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 Jan 4 at 17:50
pendermathpendermath
52910
$endgroup$
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 Jan 4 at 17:50
pendermathpendermath
52910
edited Jan 4 at 20:18
answered Jan 4 at 17:50
pendermathpendermath
52910
answered Jan 4 at 17:50
pendermathpendermath
52910
answered Jan 4 at 17:50
pendermathpendermath
52910
52910
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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