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Calculate radius of circle required to completely overlap rectangle when centered within it [closed]
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Consider a rectangle with known width ($w$), height ($h$) and center ($R$). You can assume $wgeqslant h$.
A random point ($C$) is selected within the rectangle which is a distance of $x$ (horizontally) and $y$ (vertically) from $R$:
A circle is drawn centered at $C$ such that it completely overlaps the rectangle, but no more:
Is there a way to calculate the circle's radius $r$, given the known inputs $w$, $h$, $R$, $C$, $x$ and $y$?
geometry circle rectangles
asked Jul 17 at 23:41
Luke
1112
closed as off-topic by amWhy, Xander Henderson, Isaac Browne, Parcly Taxel, user223391 Jul 19 at 1:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Isaac Browne, Parcly Taxel, Community
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Consider a rectangle with known width ($w$), height ($h$) and center ($R$). You can assume $wgeqslant h$.
A random point ($C$) is selected within the rectangle which is a distance of $x$ (horizontally) and $y$ (vertically) from $R$:
A circle is drawn centered at $C$ such that it completely overlaps the rectangle, but no more:
Is there a way to calculate the circle's radius $r$, given the known inputs $w$, $h$, $R$, $C$, $x$ and $y$?
geometry circle rectangles
asked Jul 17 at 23:41
Luke
1112
closed as off-topic by amWhy, Xander Henderson, Isaac Browne, Parcly Taxel, user223391 Jul 19 at 1:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Isaac Browne, Parcly Taxel, Community
Yes. The radius is simply the maximum distance of the rectangle's four corners to $C$. If all four corners are in the circle, then the entire rectangle is covered by the circle.
– hardmath
Jul 18 at 2:25
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider a rectangle with known width ($w$), height ($h$) and center ($R$). You can assume $wgeqslant h$.
A random point ($C$) is selected within the rectangle which is a distance of $x$ (horizontally) and $y$ (vertically) from $R$:
A circle is drawn centered at $C$ such that it completely overlaps the rectangle, but no more:
Is there a way to calculate the circle's radius $r$, given the known inputs $w$, $h$, $R$, $C$, $x$ and $y$?
geometry circle rectangles
asked Jul 17 at 23:41
Luke
1112
Consider a rectangle with known width ($w$), height ($h$) and center ($R$). You can assume $wgeqslant h$.
A random point ($C$) is selected within the rectangle which is a distance of $x$ (horizontally) and $y$ (vertically) from $R$:
A circle is drawn centered at $C$ such that it completely overlaps the rectangle, but no more:
Is there a way to calculate the circle's radius $r$, given the known inputs $w$, $h$, $R$, $C$, $x$ and $y$?
geometry circle rectangles
asked Jul 17 at 23:41
Luke
1112
asked Jul 17 at 23:41
Luke
1112
asked Jul 17 at 23:41
Luke
1112
asked Jul 17 at 23:41
Luke
1112
1112
closed as off-topic by amWhy, Xander Henderson, Isaac Browne, Parcly Taxel, user223391 Jul 19 at 1:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Isaac Browne, Parcly Taxel, Community
closed as off-topic by amWhy, Xander Henderson, Isaac Browne, Parcly Taxel, user223391 Jul 19 at 1:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Isaac Browne, Parcly Taxel, Community
Yes. The radius is simply the maximum distance of the rectangle's four corners to $C$. If all four corners are in the circle, then the entire rectangle is covered by the circle.
– hardmath
Jul 18 at 2:25
add a comment |Â
Yes. The radius is simply the maximum distance of the rectangle's four corners to $C$. If all four corners are in the circle, then the entire rectangle is covered by the circle.
– hardmath
Jul 18 at 2:25
Yes. The radius is simply the maximum distance of the rectangle's four corners to $C$. If all four corners are in the circle, then the entire rectangle is covered by the circle.
– hardmath
Jul 18 at 2:25
Yes. The radius is simply the maximum distance of the rectangle's four corners to $C$. If all four corners are in the circle, then the entire rectangle is covered by the circle.
– hardmath
Jul 18 at 2:25
add a comment |Â
1 Answer
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Yes, there is. The radius of the circle is merely the length from $C$ to the vertex of the rectangle farthest away. Just draw a line from $C$ to this vertex and make that the hypotenuse of a right triangle. The legs of the triangle would be $|y|+frach2$, and $|x|+fracw2$. Using the Pythagorean Theorem, this is $$sqrt(+frach2)^2+()^2$$.
answered Jul 17 at 23:50
RayDansh
884215
Yes, though I would add the proviso that $x, y geq 0$, without loss of generality, or replace them by $|x|$ and $|y|$, respectively.
– Brian Tung
Jul 17 at 23:52
@BrianTung Good catch. I will update my answer.
– RayDansh
Jul 17 at 23:53
Oh yes, so it is! Thank you very much
– Luke
Jul 18 at 1:02
Can't believe I didn't spot this while working on those diagrams haha
– Luke
Jul 18 at 11:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Yes, there is. The radius of the circle is merely the length from $C$ to the vertex of the rectangle farthest away. Just draw a line from $C$ to this vertex and make that the hypotenuse of a right triangle. The legs of the triangle would be $|y|+frach2$, and $|x|+fracw2$. Using the Pythagorean Theorem, this is $$sqrt(+frach2)^2+()^2$$.
answered Jul 17 at 23:50
RayDansh
884215
Yes, though I would add the proviso that $x, y geq 0$, without loss of generality, or replace them by $|x|$ and $|y|$, respectively.
– Brian Tung
Jul 17 at 23:52
@BrianTung Good catch. I will update my answer.
– RayDansh
Jul 17 at 23:53
Oh yes, so it is! Thank you very much
– Luke
Jul 18 at 1:02
Can't believe I didn't spot this while working on those diagrams haha
– Luke
Jul 18 at 11:37
add a comment |Â
up vote
3
down vote
accepted
Yes, there is. The radius of the circle is merely the length from $C$ to the vertex of the rectangle farthest away. Just draw a line from $C$ to this vertex and make that the hypotenuse of a right triangle. The legs of the triangle would be $|y|+frach2$, and $|x|+fracw2$. Using the Pythagorean Theorem, this is $$sqrt(+frach2)^2+()^2$$.
answered Jul 17 at 23:50
RayDansh
884215
Yes, though I would add the proviso that $x, y geq 0$, without loss of generality, or replace them by $|x|$ and $|y|$, respectively.
– Brian Tung
Jul 17 at 23:52
@BrianTung Good catch. I will update my answer.
– RayDansh
Jul 17 at 23:53
Oh yes, so it is! Thank you very much
– Luke
Jul 18 at 1:02
Can't believe I didn't spot this while working on those diagrams haha
– Luke
Jul 18 at 11:37
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes, there is. The radius of the circle is merely the length from $C$ to the vertex of the rectangle farthest away. Just draw a line from $C$ to this vertex and make that the hypotenuse of a right triangle. The legs of the triangle would be $|y|+frach2$, and $|x|+fracw2$. Using the Pythagorean Theorem, this is $$sqrt(+frach2)^2+()^2$$.
answered Jul 17 at 23:50
RayDansh
884215
Yes, there is. The radius of the circle is merely the length from $C$ to the vertex of the rectangle farthest away. Just draw a line from $C$ to this vertex and make that the hypotenuse of a right triangle. The legs of the triangle would be $|y|+frach2$, and $|x|+fracw2$. Using the Pythagorean Theorem, this is $$sqrt(+frach2)^2+()^2$$.
answered Jul 17 at 23:50
RayDansh
884215
edited Jul 17 at 23:53
answered Jul 17 at 23:50
RayDansh
884215
answered Jul 17 at 23:50
RayDansh
884215
answered Jul 17 at 23:50
RayDansh
884215
884215
Yes, though I would add the proviso that $x, y geq 0$, without loss of generality, or replace them by $|x|$ and $|y|$, respectively.
– Brian Tung
Jul 17 at 23:52
@BrianTung Good catch. I will update my answer.
– RayDansh
Jul 17 at 23:53
Oh yes, so it is! Thank you very much
– Luke
Jul 18 at 1:02
Can't believe I didn't spot this while working on those diagrams haha
– Luke
Jul 18 at 11:37
add a comment |Â
Yes, though I would add the proviso that $x, y geq 0$, without loss of generality, or replace them by $|x|$ and $|y|$, respectively.
– Brian Tung
Jul 17 at 23:52
@BrianTung Good catch. I will update my answer.
– RayDansh
Jul 17 at 23:53
Oh yes, so it is! Thank you very much
– Luke
Jul 18 at 1:02
Can't believe I didn't spot this while working on those diagrams haha
– Luke
Jul 18 at 11:37
Yes, though I would add the proviso that $x, y geq 0$, without loss of generality, or replace them by $|x|$ and $|y|$, respectively.
– Brian Tung
Jul 17 at 23:52
Yes, though I would add the proviso that $x, y geq 0$, without loss of generality, or replace them by $|x|$ and $|y|$, respectively.
– Brian Tung
Jul 17 at 23:52
@BrianTung Good catch. I will update my answer.
– RayDansh
Jul 17 at 23:53
@BrianTung Good catch. I will update my answer.
– RayDansh
Jul 17 at 23:53
Oh yes, so it is! Thank you very much
– Luke
Jul 18 at 1:02
Oh yes, so it is! Thank you very much
– Luke
Jul 18 at 1:02
Can't believe I didn't spot this while working on those diagrams haha
– Luke
Jul 18 at 11:37
Can't believe I didn't spot this while working on those diagrams haha
– Luke
Jul 18 at 11:37
add a comment |Â
Yes. The radius is simply the maximum distance of the rectangle's four corners to $C$. If all four corners are in the circle, then the entire rectangle is covered by the circle.
– hardmath
Jul 18 at 2:25