Finding the diameter of a circle from the length and position of a chord
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I have a circle of unknown diameter. I do, however, know the length of a chord and the distance between the centre of the circle and the centre of the chord.
Please see picture here where I have added some sample values: I want to determine the value of d
UPDATE: Diagram illustrating maxmilgram's solution below here
The solution in summary is: because we know lines $MN$ and $NP$ an well as angle $∠PNM$ we can solve using simple trig!
geometry trigonometry circle
asked Nov 20 at 9:16
alifen
33
add a comment |
up vote
0
down vote
favorite
I have a circle of unknown diameter. I do, however, know the length of a chord and the distance between the centre of the circle and the centre of the chord.
Please see picture here where I have added some sample values: I want to determine the value of d
UPDATE: Diagram illustrating maxmilgram's solution below here
The solution in summary is: because we know lines $MN$ and $NP$ an well as angle $∠PNM$ we can solve using simple trig!
geometry trigonometry circle
asked Nov 20 at 9:16
alifen
33
In your diagram you know the length of the chord and the distance from center of chord to center of circle. Is that an equivalent description of the knowns?
– coffeemath
Nov 20 at 9:24
@coffeemath - yes, it is - thanks!
– alifen
Nov 20 at 9:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a circle of unknown diameter. I do, however, know the length of a chord and the distance between the centre of the circle and the centre of the chord.
Please see picture here where I have added some sample values: I want to determine the value of d
UPDATE: Diagram illustrating maxmilgram's solution below here
The solution in summary is: because we know lines $MN$ and $NP$ an well as angle $∠PNM$ we can solve using simple trig!
geometry trigonometry circle
asked Nov 20 at 9:16
alifen
33
I have a circle of unknown diameter. I do, however, know the length of a chord and the distance between the centre of the circle and the centre of the chord.
Please see picture here where I have added some sample values: I want to determine the value of d
UPDATE: Diagram illustrating maxmilgram's solution below here
The solution in summary is: because we know lines $MN$ and $NP$ an well as angle $∠PNM$ we can solve using simple trig!
geometry trigonometry circle
geometry trigonometry circle
asked Nov 20 at 9:16
alifen
33
asked Nov 20 at 9:16
alifen
33
edited Nov 20 at 13:51
asked Nov 20 at 9:16
alifen
33
asked Nov 20 at 9:16
alifen
33
asked Nov 20 at 9:16
alifen
33
33
In your diagram you know the length of the chord and the distance from center of chord to center of circle. Is that an equivalent description of the knowns?
– coffeemath
Nov 20 at 9:24
@coffeemath - yes, it is - thanks!
– alifen
Nov 20 at 9:41
add a comment |
In your diagram you know the length of the chord and the distance from center of chord to center of circle. Is that an equivalent description of the knowns?
– coffeemath
Nov 20 at 9:24
@coffeemath - yes, it is - thanks!
– alifen
Nov 20 at 9:41
In your diagram you know the length of the chord and the distance from center of chord to center of circle. Is that an equivalent description of the knowns?
– coffeemath
Nov 20 at 9:24
In your diagram you know the length of the chord and the distance from center of chord to center of circle. Is that an equivalent description of the knowns?
– coffeemath
Nov 20 at 9:24
@coffeemath - yes, it is - thanks!
– alifen
Nov 20 at 9:41
@coffeemath - yes, it is - thanks!
– alifen
Nov 20 at 9:41
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Lets call the center point of the circle $M$, the two endpoints of the chord (="diamenter line") $P$ and $Q$ and the midpoint of the chord $N$.
Now observe that the distance between $M$ and $P$ is $r=d/2$, the distance between $N$ and $M$ is $h$ and the distance between $N$ and $P$ is $x/2$. Furthermore the angle $angle PMN$ is $90°$. Can you take it from there?
answered Nov 20 at 9:24
maxmilgram
4227
Also, note that EVERY chord is parallel to a diameter!
– maxmilgram
Nov 20 at 9:37
This is where I got to - a right-angle triangle where I know one side (h) and one (right) angle: but that isn't enough - as far as I understand - for trig to help...
– alifen
Nov 20 at 9:39
You know the length of the other side as well, as described in my answer.
– maxmilgram
Nov 20 at 9:48
Sorry, yes - my mistake. Could you explain how the line NP is x/2? It is obviously right, I just can't work out why! Thanks so much
– alifen
Nov 20 at 9:52
1
The bisection of $PQ$ is by definition the set of all points with the same distance from $P$ and $Q$. Since $P$ and $Q$ are both points are on the circle they have the same distance from the center point. Thus the center point lies on the bisection. The bisection is orthogonal to the line and cuts the distance in half. en.wikipedia.org/wiki/Bisection#Line_segment_bisector
– maxmilgram
Nov 20 at 10:03
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Lets call the center point of the circle $M$, the two endpoints of the chord (="diamenter line") $P$ and $Q$ and the midpoint of the chord $N$.
Now observe that the distance between $M$ and $P$ is $r=d/2$, the distance between $N$ and $M$ is $h$ and the distance between $N$ and $P$ is $x/2$. Furthermore the angle $angle PMN$ is $90°$. Can you take it from there?
answered Nov 20 at 9:24
maxmilgram
4227
Also, note that EVERY chord is parallel to a diameter!
– maxmilgram
Nov 20 at 9:37
This is where I got to - a right-angle triangle where I know one side (h) and one (right) angle: but that isn't enough - as far as I understand - for trig to help...
– alifen
Nov 20 at 9:39
You know the length of the other side as well, as described in my answer.
– maxmilgram
Nov 20 at 9:48
Sorry, yes - my mistake. Could you explain how the line NP is x/2? It is obviously right, I just can't work out why! Thanks so much
– alifen
Nov 20 at 9:52
1
The bisection of $PQ$ is by definition the set of all points with the same distance from $P$ and $Q$. Since $P$ and $Q$ are both points are on the circle they have the same distance from the center point. Thus the center point lies on the bisection. The bisection is orthogonal to the line and cuts the distance in half. en.wikipedia.org/wiki/Bisection#Line_segment_bisector
– maxmilgram
Nov 20 at 10:03
|
show 3 more comments
up vote
0
down vote
accepted
Lets call the center point of the circle $M$, the two endpoints of the chord (="diamenter line") $P$ and $Q$ and the midpoint of the chord $N$.
Now observe that the distance between $M$ and $P$ is $r=d/2$, the distance between $N$ and $M$ is $h$ and the distance between $N$ and $P$ is $x/2$. Furthermore the angle $angle PMN$ is $90°$. Can you take it from there?
answered Nov 20 at 9:24
maxmilgram
4227
Also, note that EVERY chord is parallel to a diameter!
– maxmilgram
Nov 20 at 9:37
This is where I got to - a right-angle triangle where I know one side (h) and one (right) angle: but that isn't enough - as far as I understand - for trig to help...
– alifen
Nov 20 at 9:39
You know the length of the other side as well, as described in my answer.
– maxmilgram
Nov 20 at 9:48
Sorry, yes - my mistake. Could you explain how the line NP is x/2? It is obviously right, I just can't work out why! Thanks so much
– alifen
Nov 20 at 9:52
1
The bisection of $PQ$ is by definition the set of all points with the same distance from $P$ and $Q$. Since $P$ and $Q$ are both points are on the circle they have the same distance from the center point. Thus the center point lies on the bisection. The bisection is orthogonal to the line and cuts the distance in half. en.wikipedia.org/wiki/Bisection#Line_segment_bisector
– maxmilgram
Nov 20 at 10:03
|
show 3 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Lets call the center point of the circle $M$, the two endpoints of the chord (="diamenter line") $P$ and $Q$ and the midpoint of the chord $N$.
Now observe that the distance between $M$ and $P$ is $r=d/2$, the distance between $N$ and $M$ is $h$ and the distance between $N$ and $P$ is $x/2$. Furthermore the angle $angle PMN$ is $90°$. Can you take it from there?
answered Nov 20 at 9:24
maxmilgram
4227
Lets call the center point of the circle $M$, the two endpoints of the chord (="diamenter line") $P$ and $Q$ and the midpoint of the chord $N$.
Now observe that the distance between $M$ and $P$ is $r=d/2$, the distance between $N$ and $M$ is $h$ and the distance between $N$ and $P$ is $x/2$. Furthermore the angle $angle PMN$ is $90°$. Can you take it from there?
answered Nov 20 at 9:24
maxmilgram
4227
answered Nov 20 at 9:24
maxmilgram
4227
answered Nov 20 at 9:24
maxmilgram
4227
answered Nov 20 at 9:24
maxmilgram
4227
4227
Also, note that EVERY chord is parallel to a diameter!
– maxmilgram
Nov 20 at 9:37
This is where I got to - a right-angle triangle where I know one side (h) and one (right) angle: but that isn't enough - as far as I understand - for trig to help...
– alifen
Nov 20 at 9:39
You know the length of the other side as well, as described in my answer.
– maxmilgram
Nov 20 at 9:48
Sorry, yes - my mistake. Could you explain how the line NP is x/2? It is obviously right, I just can't work out why! Thanks so much
– alifen
Nov 20 at 9:52
1
The bisection of $PQ$ is by definition the set of all points with the same distance from $P$ and $Q$. Since $P$ and $Q$ are both points are on the circle they have the same distance from the center point. Thus the center point lies on the bisection. The bisection is orthogonal to the line and cuts the distance in half. en.wikipedia.org/wiki/Bisection#Line_segment_bisector
– maxmilgram
Nov 20 at 10:03
|
show 3 more comments
Also, note that EVERY chord is parallel to a diameter!
– maxmilgram
Nov 20 at 9:37
This is where I got to - a right-angle triangle where I know one side (h) and one (right) angle: but that isn't enough - as far as I understand - for trig to help...
– alifen
Nov 20 at 9:39
You know the length of the other side as well, as described in my answer.
– maxmilgram
Nov 20 at 9:48
Sorry, yes - my mistake. Could you explain how the line NP is x/2? It is obviously right, I just can't work out why! Thanks so much
– alifen
Nov 20 at 9:52
1
The bisection of $PQ$ is by definition the set of all points with the same distance from $P$ and $Q$. Since $P$ and $Q$ are both points are on the circle they have the same distance from the center point. Thus the center point lies on the bisection. The bisection is orthogonal to the line and cuts the distance in half. en.wikipedia.org/wiki/Bisection#Line_segment_bisector
– maxmilgram
Nov 20 at 10:03
Also, note that EVERY chord is parallel to a diameter!
– maxmilgram
Nov 20 at 9:37
Also, note that EVERY chord is parallel to a diameter!
– maxmilgram
Nov 20 at 9:37
This is where I got to - a right-angle triangle where I know one side (h) and one (right) angle: but that isn't enough - as far as I understand - for trig to help...
– alifen
Nov 20 at 9:39
This is where I got to - a right-angle triangle where I know one side (h) and one (right) angle: but that isn't enough - as far as I understand - for trig to help...
– alifen
Nov 20 at 9:39
You know the length of the other side as well, as described in my answer.
– maxmilgram
Nov 20 at 9:48
You know the length of the other side as well, as described in my answer.
– maxmilgram
Nov 20 at 9:48
Sorry, yes - my mistake. Could you explain how the line NP is x/2? It is obviously right, I just can't work out why! Thanks so much
– alifen
Nov 20 at 9:52
Sorry, yes - my mistake. Could you explain how the line NP is x/2? It is obviously right, I just can't work out why! Thanks so much
– alifen
Nov 20 at 9:52
1
1
The bisection of $PQ$ is by definition the set of all points with the same distance from $P$ and $Q$. Since $P$ and $Q$ are both points are on the circle they have the same distance from the center point. Thus the center point lies on the bisection. The bisection is orthogonal to the line and cuts the distance in half. en.wikipedia.org/wiki/Bisection#Line_segment_bisector
– maxmilgram
Nov 20 at 10:03
The bisection of $PQ$ is by definition the set of all points with the same distance from $P$ and $Q$. Since $P$ and $Q$ are both points are on the circle they have the same distance from the center point. Thus the center point lies on the bisection. The bisection is orthogonal to the line and cuts the distance in half. en.wikipedia.org/wiki/Bisection#Line_segment_bisector
– maxmilgram
Nov 20 at 10:03
|
show 3 more comments
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In your diagram you know the length of the chord and the distance from center of chord to center of circle. Is that an equivalent description of the knowns?
– coffeemath
Nov 20 at 9:24
@coffeemath - yes, it is - thanks!
– alifen
Nov 20 at 9:41