Find a circle that is tangent to a line and a circle.
$begingroup$
I need to find the circle that will be tangent to a line at a given point and a circle. The diagram below hopefully makes it clearer.
The known data are:
points P2,P3,P4,P5
Circle C2's radius r
Angle C
The distance from P3 to P2, a
I need to find P1 or b (which makes it easy enough to work out P1). There are actually two solutions to this problem. Circle C1 could also be tangent on the far side of C2 (c=b-r).
As some background info I am using this to find the medial axis of a polygon. By stepping P3 along each line/arc of the polygon I can generate an approximation of the medial axis. I have already worked out the solution for two lines.
geometry circles
edited Dec 26 '18 at 9:34
Glorfindel
3,41381930
asked May 24 '12 at 13:38
Les NewellLes Newell
61
$endgroup$
add a comment |
$begingroup$
I need to find the circle that will be tangent to a line at a given point and a circle. The diagram below hopefully makes it clearer.
The known data are:
points P2,P3,P4,P5
Circle C2's radius r
Angle C
The distance from P3 to P2, a
I need to find P1 or b (which makes it easy enough to work out P1). There are actually two solutions to this problem. Circle C1 could also be tangent on the far side of C2 (c=b-r).
As some background info I am using this to find the medial axis of a polygon. By stepping P3 along each line/arc of the polygon I can generate an approximation of the medial axis. I have already worked out the solution for two lines.
geometry circles
edited Dec 26 '18 at 9:34
Glorfindel
3,41381930
asked May 24 '12 at 13:38
Les NewellLes Newell
61
$endgroup$
add a comment |
$begingroup$
I need to find the circle that will be tangent to a line at a given point and a circle. The diagram below hopefully makes it clearer.
The known data are:
points P2,P3,P4,P5
Circle C2's radius r
Angle C
The distance from P3 to P2, a
I need to find P1 or b (which makes it easy enough to work out P1). There are actually two solutions to this problem. Circle C1 could also be tangent on the far side of C2 (c=b-r).
As some background info I am using this to find the medial axis of a polygon. By stepping P3 along each line/arc of the polygon I can generate an approximation of the medial axis. I have already worked out the solution for two lines.
geometry circles
edited Dec 26 '18 at 9:34
Glorfindel
3,41381930
asked May 24 '12 at 13:38
Les NewellLes Newell
61
$endgroup$
I need to find the circle that will be tangent to a line at a given point and a circle. The diagram below hopefully makes it clearer.
The known data are:
points P2,P3,P4,P5
Circle C2's radius r
Angle C
The distance from P3 to P2, a
I need to find P1 or b (which makes it easy enough to work out P1). There are actually two solutions to this problem. Circle C1 could also be tangent on the far side of C2 (c=b-r).
As some background info I am using this to find the medial axis of a polygon. By stepping P3 along each line/arc of the polygon I can generate an approximation of the medial axis. I have already worked out the solution for two lines.
geometry circles
geometry circles
edited Dec 26 '18 at 9:34
Glorfindel
3,41381930
asked May 24 '12 at 13:38
Les NewellLes Newell
61
edited Dec 26 '18 at 9:34
Glorfindel
3,41381930
asked May 24 '12 at 13:38
Les NewellLes Newell
61
edited Dec 26 '18 at 9:34
Glorfindel
3,41381930
edited Dec 26 '18 at 9:34
Glorfindel
3,41381930
edited Dec 26 '18 at 9:34
Glorfindel
3,41381930
3,41381930
asked May 24 '12 at 13:38
Les NewellLes Newell
61
asked May 24 '12 at 13:38
Les NewellLes Newell
61
asked May 24 '12 at 13:38
Les NewellLes Newell
61
61
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The set of points that are equidistant from the circle $C_2$ and the line $P_4P_5$ is a parabola. Translate the line $P_4P_5$ (in the opposite direction than the $P_2$) by the radius $r$ of the circle (i.e. $|P_4'P_4| = r$, $|P_5'P_5| = r$ and $P_4P_5 | P_4'P_5'$) to get a new line $P_4'P_5'$ and observe that your solution will be equidistant from the line $P_4'P_5'$ and the point $P_2$ (if before the distance was $b$, then now it is $b+r$). Precisely, the $P_2$ will be the focus and $P_4'P_5'$ will be the directrix. To find the point $P_1$ of parabola just construct the perpendicular bisector of $P_3'P_2$, it should intersect $P_3'P_3$ exactly in $P_1$.
Edit: Added some pictures (the parabola below is just an approximation, not a real parabola).
$hspace{50pt}$
$hspace{50pt}$
answered May 24 '12 at 14:07
dtldarekdtldarek
32.5k746102
$endgroup$
$begingroup$
Thanks. Translating the line makes everything easy. Translating it the other way gives the second solution.
$endgroup$
– Les Newell
May 24 '12 at 14:43
$begingroup$
@LesNewell Yes, exactly ;-) Just that this direction of translation is easier to imagine (and so was my choice).
$endgroup$
– dtldarek
May 24 '12 at 14:55
$begingroup$
Thank you for your detailed help. It would be nice to try to calculate the parabola directly but things get complicated as the circle is actually just an arc and it is joined to other arcs/lines.
$endgroup$
– Les Newell
May 24 '12 at 16:09
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The set of points that are equidistant from the circle $C_2$ and the line $P_4P_5$ is a parabola. Translate the line $P_4P_5$ (in the opposite direction than the $P_2$) by the radius $r$ of the circle (i.e. $|P_4'P_4| = r$, $|P_5'P_5| = r$ and $P_4P_5 | P_4'P_5'$) to get a new line $P_4'P_5'$ and observe that your solution will be equidistant from the line $P_4'P_5'$ and the point $P_2$ (if before the distance was $b$, then now it is $b+r$). Precisely, the $P_2$ will be the focus and $P_4'P_5'$ will be the directrix. To find the point $P_1$ of parabola just construct the perpendicular bisector of $P_3'P_2$, it should intersect $P_3'P_3$ exactly in $P_1$.
Edit: Added some pictures (the parabola below is just an approximation, not a real parabola).
$hspace{50pt}$
$hspace{50pt}$
answered May 24 '12 at 14:07
dtldarekdtldarek
32.5k746102
$endgroup$
$begingroup$
Thanks. Translating the line makes everything easy. Translating it the other way gives the second solution.
$endgroup$
– Les Newell
May 24 '12 at 14:43
$begingroup$
@LesNewell Yes, exactly ;-) Just that this direction of translation is easier to imagine (and so was my choice).
$endgroup$
– dtldarek
May 24 '12 at 14:55
$begingroup$
Thank you for your detailed help. It would be nice to try to calculate the parabola directly but things get complicated as the circle is actually just an arc and it is joined to other arcs/lines.
$endgroup$
– Les Newell
May 24 '12 at 16:09
add a comment |
$begingroup$
The set of points that are equidistant from the circle $C_2$ and the line $P_4P_5$ is a parabola. Translate the line $P_4P_5$ (in the opposite direction than the $P_2$) by the radius $r$ of the circle (i.e. $|P_4'P_4| = r$, $|P_5'P_5| = r$ and $P_4P_5 | P_4'P_5'$) to get a new line $P_4'P_5'$ and observe that your solution will be equidistant from the line $P_4'P_5'$ and the point $P_2$ (if before the distance was $b$, then now it is $b+r$). Precisely, the $P_2$ will be the focus and $P_4'P_5'$ will be the directrix. To find the point $P_1$ of parabola just construct the perpendicular bisector of $P_3'P_2$, it should intersect $P_3'P_3$ exactly in $P_1$.
Edit: Added some pictures (the parabola below is just an approximation, not a real parabola).
$hspace{50pt}$
$hspace{50pt}$
answered May 24 '12 at 14:07
dtldarekdtldarek
32.5k746102
$endgroup$
$begingroup$
Thanks. Translating the line makes everything easy. Translating it the other way gives the second solution.
$endgroup$
– Les Newell
May 24 '12 at 14:43
$begingroup$
@LesNewell Yes, exactly ;-) Just that this direction of translation is easier to imagine (and so was my choice).
$endgroup$
– dtldarek
May 24 '12 at 14:55
$begingroup$
Thank you for your detailed help. It would be nice to try to calculate the parabola directly but things get complicated as the circle is actually just an arc and it is joined to other arcs/lines.
$endgroup$
– Les Newell
May 24 '12 at 16:09
add a comment |
$begingroup$
The set of points that are equidistant from the circle $C_2$ and the line $P_4P_5$ is a parabola. Translate the line $P_4P_5$ (in the opposite direction than the $P_2$) by the radius $r$ of the circle (i.e. $|P_4'P_4| = r$, $|P_5'P_5| = r$ and $P_4P_5 | P_4'P_5'$) to get a new line $P_4'P_5'$ and observe that your solution will be equidistant from the line $P_4'P_5'$ and the point $P_2$ (if before the distance was $b$, then now it is $b+r$). Precisely, the $P_2$ will be the focus and $P_4'P_5'$ will be the directrix. To find the point $P_1$ of parabola just construct the perpendicular bisector of $P_3'P_2$, it should intersect $P_3'P_3$ exactly in $P_1$.
Edit: Added some pictures (the parabola below is just an approximation, not a real parabola).
$hspace{50pt}$
$hspace{50pt}$
answered May 24 '12 at 14:07
dtldarekdtldarek
32.5k746102
$endgroup$
The set of points that are equidistant from the circle $C_2$ and the line $P_4P_5$ is a parabola. Translate the line $P_4P_5$ (in the opposite direction than the $P_2$) by the radius $r$ of the circle (i.e. $|P_4'P_4| = r$, $|P_5'P_5| = r$ and $P_4P_5 | P_4'P_5'$) to get a new line $P_4'P_5'$ and observe that your solution will be equidistant from the line $P_4'P_5'$ and the point $P_2$ (if before the distance was $b$, then now it is $b+r$). Precisely, the $P_2$ will be the focus and $P_4'P_5'$ will be the directrix. To find the point $P_1$ of parabola just construct the perpendicular bisector of $P_3'P_2$, it should intersect $P_3'P_3$ exactly in $P_1$.
Edit: Added some pictures (the parabola below is just an approximation, not a real parabola).
$hspace{50pt}$
$hspace{50pt}$
answered May 24 '12 at 14:07
dtldarekdtldarek
32.5k746102
edited May 24 '12 at 14:53
answered May 24 '12 at 14:07
dtldarekdtldarek
32.5k746102
answered May 24 '12 at 14:07
dtldarekdtldarek
32.5k746102
answered May 24 '12 at 14:07
dtldarekdtldarek
32.5k746102
32.5k746102
$begingroup$
Thanks. Translating the line makes everything easy. Translating it the other way gives the second solution.
$endgroup$
– Les Newell
May 24 '12 at 14:43
$begingroup$
@LesNewell Yes, exactly ;-) Just that this direction of translation is easier to imagine (and so was my choice).
$endgroup$
– dtldarek
May 24 '12 at 14:55
$begingroup$
Thank you for your detailed help. It would be nice to try to calculate the parabola directly but things get complicated as the circle is actually just an arc and it is joined to other arcs/lines.
$endgroup$
– Les Newell
May 24 '12 at 16:09
add a comment |
$begingroup$
Thanks. Translating the line makes everything easy. Translating it the other way gives the second solution.
$endgroup$
– Les Newell
May 24 '12 at 14:43
$begingroup$
@LesNewell Yes, exactly ;-) Just that this direction of translation is easier to imagine (and so was my choice).
$endgroup$
– dtldarek
May 24 '12 at 14:55
$begingroup$
Thank you for your detailed help. It would be nice to try to calculate the parabola directly but things get complicated as the circle is actually just an arc and it is joined to other arcs/lines.
$endgroup$
– Les Newell
May 24 '12 at 16:09
$begingroup$
Thanks. Translating the line makes everything easy. Translating it the other way gives the second solution.
$endgroup$
– Les Newell
May 24 '12 at 14:43
$begingroup$
Thanks. Translating the line makes everything easy. Translating it the other way gives the second solution.
$endgroup$
– Les Newell
May 24 '12 at 14:43
$begingroup$
@LesNewell Yes, exactly ;-) Just that this direction of translation is easier to imagine (and so was my choice).
$endgroup$
– dtldarek
May 24 '12 at 14:55
$begingroup$
@LesNewell Yes, exactly ;-) Just that this direction of translation is easier to imagine (and so was my choice).
$endgroup$
– dtldarek
May 24 '12 at 14:55
$begingroup$
Thank you for your detailed help. It would be nice to try to calculate the parabola directly but things get complicated as the circle is actually just an arc and it is joined to other arcs/lines.
$endgroup$
– Les Newell
May 24 '12 at 16:09
$begingroup$
Thank you for your detailed help. It would be nice to try to calculate the parabola directly but things get complicated as the circle is actually just an arc and it is joined to other arcs/lines.
$endgroup$
– Les Newell
May 24 '12 at 16:09
add a comment |
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