How to find intersection points of circle and line given only a radius and end points
$begingroup$
I need to find the point of intersection between a line and a circle. here is the information i have
1. current Location.
2. destination
3. radius of a circle that the destination resides in. Let me draw it out so you can see it as well as show you what it would look like on google maps:
here is my drawing of what i need :
see the question marks in red. that is the point of intersection. i need that point. so (x,y) & (a,b) are given. i also have the radius for the circle which is 8.
my coordinate system looks like this:
what i have tried is tangent . here is what i came up with:
tan @ = (b-y)/(x-b) since that would give me the tangent between the line orange line below and then a invisible triangle i drew. but im a little lost from here and could use some help. In the end i want points (?,?) of the intersection.
linear-algebra intersection-theory
asked Jan 3 at 11:58
j2emanuej2emanue
1064
$endgroup$
add a comment |
$begingroup$
I need to find the point of intersection between a line and a circle. here is the information i have
1. current Location.
2. destination
3. radius of a circle that the destination resides in. Let me draw it out so you can see it as well as show you what it would look like on google maps:
here is my drawing of what i need :
see the question marks in red. that is the point of intersection. i need that point. so (x,y) & (a,b) are given. i also have the radius for the circle which is 8.
my coordinate system looks like this:
what i have tried is tangent . here is what i came up with:
tan @ = (b-y)/(x-b) since that would give me the tangent between the line orange line below and then a invisible triangle i drew. but im a little lost from here and could use some help. In the end i want points (?,?) of the intersection.
linear-algebra intersection-theory
asked Jan 3 at 11:58
j2emanuej2emanue
1064
$endgroup$
$begingroup$
So you have set up the coordinate system. Do you know the equation of the circle? When you have the equation of the circle and the line, you can pair them up and solve the system numerically.
$endgroup$
– Matti P.
Jan 3 at 12:09
add a comment |
$begingroup$
I need to find the point of intersection between a line and a circle. here is the information i have
1. current Location.
2. destination
3. radius of a circle that the destination resides in. Let me draw it out so you can see it as well as show you what it would look like on google maps:
here is my drawing of what i need :
see the question marks in red. that is the point of intersection. i need that point. so (x,y) & (a,b) are given. i also have the radius for the circle which is 8.
my coordinate system looks like this:
what i have tried is tangent . here is what i came up with:
tan @ = (b-y)/(x-b) since that would give me the tangent between the line orange line below and then a invisible triangle i drew. but im a little lost from here and could use some help. In the end i want points (?,?) of the intersection.
linear-algebra intersection-theory
asked Jan 3 at 11:58
j2emanuej2emanue
1064
$endgroup$
I need to find the point of intersection between a line and a circle. here is the information i have
1. current Location.
2. destination
3. radius of a circle that the destination resides in. Let me draw it out so you can see it as well as show you what it would look like on google maps:
here is my drawing of what i need :
see the question marks in red. that is the point of intersection. i need that point. so (x,y) & (a,b) are given. i also have the radius for the circle which is 8.
my coordinate system looks like this:
what i have tried is tangent . here is what i came up with:
tan @ = (b-y)/(x-b) since that would give me the tangent between the line orange line below and then a invisible triangle i drew. but im a little lost from here and could use some help. In the end i want points (?,?) of the intersection.
linear-algebra intersection-theory
linear-algebra intersection-theory
asked Jan 3 at 11:58
j2emanuej2emanue
1064
asked Jan 3 at 11:58
j2emanuej2emanue
1064
edited Jan 4 at 3:26
j2emanue
asked Jan 3 at 11:58
j2emanuej2emanue
1064
asked Jan 3 at 11:58
j2emanuej2emanue
1064
asked Jan 3 at 11:58
j2emanuej2emanue
1064
1064
$begingroup$
So you have set up the coordinate system. Do you know the equation of the circle? When you have the equation of the circle and the line, you can pair them up and solve the system numerically.
$endgroup$
– Matti P.
Jan 3 at 12:09
add a comment |
$begingroup$
So you have set up the coordinate system. Do you know the equation of the circle? When you have the equation of the circle and the line, you can pair them up and solve the system numerically.
$endgroup$
– Matti P.
Jan 3 at 12:09
$begingroup$
So you have set up the coordinate system. Do you know the equation of the circle? When you have the equation of the circle and the line, you can pair them up and solve the system numerically.
$endgroup$
– Matti P.
Jan 3 at 12:09
$begingroup$
So you have set up the coordinate system. Do you know the equation of the circle? When you have the equation of the circle and the line, you can pair them up and solve the system numerically.
$endgroup$
– Matti P.
Jan 3 at 12:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So give two points $(a, b)$ and $(x, y)$ and a radius $r$, we want a point on the segment between these two points which lies a distance of $r$ away from $(a, b)$.
The line segment is parametrised by
$$
p(t) = (a, b) + t(x-a, y-b)
$$
for $t$ between $0$ and $1$. In other words, for each value of $t$ you get a point on the segment, and each point on the segment corresponds to a value of $t$. We are intereste in the point which lies $r$ away from the point $(a, b)$, which by the Pythagorean theorem means
$$
r = |p(t) - (a, b)|\
= |t(x-a, y-b)|\
= tsqrt{(x-a)^2 + (y-b)^2}\
t = frac{r}{sqrt{(x-a)^2 + (y-b)^2}}
$$
Inserting that $t$ into the formula for $p(t)$ gives you the exact coordinates of your point:
$$
pleft(frac{r}{sqrt{(x-a)^2 + (y-b)^2}}right) = (a, b) + frac{r}{sqrt{(x-a)^2 + (y-b)^2}}(x-a, y-b)
$$
Edit for example
Say $(x, y) = (11, 12)$ and $(a, b) = (2, 5)$, with $r = 8$. In that case, we get
$$
t = frac{8}{sqrt{(11-2)^2 + (12-5)^2}} = frac{8}{sqrt{81 + 49}} approx 0.702
$$
which gives us
$$
p(t) approx p(0.702) = (2, 5) + 0.702(11-2,12-5)\
approx (2, 5) + (6.315, 4.912)\
= (8.315, 9.912)
$$
answered Jan 3 at 12:10
ArthurArthur
113k7113195
$endgroup$
$begingroup$
your producing to me a final result of p(t). i was looking for a way to get the points. can you show me how to use your formula ? lets say i have (11,12) being x,y and (13,14) and radius of 8 how would i plug those into your formula ?
$endgroup$
– j2emanue
Jan 3 at 12:18
$begingroup$
@j2emanue Those points are way too close to one another, and they would both be inside the circle. But i'll make an example.
$endgroup$
– Arthur
Jan 3 at 12:20
add a comment |
Your Answer
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
So give two points $(a, b)$ and $(x, y)$ and a radius $r$, we want a point on the segment between these two points which lies a distance of $r$ away from $(a, b)$.
The line segment is parametrised by
$$
p(t) = (a, b) + t(x-a, y-b)
$$
for $t$ between $0$ and $1$. In other words, for each value of $t$ you get a point on the segment, and each point on the segment corresponds to a value of $t$. We are intereste in the point which lies $r$ away from the point $(a, b)$, which by the Pythagorean theorem means
$$
r = |p(t) - (a, b)|\
= |t(x-a, y-b)|\
= tsqrt{(x-a)^2 + (y-b)^2}\
t = frac{r}{sqrt{(x-a)^2 + (y-b)^2}}
$$
Inserting that $t$ into the formula for $p(t)$ gives you the exact coordinates of your point:
$$
pleft(frac{r}{sqrt{(x-a)^2 + (y-b)^2}}right) = (a, b) + frac{r}{sqrt{(x-a)^2 + (y-b)^2}}(x-a, y-b)
$$
Edit for example
Say $(x, y) = (11, 12)$ and $(a, b) = (2, 5)$, with $r = 8$. In that case, we get
$$
t = frac{8}{sqrt{(11-2)^2 + (12-5)^2}} = frac{8}{sqrt{81 + 49}} approx 0.702
$$
which gives us
$$
p(t) approx p(0.702) = (2, 5) + 0.702(11-2,12-5)\
approx (2, 5) + (6.315, 4.912)\
= (8.315, 9.912)
$$
answered Jan 3 at 12:10
ArthurArthur
113k7113195
$endgroup$
$begingroup$
your producing to me a final result of p(t). i was looking for a way to get the points. can you show me how to use your formula ? lets say i have (11,12) being x,y and (13,14) and radius of 8 how would i plug those into your formula ?
$endgroup$
– j2emanue
Jan 3 at 12:18
$begingroup$
@j2emanue Those points are way too close to one another, and they would both be inside the circle. But i'll make an example.
$endgroup$
– Arthur
Jan 3 at 12:20
add a comment |
$begingroup$
So give two points $(a, b)$ and $(x, y)$ and a radius $r$, we want a point on the segment between these two points which lies a distance of $r$ away from $(a, b)$.
The line segment is parametrised by
$$
p(t) = (a, b) + t(x-a, y-b)
$$
for $t$ between $0$ and $1$. In other words, for each value of $t$ you get a point on the segment, and each point on the segment corresponds to a value of $t$. We are intereste in the point which lies $r$ away from the point $(a, b)$, which by the Pythagorean theorem means
$$
r = |p(t) - (a, b)|\
= |t(x-a, y-b)|\
= tsqrt{(x-a)^2 + (y-b)^2}\
t = frac{r}{sqrt{(x-a)^2 + (y-b)^2}}
$$
Inserting that $t$ into the formula for $p(t)$ gives you the exact coordinates of your point:
$$
pleft(frac{r}{sqrt{(x-a)^2 + (y-b)^2}}right) = (a, b) + frac{r}{sqrt{(x-a)^2 + (y-b)^2}}(x-a, y-b)
$$
Edit for example
Say $(x, y) = (11, 12)$ and $(a, b) = (2, 5)$, with $r = 8$. In that case, we get
$$
t = frac{8}{sqrt{(11-2)^2 + (12-5)^2}} = frac{8}{sqrt{81 + 49}} approx 0.702
$$
which gives us
$$
p(t) approx p(0.702) = (2, 5) + 0.702(11-2,12-5)\
approx (2, 5) + (6.315, 4.912)\
= (8.315, 9.912)
$$
answered Jan 3 at 12:10
ArthurArthur
113k7113195
$endgroup$
$begingroup$
your producing to me a final result of p(t). i was looking for a way to get the points. can you show me how to use your formula ? lets say i have (11,12) being x,y and (13,14) and radius of 8 how would i plug those into your formula ?
$endgroup$
– j2emanue
Jan 3 at 12:18
$begingroup$
@j2emanue Those points are way too close to one another, and they would both be inside the circle. But i'll make an example.
$endgroup$
– Arthur
Jan 3 at 12:20
add a comment |
$begingroup$
So give two points $(a, b)$ and $(x, y)$ and a radius $r$, we want a point on the segment between these two points which lies a distance of $r$ away from $(a, b)$.
The line segment is parametrised by
$$
p(t) = (a, b) + t(x-a, y-b)
$$
for $t$ between $0$ and $1$. In other words, for each value of $t$ you get a point on the segment, and each point on the segment corresponds to a value of $t$. We are intereste in the point which lies $r$ away from the point $(a, b)$, which by the Pythagorean theorem means
$$
r = |p(t) - (a, b)|\
= |t(x-a, y-b)|\
= tsqrt{(x-a)^2 + (y-b)^2}\
t = frac{r}{sqrt{(x-a)^2 + (y-b)^2}}
$$
Inserting that $t$ into the formula for $p(t)$ gives you the exact coordinates of your point:
$$
pleft(frac{r}{sqrt{(x-a)^2 + (y-b)^2}}right) = (a, b) + frac{r}{sqrt{(x-a)^2 + (y-b)^2}}(x-a, y-b)
$$
Edit for example
Say $(x, y) = (11, 12)$ and $(a, b) = (2, 5)$, with $r = 8$. In that case, we get
$$
t = frac{8}{sqrt{(11-2)^2 + (12-5)^2}} = frac{8}{sqrt{81 + 49}} approx 0.702
$$
which gives us
$$
p(t) approx p(0.702) = (2, 5) + 0.702(11-2,12-5)\
approx (2, 5) + (6.315, 4.912)\
= (8.315, 9.912)
$$
answered Jan 3 at 12:10
ArthurArthur
113k7113195
$endgroup$
So give two points $(a, b)$ and $(x, y)$ and a radius $r$, we want a point on the segment between these two points which lies a distance of $r$ away from $(a, b)$.
The line segment is parametrised by
$$
p(t) = (a, b) + t(x-a, y-b)
$$
for $t$ between $0$ and $1$. In other words, for each value of $t$ you get a point on the segment, and each point on the segment corresponds to a value of $t$. We are intereste in the point which lies $r$ away from the point $(a, b)$, which by the Pythagorean theorem means
$$
r = |p(t) - (a, b)|\
= |t(x-a, y-b)|\
= tsqrt{(x-a)^2 + (y-b)^2}\
t = frac{r}{sqrt{(x-a)^2 + (y-b)^2}}
$$
Inserting that $t$ into the formula for $p(t)$ gives you the exact coordinates of your point:
$$
pleft(frac{r}{sqrt{(x-a)^2 + (y-b)^2}}right) = (a, b) + frac{r}{sqrt{(x-a)^2 + (y-b)^2}}(x-a, y-b)
$$
Edit for example
Say $(x, y) = (11, 12)$ and $(a, b) = (2, 5)$, with $r = 8$. In that case, we get
$$
t = frac{8}{sqrt{(11-2)^2 + (12-5)^2}} = frac{8}{sqrt{81 + 49}} approx 0.702
$$
which gives us
$$
p(t) approx p(0.702) = (2, 5) + 0.702(11-2,12-5)\
approx (2, 5) + (6.315, 4.912)\
= (8.315, 9.912)
$$
answered Jan 3 at 12:10
ArthurArthur
113k7113195
edited Jan 3 at 12:25
answered Jan 3 at 12:10
ArthurArthur
113k7113195
answered Jan 3 at 12:10
ArthurArthur
113k7113195
answered Jan 3 at 12:10
ArthurArthur
113k7113195
113k7113195
$begingroup$
your producing to me a final result of p(t). i was looking for a way to get the points. can you show me how to use your formula ? lets say i have (11,12) being x,y and (13,14) and radius of 8 how would i plug those into your formula ?
$endgroup$
– j2emanue
Jan 3 at 12:18
$begingroup$
@j2emanue Those points are way too close to one another, and they would both be inside the circle. But i'll make an example.
$endgroup$
– Arthur
Jan 3 at 12:20
add a comment |
$begingroup$
your producing to me a final result of p(t). i was looking for a way to get the points. can you show me how to use your formula ? lets say i have (11,12) being x,y and (13,14) and radius of 8 how would i plug those into your formula ?
$endgroup$
– j2emanue
Jan 3 at 12:18
$begingroup$
@j2emanue Those points are way too close to one another, and they would both be inside the circle. But i'll make an example.
$endgroup$
– Arthur
Jan 3 at 12:20
$begingroup$
your producing to me a final result of p(t). i was looking for a way to get the points. can you show me how to use your formula ? lets say i have (11,12) being x,y and (13,14) and radius of 8 how would i plug those into your formula ?
$endgroup$
– j2emanue
Jan 3 at 12:18
$begingroup$
your producing to me a final result of p(t). i was looking for a way to get the points. can you show me how to use your formula ? lets say i have (11,12) being x,y and (13,14) and radius of 8 how would i plug those into your formula ?
$endgroup$
– j2emanue
Jan 3 at 12:18
$begingroup$
@j2emanue Those points are way too close to one another, and they would both be inside the circle. But i'll make an example.
$endgroup$
– Arthur
Jan 3 at 12:20
$begingroup$
@j2emanue Those points are way too close to one another, and they would both be inside the circle. But i'll make an example.
$endgroup$
– Arthur
Jan 3 at 12:20
add a comment |
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$begingroup$
So you have set up the coordinate system. Do you know the equation of the circle? When you have the equation of the circle and the line, you can pair them up and solve the system numerically.
$endgroup$
– Matti P.
Jan 3 at 12:09