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INTRODUCTION
There are three relationships within coordinate geometry that can exist between a line and a circle. They can intersect at one or two points or they will not intersect. If there is only one intersection then the line must be tangent to the circle at that point of intersection. To find where the intersection exists, one can use either the solution of two equations or by the triangle method.
To solve the problem in a simple mode, one can solve for the equation of a line and the equation of the circle simultaneously. Since the equation of a circle is a second-degree equation, the solution is through the quadratic equation. The development of the solution by simultaneous equations begins by writing the equation of the line and circle in the following forms (Hashimi, 1988):
( ) ( )
( ) ( )
y y m x x y m x x y
R x h y k
1 1 1 1 2 2 2
where: x and y are the coordinates of the point of intersection x 1 and y 1 are the coordinates of a point on the line h and k are the translation between the center of the circle and the origin of the coordinate system in the x and y directions respectively. m is the slope of the line.
As a reminder, the slope is the cotangent of the azimuth of the line. Write the equation of the line and expand the equation by taking the squares and equating the equation to zero.
( x h) ( y k) R
x hx h y ky k R
(^2 2 )
2 2 2 2 2
LINE-CIRCLE INTERSECTION
Surveying Engineering Department Ferris State University
Substitute into this equation the relationship developed for the y-coordinate above, yielding:
x hx h (^) [ m x ( x (^) ) y (^) ] k m x[ ( x (^) ) y (^) ] k R x hx h m x m xx mxy m x mx y y kmx kmx ky k R
2 2 1 12 1 1 2 2 2 2 2 2 2 1 1 2 12 1 1 2 1 1 2 2
Recall that the quadratic e quation has the following form:
aX 2 + bX + c= 0 (3)
Substitute (2) into (3) and factoring results in
(^1 ) 2 ( ) 2 2 2 0
2 2 2 1 1 2 2 2 2 12 2 1 1 1 1
m x h m x my mk x h k y m x R mx y kmx ky
The solution is
x = −^ b^ ±^ ba^ − ac
where:
( )
a m b h m x my mk c h k y m x R mx y kmx ky
2 (^21 ) (^2 2 2 212 21 1 1 )
Example: If the offset to the center of the circle with a radius of 2 units are defined a 3 units in the x-direction and 7 units in the y-direction (i.e., h = 3 and k = 7) and if a line passing through point (1,4) has an azimuth of 62 o^ 11’ 40”, what are the coordinates of the intersection of this line with the circle (Hashimi, 1988).
Solution: Recall that the slope (m) is defined by the cotangent of the azimuth (m = cot Az) therefore the parameters a, b, and c can be found using the relationships shown in (6).
a 1 cot^2 Az 1 cot^262 o 11 ' 40 "
The distance AO is found using the Pythagorean Theorem
( ) ( ) ( ) ( )
D AO = X O − X A + YO −YA
2 2 3 1 2 7 42 3 606.
The azimuth between A and O is found to be:
AZ (^) AO XY^ O XYA O A o
^
^
tan− tan− ' "
The azimuth from A to P 1 is given as 62o^ 11’ 40”. Also, the azimuth from A to P 2 is equal to the azimuth from A to P 1. Therefore, the angle at point A (θ) is the difference between the azimuths along the line and the azimuth from A to O.
θ = (^) ( ) −( )
o o o
Also, from the problem statement, we know that DOP1 = DOP2 = Radius of the circle = 2. Then, from the sine law given as
D (^) OP 1 D (^) AO DAO sin θ =^ sin ϕ =sinγ
These relationships can be rearranged and relationships for ϕ and γ can be developed.
ϕ = θ
sin −^ sin^ sin− sin^ '^ "^. ' "
1 1 1
D OP DAO
o
o
γ = θ
sin −^ sin^ sin− sin^ '^ "^. ' "
1 1 1
D OP DAO
o
o
With ϕ and γ known, α and β can be found using the relationship that a triangle contains 180 o.
( )
( )
α α
β β
o o o o
p o o o
Then, compute the distance from A to P 1 and P 2 by using the relationships from the sine law:
D (^) AP 1 D (^) AO D (^) AP 2 DAO sin α =^ sin ϕ ;^ sin β =sinγ
from which the distances are found as
D D D
D D
D
AP AO o o AP
AP AO o o AP
1 1
2 2
sin sin^
sin ' " sin^ '^ " .
sin sin^
sin ' " sin^ '^ " .
ϕ α
γ β
The X and Y coordinates of the points of intersection can be computed as normal.
X X D AZ Y Y D AZ
X X D AZ Y Y D AZ
P A AP AP O P A AP AP O
P A AP AP O P A AP AP O
1 1 1 1 1 1
12 2 1 2 2 1
sin. sin ' ". cos. cos ' ".
sin. sin ' ". cos. cos ' ".
Check by inversing between the points of intersection (P 1 and P 2 ) and the center of the circle (O). These values should equal the radius of the circle, which is known.
Line-Circle Intersection
Given the following data, compute the coordinates of point P Coordinates of the PI Coordinates of the center of the circle (O) Arc length from the PC to point P Central angle from the PC to point P Coordinates of point 1 on the line and the point on curve (PC): X1 5946.976 Y1 4801. XPC 5429.259 YPC 4447. Azimuth from point 1 to the point of intersection: AZ_1P 231.3121 (In ddd.mmss format)^ a12 floor AZ_1P( ) amn ( AZ_1P a12) 100. A1P a12 (^ floor amn 60 (^ )) ( amn floor amn( )). 3600100 A1P = 231.5225 (in decimal degrees format) Forward azimuth of the back tangent of the curve: AZ_BT 220.2508(In ddd.mmss format)^ a12 floor AZ_BT( ) amn ( AZ_BT a12) 100. ABT a12 (^ floor amn 60 (^ )) ( amn floor amn( )). 3600100 ABT = 220.418889 (in decimal degrees format) Radius of curve: (^) Rad 598. Central angle of the curve (∆): CA 74.3917 (In ddd.mmss format)^ a12 floor CA( ) amn ( CA a12) 100. ∆ a12 (^ floor amn 60 (^ )) ( amn floor amn( )). 3600100 ∆ = 74.655 (in decimal degrees format) rd (^180) π
SOLUTION Tangent Distance: (^) T Rad tan. (^) 2 rd∆. T =456. Azimuth from PC to Center of Curve: AZpco ABT 90 AZpco =130.
Coordinates of center of circle XO XPC Rad sin^.^ AZpcord YO YPC Rad cos^. AZpcord XO = 5.885063310^3 YO =4.059687 10^3
Coordinates of the PI: XPI XPC T sin^.^ ABTrd YPI XPC T cos ABT rd . XPI = 5.133252810^3 YPI =5.081685110^3 num XO X den YO Y
The azimuth between point 1 on the line and the center of the circle is computed using the arc tangent function as
AZ atan numden AZ =0. AZ1O AZ π α A1P AZ1O.^180 π α =46. D1O ( XO X1) 2 ( YO Y1) 2 D1O =744. β asin
sin (^) rdα Rad D1O From the sine law, the the. β can be computed as: β =1. Then, the angle (^) γ is found by subtracting the sum of the other two angles in the triangle, α and^ β γ 180 α β.^180 π γ =68. The azimuth between the center of the circle (O) and the point of intersection (P) is: AZOP ( AZ1O π) γ.^180 π 2 .π AZOP =5. The coordinates of the point of intersection are now computed as normal:
XP XO Rad sin AZOP.^ ( ) YP YO Rad cos AZOP. ( ) XP = 5.348687110^3 YP =4.32565910^3
Example 2: Given: X 1 = 5313.674 Y 1 = 4200. XPI = 4977.455 YPI = 3951. Distance from 1 to point P = 327.387’ Forward Azimuth of Back Tangent = 137o^ 54’ 22” Radius of the circular curve = 819.524’ Central angle (∆) = 48o^ 39’ 53”
Compute: - The coordinates of point P
- Coordinates of the PC
- Coordinates of the center of the circle (O)
- Arc length from the PC to point P
- Distance and azimuth from the PC to P, and
- Central angle from the PC to point P
Solution:
Solution to parts of the horizontal circle:
T R
o = = =
tan. tan '^ "
. '
2 819 524^
X X T AZ
X
Y Y T AZ
Y
PC PI PI PC o PC
PC PI PI PC p PC
−
−
sin. '. sin ' "
. '
cos. '. cos ' "
. '
AZ (^) PC O AZPC PI o^ ( o^ ) o o
− =^ − −^ =^ −
X X R AZ
X
Y Y R AZ
Y
O PC PC O o O
O PC PC O o O
−
−
sin. '. sin ' "
. '
cos. '. cos ' "
. '
( ) ( ) ( ) ( )
D O − = X − X O + Y −YO
1 1 2 1 2 5313 674 5337 159 2 4200 812 47758082 575 475
( ) ( ) cos.^..^.. ' "
α α
= +^ −^ = +^ −
− − − − −
D D D
D D
O P O P O P O o
2 21 12 1
2 2 2 2
( ) ( )
AZ XY^ XY
AZ
AZ AZ
AZ
O OO O o
O P O o^ o O P o
− −^ − −
− − −
^
1 1 11 1 1
1
tan tan ..^ .. ' "
' " ' " ' "
α
X X D AZ
X
Y Y D AZ
Y
P O O P O P o P
P O O P O P o P
− −
− −
sin.. sin ' "
. '
cos.. cos ' "
. '
