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Given the radius and its $2$ tangent lines and their point of intersection of a circle.

A similar question is How to calculate the two tangent points to a circle with radius R from two lines given by three points

But how do I find coordinates of the center of the circle. I believe there must be $2$ solutions—each side of the intersection point. Please let me know what I am doing wrong am here, or if I am missing something?

Thanks in advance. enter image description here

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Maybe its just me, but I don't fully get it. A picture might help... –  imranfat May 28 '13 at 19:39
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I am presuming that you are referring to a picture similar to the one in the post you referenced. Yes, there would be two such circles, but since the intersecting lines form vertical angles, the situation on both sides of the intersection point is the same, so you only need to find one circle and place the center of the other symmetrically about the intersection point of the lines. As to finding the center of the circle, it will fall on the line bisecting the angle between the lines. There are radii of the circle perpendicular to each tangent line, of which you know the length (continued) –  RecklessReckoner May 28 '13 at 19:45
    
so you know two right triangles with a common side and one leg of each has a length equal to the specified radius. This should permit you to locate the circle's center. –  RecklessReckoner May 28 '13 at 19:46
    
Thanks for the help, but one addition to the problem is - What if I don't know tangent points where circle and line meets, that too I need to determine. Once I get those points I can find the center using the logic you mentioned. Thanks again. –  Nitesh May 29 '13 at 3:39
    
Thank you guys, I found the solution. Writing C for the center of the circle and θ for the angle at the intersection point of tangents, denoted as P. Lets assume a point B, where line and circle meets. Note that CB/PB=tan(θ/2). Since CB=r, we have PB=rcot(θ/2). Now we have length of 2 sides of a right angle triangle, so we can easily use pythogoras theoram to compute third side. So this will be distance from the intersection point P to center C. This solution was originally given from math.stackexchange.com/users/409/blue –  Nitesh May 29 '13 at 5:02

3 Answers 3

You have equation of two tangents. STEP $1$:Draw perpendicular(normal) from centre to tangents it is equal to radius.since you have two tangents you will draw two perpendicular(normal).

STEP $2$ :so you will get two equation consists of coordinates of centre.

STEP $3$: Then solve those equation you will get coordinates of centre.If you dont know how to draw perpendicular let me know and give data I'll solve it.

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Center is unknown so how I will draw perpendicular from center to tangents ? –  Nitesh May 29 '13 at 3:45
    
@Nitesh you just take centre $(x_1,y_1)$ and then draw perpendicular.You will get one equation from one tangents then follow STEP 3 –  iostream007 May 29 '13 at 4:44
    
Hi, Thanks a lot for help. I found the solution. Writing C for the center of the circle and θ for the angle at the intersection point of tangents, denoted as P. Lets assume a point B, where line and circle meets. Note that CB/PB=tan(θ/2). Since CB=r, we have PB=rcot(θ/2). Now we have length of 2 sides of a right angle triangle, so we can easily use pythogoras theoram to compute third side. So this will be distance from the intersection point P to center C. This solution was originally given from math.stackexchange.com/users/409/blue –  Nitesh May 29 '13 at 5:15
    
but one thing i want to tell you that finding out one coordinate from distance formula is little bit messy because that will gives a quadratic eqn in x and y.If you have any data of this question give it i'll solve it it is very easy to find centre coordinates from my solution –  iostream007 May 29 '13 at 5:21

The question is tagged (geometry) not (analytic-geometry), so I assume you need to solve the problem by a construction, not calculation. Here are first four steps:

  1. Let your given lines be denoted as $a$ and $b$.
  2. Construct lines $f$ and $g$ on both sides of the given line $a$, parallel to $a$, at a distance $R$ from $a$.
  3. Similary construct lines $k$ and $l$, parallel to $b$, $R$ apart from it.
  4. Find four intersection points, say $P_1$ through $P_4$, of $f$ and $g$ with $k$ and $l$. These are center points of four circles with radius $R$, tangent to the two given lines.

Here additional question comes: are you given two lines, as the question states, or two line segments, as in the title? If there are two tangent lines, you're done at step 4. However if you have two line segments, you need to check, if the tangency points are in the segments. Then:

  1. For each $P_i$ draw two lines, one perpendicular to $a$ and the other one to $b$. Check if they meet $a$ and $b$, respectively, inside the given segments. If so, the $P_i$ is a center point of the circle sought. Otherwise discard it.

Note that the solution may contain from zero up to four circles, depending on the line segments configuration.

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But this also hints at the analytic solution, no? –  mojuba Jul 29 at 20:35
    
@mojuba Yes, the analytic solution should start from the same given data, test the same conditions and eventually find the same answer as a classic construction. –  CiaPan Aug 3 at 9:16

There is also another possibility to solve this problem.

  1. If you know the three points A, B, C (B is the crossing point) you can determine the directional vectors of the lines: v1 = A - B and v2 = C - B. It is better to make them unit vectors by dividing by the lengths.

  2. From the two directional vectors you can find the angle alpha between them from the dot product definition.

  3. Then you can find the distance from the middle point B to the centre of the circle O as |BO| = R / sin(alpha).

  4. and the coordinates of the centre O as: O = B + BO, where BO is a unit vector. In the same time you have enough data to find the distance from B to your tangent points and their coordinates.

This approach will work for both 2- and 3-dimensional rectangular CS.

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