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I need to find the equation for a circle (mainly its radius) which is tangent to the following three lines:

$y = 0$

$y = \tan(70)x$

$y = -1.428148x + 0.790201$

For the last tangent line equation, I know that it is tangent at the point $(0.371272, 0.259968)$ However, for the other two I do not know the exact point of tangency, only that the circle is tangent to them.

enter image description here

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  • $\begingroup$ And you are definitely given that the third tangent point is along the angle bisector between $y=0$ and $y=\tan(70^\circ)x$? $\endgroup$ – abiessu Jan 29 '14 at 21:42
  • $\begingroup$ Yes, the angle is 70 degrees from the horizontal. $\endgroup$ – Hiero Jan 29 '14 at 23:04
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Hint : Find the point

(u / 0.700209 u )

(0.700209 is the slope of the line orthogonal to the third line)

which has distance 0.700209 u from the tangency point.

Then, the radius of the circle is 0.700209 u.

The equation, you have to solve is :

$$\sqrt{(u - 0.371272)^2 + (0.700209u - 0.259968)^2} = 0.700209u$$

which gives u = 0.870666.

Then r = 0.700209 * 0.870666 = 0.609648.

The other solution for u corresponds with the circle inside the triangle, so you have to choose the larger u to get the outside circle.

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  • $\begingroup$ I'm not sure how to incorporate (u,0.700209 u) into the solution since there are so many unknowns. It appears that the only way to go about this is to build a system of equations, like ccorn denoted in the reply below. $\endgroup$ – Hiero Jan 30 '14 at 14:02
  • $\begingroup$ The line between the tangency point and the midpoint of the circle must be orthogonal to the third tangent and contain the origin. So, if the x-value of the midpoint of the circle is known, the y-value is also known because the midpoint must lie on the described line. $\endgroup$ – Peter Jan 30 '14 at 14:19
  • $\begingroup$ Since the tangency point is known, the distance from this point can be calculated depending on u. The distance must be equal to the y-value of the midpoint of the circle. So you get an equation with only one unknown, namely u. $\endgroup$ – Peter Jan 30 '14 at 14:22
  • $\begingroup$ I tried to modify the distance formula with your hint in mind... My equation is: r=((r-0.371272)^2+(0.700209*r-0.259968)^2)^0.5 after solving the equation I got r = 0.20409 and 2.05294 which are both incorrect. Did I set it up incorrectly or should I be going about this from a different angle? $\endgroup$ – Hiero Jan 30 '14 at 19:48
  • $\begingroup$ I get r = 0.609648, is that right ? $\endgroup$ – Peter Jan 30 '14 at 21:45
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Given three lines, find a tangent circle: This is like finding the incircle of a triangle.

Pick two lines, construct an angle-bisecting line. Pick another pair of the original lines, construct another angle-bisecting line. The intersection of the bisecting lines gives the center of the circle. The rest is left to you.

However, an angle-bisecting line can be rotated $90$ degrees to give another valid bisecting line. Therefore several solutions (in this case four) are possible. The figure below shows an example. In your case, what you are looking for seems to be the point labeled $F$.

enter image description here

Update: Here is an algebraic solution.

Note that for a line $g_i$ (here $i\in\{1,2,3\}$) described by the normal form $$\begin{align} g_i: n_{i,x} x + n_{i,y} y + c_i &= 0 & n_{i,x}^2 + n_{i,y}^2 &= 1 \end{align}$$ the oriented (i. e. signed) distance of a point $C=(x_{\text{C}},y_{\text{C}})$ to the line $g_i$ is given by $$d(C,g_i) = n_{i,x} x_{\text{C}} + n_{i,y} y_{\text{C}} + c_i$$

Choose $C$ to be the center of any circle tangent to $g_1$, $g_2$, $g_3$. Then $C$ clearly fulfills $$\begin{align} d(C,g_1)^2 - d(C,g_2)^2 &= 0 \\ d(C,g_1)^2 - d(C,g_3)^2 &= 0 \end{align}$$ That is, $$\begin{align} \text{one of}\quad d(C,g_1) \pm d(C,g_2) &= 0 \\\text{and}\quad \text{one of}\quad d(C,g_1) \pm d(C,g_3) &= 0 \end{align}$$ The combination of the two $\pm$ allows $4$ solutions, which is as it should be. Suppose we specify the lines $g_i$ such that the first nonzero component of $(n_{i,x},n_{i,y})$ is positive. Then, in your case, the center that you are looking for has positive distance from all $g_i$, so all signed distances are equal, therefore we have to choose all the above $\pm$ to be $-$. This gives a linear system of equations for the coordinates of $C$: $$\begin{pmatrix} n_{1,x}-n_{2,x} & n_{1,y}-n_{2,y} \\ n_{1,x}-n_{3,x} & n_{1,y}-n_{3,y} \end{pmatrix} \begin{pmatrix}x_{\text{C}}\\y_{\text{C}}\end{pmatrix} = \begin{pmatrix}c_2-c_1\\c_3-c_1\end{pmatrix}$$ This determines $(x_{\text{C}},y_{\text{C}})$. The radius $r$ of the tangent circle is then given as the distance of $C$ to any of the lines: $$r = d(C,g_i)$$ I leave it to you to fill in the numbers.

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Given 4 points. p1,p2,p3,p4

  1. The converging point of the bisector angles for p2 and p3 gives you the centre of the circle.
  2. To find the radius: Project the orthogonal normal from the centre onto the line (p2,p3)
  3. Draw the circle with the centre and the radius

(You can use a compass and a ruler to test this theory out)

enter image description here

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  • $\begingroup$ Could you clarify on which points being $P_1,P_2,P_3$ and $P_4$? Also, what do you mean by a converging point for a bisector angle? $\endgroup$ – Frenzy Li Aug 29 '16 at 15:11
  • $\begingroup$ Converging point = where two trajectories meet. bisectorAngle = mathworld.wolfram.com/AngleBisector.html. The points p1,p2,p3,p4 makes up the 3 lines described in OP $\endgroup$ – eonist Aug 29 '16 at 16:12
  • $\begingroup$ I see your point now. A better statement would be the (point of) intersection of the two bisectors. Nice answer. $\endgroup$ – Frenzy Li Aug 30 '16 at 3:55
  • $\begingroup$ It's not the intersection of two lines, intersections are heavier/more un-precise to calculate than two converging rays. we can use the "law of sin" to calculate where the rays met. In this case b/SinB = c/SinC. Anyways I added an illustration to help explain the process. $\endgroup$ – eonist Aug 30 '16 at 16:22
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Given the three tangent line equations: y=0 y=tan(70)x y=−1.428148x+0.790201

Using: r=|Ax+By+C|/(A^2+B^2)^0.5

r = k r = (-tan(70)*h+k)/(2.74748^2+1^2)^0.5 r = (1.428148*h+k-0.7900201)/(1.428148^2+1^2)^0.5

Solving the above system of equations nets 4 possible solutions.

Correct solution: h=0.870667, k=0.609648

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