Find a circle's radius with three known tangent lines 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.

 A: 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 $F$.

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.
A: Given 4 points. p1,p2,p3,p4


*

*The converging point of the bisector angles for p2 and p3  gives you the centre of the circle.  

*To find the radius: Project the orthogonal normal from the centre onto the line (p2,p3)

*Draw the circle with the centre and the radius


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

A: 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.
A: 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
