A tangent circle element between 2 intersecting vectors 2 Vectors, which are originating from one point I. I want to the replace the sharp corner (I) with an arc (circle element) with a radius of r. The arc touches the vectors at T1 & T2.
What is the fastest/easiest way to calculate the center of the arc, and the vectors (I,T1) & (I,T2). The angle between the vectors is less than 180 degrees.

 A: You can figure out the angle Θ between your vectors. Let's treat the vertex I as being located at (0,0). Also let one vector be the x-axis, and the other vector be treated as a line y=Ax, where A is determined by the angle Θ. Let the circle of radius r be located above xo, where the circle touches the x axis. The line y=Ax touches a distance xo from the origin. The perpendicular from xo through the center of the circle touches line y at distance xo+r from the origin. This point is yo = (xo+r)*sin(Θ) above xo. Also, y = r + 2*r*cos(Θ). Set these heights equal to each other.
(xo+r)sin(Θ) = r(1+2*cos(Θ))
Now solve for xo.
(xo+r)/r = (1+2*cos(Θ)) / sin(Θ)
xo/r = (1+2*cos(Θ)) / sin(Θ) - 1
xo = r * [(1+2*cos(Θ)) / sin(Θ) - 1]
A: Initially choose unit normal vectors $n_1$ and $n_2$ for the two vectors going from $I$ to the two (not yet determined) points $T_1,T_2.$ Let $C$ be the center of the desired circle. The distance $d(I,T_1)=d(I,T_2)$ can be found using trig on the right triangle $CT_1I$ with its right angle at $T_1$ and the angle at $I$ of $\frac\theta 2.$ So that distance is $r \cot \frac \theta 2.$
Next, $T_1$ as a point can be expressed as $I + (r \cot \frac \theta 2) n_1.$ $T_2$ has the same expression only using $n_2$ rather than $n_1.$
For the last thing, to get the center point $C$, we can initially move from $I$ to $T_1,$ and then turn 90 degrees counterclockwise (the "positive direction" for angles) and then move $r$ units. To get the result of turning the unit vector $n_1$ by 90 degrees, if $n_1=(a,b)$ then the vector $n_1*=(b,-a)$ is the vector $n_1$ turned by 90 degrees. Then we get the formula for $C$ as $C=T_1+r\ n*.$
A: Let's try and do this without any trigonometry.
Instead of considering vectors, I'm going to consider two lines going through the origin.
The centre of the required circle of radius r is found at the intersection of two lines  parallel to the two given lines through the origin, such that the parallel lines are at distance r from the given lines.
If one of our given lines is y=ax, we need to find the line y=ax+c at distance r from y=ax.
The line perpendicular to y=ax passing through the origin is -ay=x or y=-x/a.
The required parallel line is tangent to the circle y^2+x^2=r^2 (which, as it is centred on the origin, is not the circle we are looking for.) The point where the tangent touches the circle satisfies both the equations -ay=x and y^2+x^2=r^2. Substituting we have:
y^2+(-ay)^2=r^2                  and        (-x/a)^2+x^2=r^2
                              therefore
        y^2=r^2/(1+a^2)          and                x^2=r^2/(1+1/a^2)

Each equation has two solutions, positive/negative x and positive/negative y. Picking the correct one is left to the reader (note: only two of the four combinations will give a valid solution to the equation -ay=x.) 
Substituting these into the equation y=ax+c and solving gives the correct value of c.
Now do the same with your other line y=bx to give your line y=bx+d.
Finally, find the point that satisfies both equations y=ax+c and y=bx+d. That is, solve ax+c=bx+d. The solution is, x=(d-c)/(a-b). Now find y using one of the two line equations just mentioned.
That point is the centre of the required circle. There are four possible circles of radius r which touch your original two lines through the origin. The one you have found will depend on the signs chosen earlier (of which there were two valid possibilities for each given line, one on each side of it.)
