Plotting an arc with no center point - a practical solution please! I need a mathematical solution to a very practical problem (laying a patio).
The attached will hopefully explain.
The center of the circle for the arc we wish to have is inaccessible (ie in the house behind walls). 
There are 2 fixed points the arc must intersect.
There is an existing arc drawn using these 2 fixed points, but this circle is too small - radius 539cm
The desired arc is more shallow so would have a greater radius.
I think I need 10 or so measurements from the center of this existing circle, to the new arc. 
These points can then be plotted on the ground and the dots joined up (obviously more than 10 will me more accurate, but 10 seems a sensible number) 
I am sure this is possible, but my mathematical knowledge is not good enough, sorry. Answers or a simple formula I can apply would be very much appreciated.

Anyone out there to help?
 A: This is not a direct answer, but (maybe) a practical solution for your problem:
usually, in order to drawing a circle, you can keep one end of a rope in the center of the circle and turn around with the other end. In this case this is not possible because the center of the circle is inside the house.
Following your picture, I've drawn this:

$C$ is the center of the circle inside the house, $A$ and $B$ are the 2 fixed points the arc (red colored) must intersect.
As you are not able to draw the red arc, I propose to draw its symmetric, the dotted arc, and then take its symmetric.
So partically I propose:
1-Draw the line $AB$
2-Find the symmetric of the center $C$ respect to the segment $AB$. We will call it $C'$
3- Take a rope of length $R$ equal to the radius of the circle. Call one person to help you (we will call it Bill). 
4-Fix one end of the rope in $C'$. Keep the other end then ask to Bill to walk with the rope along the segment $AB$, starting from $A$ to $B$. 
5-Clearly the rope will be longer, so you will be able to get the rope double. Mark the extremity of the doubled rope.
6- You will obtain the desired arc.
Remark: maybe there are some others barrier which prevent my solution too.
A: If you can measure tangent angles in the diagram the following could be of use in a scaled geometrical construction:
$$ R_{old}= \dfrac{r_2^2-r_1^2}{ 2(r_2 \sin \beta - r_1 \sin \alpha )}$$
Along perpendicular bisector of given points $(1,2)$ mark old center of circle $O$ with this radius and a new center $N$ with a new desired (increased/reduced) radius passing through $(1,2)$ on bisector. Sketch not to scale.
The above result for Irregular Pie is obtained by eliminating invariant semi-chord length $L$ from Circle property that product of two line segments is constant. 
$$ r\cdot (2 R \sin \theta -r)= L^2 $$

EDIT1
Yet another practical way to find $R$ is the relation $ R = \dfrac{s}{\alpha_ d +\beta_d} $ for Diagon from the Gauss-Bonnet theorem. Arc length $s=1P2 $ and corner angles are to be measured.
