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Document number: 8900
Date: Tue 22 Nov 1864
Recipient: TALBOT William Henry Fox
Author: TALBOT Charles Henry
Collection: British Library, London, Manuscripts - Fox Talbot Collection
Collection number: envelope 22950
Collection number historic: LA64-100
Last updated: 7th March 2012

Llandaff.
Tuesday evening. Novr 22nd

My dear Father.

I send you some geometrical theorems as I think you may like them. A problem proposed to us at Cambridge one day was to insert a circle in a gothic arch such with a base [diagram] such as that. I remember you found a way of doing it. I have found another. I give you the figures on a separate sheet of paper for convenience sake.

& I have got to a 4 different infinities on paper

In the first place I must remind you of a theorem we proved

Fig. I

If 2 circles intersect, of the 2 sets of circles which can at the same time touch both of the given circles. The locus of the centres of one set is a hyperbola and of the other an ellipse (confocal) with the given centres as foci

HH is hyperbola. EE ellipse. OO foci.

Now (figure 2) Let ABC be an equilateral arch struck from centres A,C, it is required to insert inscribe a circle in ABC. Complete the circle CAD. Now the required centre is on an since these ˘Ás intersect the required centre is on an hyperbola which is the straight line DE ˇŃ r to DC.

Again because CBD a circle cuts CAD a circle of infinite radius in the required centre is on an hyperbola C.F which hyperbola becomes a parabola since its other focus has moved to infinity. its focus is A and its [illegible] CD. Therefore the required centre is O the pt of intersection of these 2 curves. Also the tangent GH to this circle CBGD through G is the direction of the parabola.

From this we at once get an easy method. (Figure 3).

ABC as before. draw GH the common tangent to the 2 circles. GH is the direction of the parabola whose focus is A. bisect GH in K. join AK & bisect it in L. BE is as before. draw. LO perpendicular to AK to cut BE in O. O is the centre for if AO. be joined. AO is equal to OK. Therefore O is a point on the parabola & is therefore the centre required.

Now I come to a much more general theorem. problem. viz. (Figure 4.) To find the centre of the circle inscribed in any circular triangle, formed by 3 circles. As I have drawn in it there are 7 triangles they are. ABC. AEB. BFC. CDA DEA. EFB. FDC. I have given the hyperbol©ˇ & ellipses in dotted lines and the centres as big black dots to avoid confusion. You will easily make it out. You see there are 2 hyperbol©ˇ & 2 ellipses. The theorem of the inscribed & escribed [sic] circles of a rectilinear triangle is merely a case of this. I give the figure (figure 5), the letters are those of the curvilinear circular triangles and are some of them at infinity. I should have drawn the secondary lines dotted but being late, I forgot it. please keep me my letter as it is too elaborate to copy the figure.

Your affect son
Charles.

P.S. I have another extension to send.

N.B. The points D

[envelope]
H Fox Talbot Esq
Lacock Abbey
Chippenham

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