sphere, the four angles above mentioned are constant; therefore is constant, and the ratio of FN. FM to FG2 is constant ; Р which, by conic sections, is a property of an ellipse: consequently the figure N G M D is an ellipse. Q. E. D. A n If the given circle of the sphere be in the position A S B V, so that it does not intersect the plane of projection, and is not parallel to it; the projected figure is an ellipse. Let a plane parallel to that of the circle cut, in a sb d, the oblique cone formed by lines drawn from E to the circumference of AS BV, so that it may intersect the plane of projection; this section of the cone will be a circle. For, K being the centre of the circle ASB V, let any plane E b m B P K EK S passing through the axis EK cut the section asbd inks; this line (Geom. Planes, Prop. 14.) will be parallel to K S, and the triangles E K S, Eks will be similar to one another: therefore EK: Ek:: Ks: ks. But the ratio of EK to Ek is constant, and Ks is constant; therefore ks is constant, and a sbd is a circle. Now if ng md represent the projection of A S B V, and dfg be the intersection of this projected figure with the plane of the circle a s b d, we shall have, reasoning as in the former case, a constant ratio between nf. fm and fg2: consequently, the projected figure ng md is an ellipse. Cor. 1. When the projecting point is on the surface of the sphere, as at E' in the preceding figure, the projection of any circle of the sphere, as A G B D not passing through E', is a circle. For draw A R parallel to QT; then (Euc. 30. 3.) the arc E'A will be equal to the arc ER, and consequently (Euc. 26. 3.) the angle EAR to the angle ABE'; but E' AR=AN'T (Euc. 29. 1.); therefore FN'A FBM', and the opposite angles at F being equal to one another, the angle F A N′ = F M'B. It follows that, in the present case, the products corresponding to those represented by p and q above are equal to one another, and FN. F M' becomes equal to F G2, which is a property of a circle (Euc. 35. 3.); therefore when E is on the surface of the sphere, &c. Scholium. If the plane of a circle of the sphere were parallel to the plane of projection, it is evident, whether E be on the surface of the sphere, or within it, or on its exterior, that lines drawn from thence to every part of the circumference would form the convex surface of an upright cone having a circular base; and the surface of this cone, produced if necessary, being cut by the plane of projection, the section, that is, the projected figure, will (Geom., Cor. 7. Def. Cyl.) be a circle. The observation will evidently hold good whether the plane of projection pass through the centre of the sphere or be elsewhere situated. Cor. 2. When the point E is infinitely remote, so that all lines drawn to it from the circumference of any circle of the sphere may be considered as parallel to one another, those lines will constitute the convex surface of a cylinder whose base is a circle forming any angle whatever with the axis, that is, with a line drawn through the centre of the circle parallel to the lines before mentioned. In this situation of the eye the section of the cylinder made by the plane of projection, that is, the projected figure, will, except when the circle of the sphere is perpendicular to the axis, be an ellipse. For, in the first figure to this proposition, if A G B D be the given circle of the sphere, the points A and B, the extremities of its diameter, will be projected respectively in the points N", M", where perpendiculars from A and B meet QT: therefore, in the triangles AFN", BFM", the angles at N'' and M' are right angles; and since the opposite angles at F are equal to one another, we have (Pl. Trigon., Art. 56.) A F Cos. FF N', and F B cos. FF M'; consequently, A F. F B = F N". FM" FN." FM' = F G2, But A F. F BF G2; therefore that is, F N". F M" has a constant ratio to F G2, which is a property of an ellipse. Therefore the section N' G M'D is an ellipse. 27. Scholium. The orthographical projection of an ellipse, on any plane, is also an ellipse. For let it be supposed that the figure DAG B is an ellipse: then by conic sections, t being the semi-transverse, and c the semi-conjugate axis, A F. F B = F G2; or the rectangle A F. F B has a constant ratio to FG2. Now from the second corollary we have AF. FB c2 t2 FN". F M" ; therefore AF. F B (= F G2) = c' FN." FM" t' cos.2 F hence the rectangle F N." FM" has still a constant ratio to F G2, or the projected figure N' G M" D is an ellipse. C : PROPOSITION III. 28. If a plane touch a sphere at any point, and from the centre of the sphere lines be drawn through the circumference of a small circle whose plane is perpendicular to the tangent plane; those lines if produced will meet the latter in points which will be in an hyperbola. Let c be the centre of the sphere, and D the point at which the tangent plane DG is in contact with it: let aa" be part of the circumference of a small circle whose plane is perpendicular to DG, and let DE be part of the circumference of a great circle parallel to the small circle; also let the plane of this great circle meet the tangent plane, to which it is perpendicular, in DF. Again, let a plane pass through CD perpendicular to the planes of the great and small circle, it will also be perpendicular to the tangent plane; and in it let the point a be situated: then a line drawn from c through a will meet the tangent plane in some point as A. Let, also, a plane pass through c and any other point a' in the circumference of the small circle, perpendicularly to CDE; it will meet the tangent plane in the line FG which (Geom., 19. Planes) will be perpendicular to the plane CDE, and consequently (Geom. 6. Planes) parallel to DA. Draw a line from c through a' to meet FG in G; and from G in the tangent plane draw Gd parallel to FD; Gd will be equal to D F, and FG to Dd. Let CD, the semidiameter of the sphere, be represented by unity; then (Euc. 47. 1.) c F21+ DF2, and (Pl. Trigon. Art. 56.) FGCF tan. GCF. But the angle GCF ACD, the arcs Da and Ha' which measure those angles being equal to one another (Geom., 2. Cyl.), and therefore tan. GCF, or tan. ACD, = AD. It follows that FG CF. AD and FG2 - CF2. AD2, that is, FG2 = AD2 + AD2. DF2, or Dd2 = AD2 + AD2. Gd2, and Dd2 — AD2 = A D2. Gd2. In like manner, for any other point, as a", in the same small circle we should have Dd'2. - AD2 = A D2. G'd'2; and consequently Gd2: G'd2: Dd2 - AD2: Dd2 — AD2. :: This, by conic sections, is a property of an hyperbola of which AD is one of the semi-axes; therefore, &c. If c be the projecting point, or the eye of the spectator, it is manifest that the representation of a portion of a small circle whose plane is perpendicular to the tangent plane, or plane of projection, will be an hyperbola consisting of two similar branches, one on each side of the plane CDA, which passes through the centre c and the point D of contact perpendicularly to the plane of the circle. Note. The fifteen following propositions relate to the stereographical projection only. PROPOSITION IV. 29. The angle contained between the tangents to two circles of the sphere, which intersect each other, when drawn from one of the points of section, is equal to the angle contained between the projections of those tangents. P M Let PS and AB be two diameters of the sphere at right angles to one another, and let the plane of projection pass through AB perpendicularly to that of the paper: then P may be the projecting point. Let HM, HN be tangents to two circles of the sphere which intersect each other at H, and let them be produced till, in M A S H B m n N and N, they meet a plane touching the sphere at P, which plane is consequently parallel to the plane of projection. Draw PM, PN, and PH, and imagine planes to pass through PM, HM, and PN, HN: these will intersect each other in PH, and the plane of projection in h m and hn; therefore hm and hn, produced if necessary, are the projections of the tangents HM, HN; and it is required to prove that the angle mhn is equal to MHN. Since N P, N H are tangents drawn to the sphere, they are tangents drawn from the point N to the circle formed by the intersection of the plane PNH with the sphere; therefore (Euc., A. Cor. 36. 3.) they are equal to one another: for the like reason M P, M H are equal to one another. Therefore the triangle MP N is equal to the triangle M HN, and the angle M P N to the angle M HN. But the triangle mhn is a section parallel to MPN, in the pyramid H MPN: therefore mhn is similar to MPN (Geom., 1. Prisms, &c.), and the angle mhn is equal to MP N, that is, to M HN. Since the angle made by two circles which intersect one another on the surface of the sphere, or on a plane, is ex pressed by the angle contained between the tangents at the point of intersection (Spher. Geom., 2. Cor. 2. Def.), it may be said that the angle contained between the planes of two circles, which intersect one another on a sphere, is equal to the angle contained between the projections of those circles. PROPOSITION V. 30. If a circle of the sphere be described about each of the intersections of two circles of the sphere as a pole, at equal distances from the intersections; the arcs intercepted upon the circles so described will be equal to one another. B P Let the straight line PP be the intersection of any two circles of the sphere, and let abAB, a'b' A'B' be parts of two circles of the sphere described about P and P' as poles at equal distances from each; the intercepted arcs ab, a'b' will be equal to one another. PA Imagine arcs of great circles to pass through the points P and a, P and b, P' and a', P' and b'; all those arcs will (Geom., 2. Cylind.) be equal to one another, and consequently (Euc., 29. 3.) all their chords, viz. the straight lines Pa, Pb, P'a', 'b', will be equal to one another: also (Euc., 28. 3.) the arcs Pa, Pb, P'a', P'b', in the diagram, will be equal to one another. Then, since the arc Pb is equal to the arc P'b', the arcs P'b and Pb' are equal to one another, and therefore the angle P'Pb is equal to PP'b' (Euc., 27. 3.). In like manner the angle P'Pa is equal to PP'a': therefore the solid angles at P and P' have two plane angles of the one equal to two plane angles of the other, and the inclinations of the plane angles to one another are equal, each of them being the inclination of the circles Pa P', PbP', to one another; consequently the third plane angles a Pb and a' P′b' are equal to one another (Geom., 23. Planes). It follows that in the two triangles a Pb, a' P'b', the chord ab is equal to the chord a'b' (Euc., 4. 1.); and the circles described about P and P' being equal, the arcs ab and a'b' are equal to one another. Scholium. It is obvious that the arcs AB and A'B' are equal to the arcs ab and a'b' and to one another. It is obvious also that the demonstration would be the same if the circle a ba described about P should inclose the point P': the equal circle about P' consequently inclosing P. Thus, if P and P' were, respectively, a pole of the primitive and of any other great circle of the sphere, the arcs intercepted on the pri |