PROBLEM XX. Fig. 98.

To draw a plane to touch two given spheres and contain a given point.

Let C be the centre of a sphere of radius ca, and O the centre of a sphere of radius od; it is required to draw a common tangent plane to these two spheres which shall also contain the point P.

Construction. Find V, the vertex of the cone which envelops the two given spheres (Prob. XIX.), and draw the traces of a plane TUF containing the line PV and touching one of the spheres (Prob. XVIII.); TUF is the plane required.

In the figure, LMN is a plane containing C and perpendicular to PV; Q is the point of intersection of PV and LMN; Q, is the rabatment of Q; and R is the point in which TUF touches the sphere.

Proof. As the plane TUF contains the point V, and touches the sphere (C) in the point R, it contains one of the generators VR of the conical surface which envelops the two spheres; therefore TUF touches the other sphere also; and as it contains P it is the plane required.

Remarks. As in Prob. XVIII. there are two planes which contain the line PV and touch one of the spheres (TN is the horizontal trace of the second plane); and as two conical surfaces may be drawn to envelop the two spheres, there are in all four planes which fulfil the conditions of this problem. The others may be found in a similar way.

It may be observed that the problem might also be solved by drawing a plane containing the point P and touching the conical surface which envelops the spheres, as in Prob. III.

PROBLEM XXI.

To draw a tangent plane to three given spheres.

Let the three planes be denoted by A, B and C. Then if the vertex of a cone enveloping A and B be determined, and also the vertex of a cone enveloping A and C, or B and C, it follows from Prob. XX. that the plane which contains these two vertices and touches one of the spheres must also touch the other two.

Now there are two conical surfaces which envelop A and B, and two which envelop A and C, and two which envelop B and C; that is six altogether. The six vertices of these cones will be found to lie on four straight lines, three on each line; and two planes can be drawn containing any one of these lines, and touching the spheres; so that in all eight planes may be drawn to touch three given spheres.

HYPERBOLOID OF REVOLUTION.

PROBLEM XXII. Fig. 99.

To find the projections of the surface generated by a straight line which revolves about an axis not in the same plane with it.

Let the horizontal plane of projection be taken at right angles to the axis, the projections of which are o, and n'a'; and let the initial position of the generating line be parallel to the vertical plane of projection, so that aB, a'b' are its projections.

The generator may be considered as indefinitely extended, but in the figure the solid is bounded by the horizontal plane of projection and the plane described by the common perpendicular of the axis and generator-that common perpendicular is the line OA, equal and parallel to oa.

As the axis is perpendicular to the horizontal plane the horizontal trace of the surface will be the circle described by the horizontal trace of AB; that is the circle through B described from the centre o.

The outline of the vertical projection will be the projection of the meridian parallel to the vertical plane; that is the curve of intersection of the generator with the meridian plane FB. It is required to find that curve.

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Construction. Take any point D on the generator. The distance of D from the axis is equal to od; and as the generator revolves about Oo, the point D moves on the circumference of a circle. Find the vertical projections of the points D, and H, in which the path of D intersects the meridian plane FB1; h and d' are points on the required curves. In the same way any number of points, as c', è and g, k, may be found, and the curve traced through them. A is the point of the generator nearest to the axis, and its path AAA, is consequently the smallest parallel of the solid. It is called the throat circle.

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