Page images

(5.) A straight line drawn from a sphere's center to its surface, is called a Radius of the Sphere.


(6.) Theorem. The common section of a sphere and a plane is a circle.

First, if the cutting plane pass through the sphere's center, it is evident (Art. 1, 2.) that the common section is bounded by a line, of which every point is equidistant from that center; therefore, (E. Def. 15. 1.) the common section is a circle.

But, secondly, let Ap HE be a sphere, of which C is the center; let it be cut by a plane MN, which does not pass through its center; and let EFH be the common section of the plane, and the sphere: EFH is a circle.

For, from C, let CG be supposed to be drawn (E*. 11. 11.) at right angles to the plane EFH; and, in that plane, let there be drawn, from G, any two straight lines GE and GF, to the sphere's surface. Then (E. Def. 3. 11.) CG is at right angles to GE and GF; let, also, the points C, E, and C, F be joined.

Then, in the two right-angled triangles CGE, CGF,

*The letter E refers to Euclid's Elements; and, of the subsequent numbers, the former denotes the Proposition, and the latter the Book, which are intended to be cited.

because the hypotenuse CE is equal to the hypotenuse

CF, (Art. 1.), and the side CG is common to both the triangles, therefore (Demonstration of E. 14. 3.) the

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small]

third side GE, in the one, is equal to the third side GF in the other:

In the same manner may any other straight line as GH, drawn in the plane EFH from G to the sphere's surface, be shewn to be equal to GE, or GF. Therefore (E. Def. 15. 1.) the figure EFH is a circle.

The proposition may also be proved, by shewing that any two straight lines in EFH, drawn through G, and terminated by the sphere's surface, being chords of two equal circular sections of the sphere made by planes passing through them and the sphere's center, and being equally distant from the common center, 'are (E. 14. 3.) equal to one another; and that each of them being (E. 3. 3.) bisected, in G, by the perpendicular CG,

therefore, the straight lines GE and GF are equal to one another.

(7.) COR. 1. If a sphere be cut by any two planes, passing through its center, the two circular sections, having a diameter of the sphere for their common diameter, shall bisect one another: they shall also, because their diameters are equal, be equal to one another all such sections of the sphere are, therefore, equal; they all bisect each other; and the center of the sphere is their common center.

(8.) COR. 2. It has appeared, from the demonstration of Art. 6, that the center G of the circular section, EFH, of a sphere, made by a plane which does not pass through the sphere's center, is in the perpendicular CG, drawn to that section from the sphere's center C: and, conversely, (E. 13. 11.) the center C of the sphere is in the straight line GC drawn at right angles to the plane of the section EFH, from its center G. Therefore CG, the straight line joining the centers of the sphere and of any such section, is perpendicular to the plane EFH of that section and, a plane which passes through the center of such a section and cuts it perpendicularly, passes also (E. Def. 4. 11 and 13. 11.) through the center of the sphere.

* Two circles, of which the diameters are equal, may be shewn to be equal, by applying the one to the other so that their centers shall coincide: for then the one figure must wholly coincide with the other. In the same manner, also, it may be shewn, that of two circles that which has the greater diameter is the greater.

[ocr errors]



Circles in a sphere are said to be

equally distant from the sphere's center, when the perpendiculars drawn to their planes from the sphere's center are equal: and that circle is said to be the farther distant, upon the plane of which the the plane of which the greater perpendicular



(10.) Theorem. Of circles in a sphere, the greatest are those which pass through the sphere's center: of the rest, those are equal, which are equally distant from that center; and those which are nearer to the center, are greater than those which are more remote.

Let PFp be a circle passing through the center C of the sphere AD, and let EH and AI be circles in the sphere, which do not pass through its center: Pp is greater than EH, or than AI: If EH and AI be equidistant from the sphere's center, they are equal: If they be not equidistant from it, that, which is the nearer to it, is greater than that which is more remote.

For, let G and O be the centers of EH and AI; join

[merged small][merged small][merged small][ocr errors][merged small]

C, G and C, E and C, O and C, I; join also G, E and O, I. Then (Art. 8.) the angles CGE and COI are right angles; wherefore (E. 16. and 19. 1.) CE is greater than GE, and CI than OI; that is, (Art. 7.) the radius of the circle Pp is greater than the radius of either of the other circles EH, or Al; therefore the circle PFp is greater than either EH, or AI.

Again, since (Art. 1.) CE is equal to CI, and since the angles at G and O are right angles, the squares of CG and GE are (E. 47. 1.) together equal to the squares of CO and OI; it is evident, therefore, that, if CG be equal to CO, that is, if the circles EH and AI be equidistant from the center C, the radius GE will be equal to the radius OI, and therefore the circle EH will be equal to AI: but, if CG be less than CO, that is, if the circle EH be nearer than AI to C, then the radius GE must be greater than the radius OI; otherwise the squares of CG, GE would not be equal to the squares of CO, OI: in this case, therefore, the circle EH is greater than AI.

(11.) COR. 1. In the same manner, it may be shewn, conversely, that the greater of two circles, in a sphere, is nearer to the sphere's center than the less, and that equal circles, which do not pass through the sphere's center, are equally distant from it.

(12.) COR. 2. Two equal circles in a sphere, which do not pass through the sphere's center, cannot have a common diameter.

« PreviousContinue »