If the distance between the centres of two circles is equal to the difference of their radii, one will be tangent to the other internally.

Let C and D be the centres of two circles, and let the distance between these centres be equal to the difference of the radii: then will the one be tangent to the other internally.

For, they will have a point A, on DC, common, and they will have no other point in common. For, if they had two points in common, the distance between their centres would be greater than the difference of their radii; which is contrary to the hypothesis:

E

A

C D

hence, one touches the other internally; which was to be proved.

Cor. 1. If two circles are tangent, either externally or internally, the point of contact will be on the straight line drawn through their centres.

Cor. 2. All circles whose centres are on the same straight line, and which pass through a common point of that line, are tangent to each other at that point. And if a straight line be drawn tangent to one of the circles at their common point, it will be tangent to them all at that point.

Scholium. From the preceding propositions, we infer that two circles may have any one of six positions with respect to each other, depending upon the distance between their

centres:

1o. When the distance between their centres is greater

than the sum of their radii, they are external, one to the other:

2o. When this distance is equal to the sum of the radii, they are tangent, externally:

3o. When this distance is less than the sum, and greater than the difference of the radii, they intersect each other:

4°. When this distance is equal to the difference of thei radii, one is tangent to the other, internally:

5o. When this distance is less than the difference of the radii, one is wholly within the other:

6o. When this distance is equal to zero, they have a common centre; or, they are concentric.

PROPOSITION XV. THEOREM.

In equal circles, radii making equal angles at the centre, intercept equal arcs of the circumference; conversely, radii which intercept equal arcs, make equal angles at the centre.

1o. In the equal circles ADB and EGF, let the angles ACD and EOG be equal: then will the arcs AMD and ENG be equal.

For, draw the chords AD

B

M

F

and EG; then will the triangles ACD and EOG have wo sides and their included angle, in the one, equal to two sides and their included angle, in the other, each to each. in all their parts; consequently, But, if the chords AD and EG and ENG are also equal (P. IV.); which was to be proved.

They are, therefore, equal AD is equal to EG. are equal, the arcs AMD

2o. Let the arcs AMD and ENG be equal: then will

the angles ACD and EOG be equal.

For, if the arcs AMD

and ENG are equal, the chords AD and EG are equal (P. IV.); consequently, the triangles ACD and EOG have their sides equal, each

to each; they are, therefore,

M

equal in all their parts: hence, the angle ACD is equal to the angle EOG; which was to be proved.

In equal circles, commensurable angles at the centre are pro portional to their intercepted arcs.

In the equal circles, whose centres are C and 0, let the angles ACB and DOE be commensurable; that is, be exactly measured by a common unit: then will they be proportional to the intercepted arcs AB and DE.

M

Z

Let the angle M be a common unit and suppose, for example, that this unit is contained 7 times in the angle ACB, and 4 times in the angle DOE ACB be divided into 7 angles, by the radii

Then, suppose

Cm, Cn, Cp,

&c.; and DOE into 4 angles, by the radii Or, Oy, and Oz, each equal to the unit M.

From the last proposition, the arcs Am, mn, &c., Dœ, ay, &c., are equal to each other; and because there are 7 of these arcs in AB, and 4 in DE, we shall have,

angle ACB : angle DOE :: arc AB : arc DE

If any other numbers than 7 and 4 had been used, the same proportion would have been found; which was to be proved.

Cor. If the intercepted arcs are commensurable, they will be proportional to the corresponding angles at the centre, as may be shown by changing the order of the couplets in the above proportion.

PROPOSITION XVII. THEOREM.

In equal circles, incommensurable angles at the centre are proportional to their intercepted arcs.

In the equal circles, whose centres are C and 0, let ACB and FOH be incom

:

mensurable then will they be proportional to the arcs AB and FH.

For, let the less angle FOH, angle ACB, So that it shall

DIOB F

be placed upon the greater take the position ACD.

Conceive the arc AB to be divided into equal parts, each less than DO: there will be at least one point of division between D and 0; and ; let I be that point; and draw CI. Then the arcs AB, AI, will be commensurable, and we shall have (P. XVI.),

angle ACB : angle ACI :: arc AB : arc AI

Comparing the two proportions, we see that the antecedents are the same in both: hence, the consequents are propor tional (B. II., P. IV., C.); hence,

angle ACD : angle ACI ::

arc 40 : arc AI

But, 40 is greater than AI: hence, if this proportion is true, the angle ACD must be greater than the angle ACI. On the contrary, it is less: hence, the fourth term of the assumed proportion cannot be greater than AD.

In a similar manner, it may be shown that the fourth term cannot be less than AD: hence, it must be equal to

Cor. 1. The intercepted arcs are proportional to the cor