Page images

5. The tangent to a circle is a straight line which touches the circumference, or meets it only in a single point.

6. Circles are said to touch mutually, if they meet but do not cut each other.

7. The point where a straight line touches a circle, or one circle touches another, is termed the point of contact.

[blocks in formation]

D is also a centre, AD must be equal to DE; that is, a greater to a less,-which is impossible. Wherefore the circle AEBF has no other centre than the point C.


A circle is bisected by its diameter.

The circle ADBE is divided into two equal portions by the diameter AB.

For let the portion ADB be reversed and applied to AEB, the straight line AB and its

middle point, or the centre C, re

maining the same. And since the
radii of the circle are all equal, or
the distance of C from any point in A
the boundary ADB is equal to its
distance from any point of the

boundary AEB, every point D of the
former must meet with a correspond-



ing point of the latter, and consequently the two portions ADB and AEB will entirely coincide.

Cor. The portion ADB limited by a diameter, is thus a semicircle, and the arc ADB is a semicircumference.


A straight line cuts the circumference of a circle only in two points.

If the straight line AB cut the circumference of a circle in D, it can only meet it again in a single point E.

For join D and the centre C; and because from the point C only two equal straight lines, such as



CD and CE, can be drawn to AB (I. 22. cor.), the circle described from C through the point D will cross AB again only at E.


The chord of an arc lies wholly within the circle.

The straight line AB which joins two points A, B in the circumference of a circle, lies wholly within the figure.

For from the centre C draw CD to any point in AB, and join CA and CB.

Because CDA is the exterior angle of the triangle CDB, it is greater than the interior CBD or CBA (I. 10.); but CBA, being equal to CAB or CAD (I. 8.), CDA is greater than CAD;



wherefore the opposite side CA is greater than CD (I. 15.),

or CD is less than CA, and consequently the point D must lie within the circle.

Cor. Hence a circle is concave towards its centre.


A straight line drawn from the centre of a circle at right angles to a chord, likewise bisects it; and, conversely, the straight line which joins the centre with the middle of a chord, is perpendicular to it.

The perpendicular let fall from the centre C upon the chord AB, cuts it into two equal parts AD, DB.

For join CA, CB: And in the triangles ACD, BCD, the side AC is equal to CB, CD is common

to both, and the right angle ADC is equal to BDC; these triangles, being of the same affection, are equal (I. 24.), and consequently the corresponding side AD equal to BD.

Again, let AD be equal to BD; the bisecting line CD is at right angles to AB.


For join CA, CB. The triangles ACD and BCD, having the sides AC, AD equal to CB, BD, and the remaining side CD common to both, are equal (I. 2.), and consequently the angle CDA is equal to CDB, and each of them a right angle.


A straight line which bisects a chord at right angles, passes through the centre of the circle.


If the perpendicular FE bisect a chord AB, it will through G the centre of the circle.







For in FE take any point D, and join DA and DB. The triangles ADC and BDC, having the side AC equal to BC, CD common, and the right angle ACD equal to BCD, are equal (I. 3.), and consequently the base AD is equal to BD. The point D is, therefore, the centre of a circle described through A and B; and thus the centres of the circles that can pass through A and B are all found in the straight line EF. The centre G of the circle AEBF must hence occur in that perpendicular.

Cor. The centre of a circle may hence be found by bisecting the chord AB by the diameter EF (I. 7.), and bisecting this again in G.


The greatest line that can be drawn within a circle, is the diameter.

[blocks in formation]
« PreviousContinue »