NOTE ON THE CASE OF "THE SAME CIRCLE."

Ths. 38 and 39 have been proved for equal circles. To see that they apply also to "the same circle," regard the circle as composed of two equal circles superimposed.

THEOREM 40. In equal circles (or, in the same circle)

(i) equal chords are equidistant from the centres. (ii) Conversely, chords that are equidistant from the centres are equal.

(i) Data ABC, DEF are equal Cs; their centres P and Q.

PG, QH are 1s from the centres P, Q upon the chords AB, DE.

Constr.

Since S G and H are rt. s,

As APG, DQH are right-angled with

hyp. AP hyp. DQ (Data), AG = DH (Proved),

=

Proof Since s G and H are rt. ▲ s,

As APG, DQH are right-angled with

hyp. AP = hyp. DQ (Data), PG = QH (Data),

But the perpendicular from the centre on a chord bisects the chord.

.. ABDE.

Th. 36.

Q. E. D.

DEF. A tangent to a circle is a straight line which, however far it may be produced, has one point, and one only, in common with the circle.

The tangent is said to touch the circle; the common point is called the point of contact.

We shall assume that at a given point on a circle there is one tangent and one only.

THEOREM 41. The tangent at any point of a circle and the radius through the point are perpendicular to one another.

Data O is the centre of O; A is a point on

To

the circumference; BC is the tangent at A. prove that BC and OA are 1 to one another. Proof If OA is not to BC, suppose OT drawn to BC.

B

B

Fig. 79.

.. the tangent AT, if produced, would cut the circle in a second point.

But by definition this is impossible.

.. it cannot be possible to draw another line OT to BC.

.. OA must be 1 to BC,

that is, the tangent at A and the radius through A are perpendicular to one another.

S. H. T. G.

Q. E. D.

6

COR. A straight line drawn through the point of contact of a tangent at right angles to the tangent will, if produced, pass through the centre of the circle.

NOTE. The method of proof used in Th. 41 is called reductio ad absurdum. The same method was used in Th. 5.

THEOREM 42. The two tangents to a circle from an external point are equal.

Data PH, PK are two tangents from a point P

Construction Join CH, CK, CP.

Proof Since PH is a tangent and CH a radius,

▲ CHP is a rt. .

Simly CKP is a rt. L.

K

Fig. 80.

Hence As CHP, CKP are right-angled with
CH=CK (radii) and the hyp. CP common.

COR. 1. The two tangents from a point are equally inclined to the line joining the point to the centre of the circle.

COR. 2. The line joining the point to the centre bisects the chord of contact at right angles. (Use Th. 34.)

DEF. Two circles are said to touch one another if they touch the same line at the same point.

THEOREM 43. If two circles touch, the point of contact lies in the straight line through the centres.

CASE I. When the circles touch internally

A

Proof [CX, CY are not to be assumed to be the same straight line; we have

to prove that they are.]

Since CA is the tangent at C to CMN and CX the radius through C

Sim1y YCA is a rt. 4, . L XCA = LYCA,

.. CX and CY must be the same straight line,

.. XY produced will pass through C.

Th. 41.

CASE II. When the circles touch externally (fig. 82).

Data The Os CMN, CPQ touch externally at C; X and Y the centres.

Proof [CX, CY are not to be assumed to be in one straight line; we have to prove that they are.]

Since CA is the tangent at C to CMN and CX the radius through C, XCA is a rt. .

Th. 41.

SimlyYCA is a rt. 4,

.. 4 XCA + 2 YCA = 2 rt. 4 S,
.. XCY is one straight line.

Th. 2.

Q. E. D.

COR. If two circles touch internally, the distance between their centres is the difference of their radii; if externally, the distance between their centres is the sum of their radii.

CONSTRUCTIONS.

To circumscribe a circle about a given triangle*.

ABC is the given triangle.

Construction [It is necessary to find a point equidistant from A, B, C.]

Let these lines meet in O.

Then O is the centre of the circle.

Proof Every point on DO is equidistant from B

and C.

Th. 34.

... OB OC.

B

Fig. 83,

Every point on OF is equidistant from B and A.

.. O is the centre of a circle through A, B, C.

To inscribe a regular hexagon in a circle.

Fig. 84.

In the circle place a chord AB equal to the radius.

Join A, B to O, the centre.

Then ▲ OAB is equilateral,

.. LAOB = 60°.

Place end to end in the circle 6 chords each equal to the radius. Each chord subtends 60° at the centre,

.. the total angle subtended by the 6 chords is 360°.

In the course of Th. 37 we showed how to find the centre of the circle through three given points, which is virtually the same problem as this.