ON THE METHOD OF LIMITS.

DEF.-If a secant of a circle alters its position in such a manner that the two points of intersection approach and ultimately coincide with one another, the secant in its limiting position is said to 'touch,' or to be a 'tangent' to, the circle (Syllabus).

DEF.-The point in which the two points of intersection ultimately coincide is called the 'point of contact,' and the tangent is said to 'touch' the circle at that point (Syllabus).

These definitions of tangency will be found to lead to the same results as Euclid's.

PROPOSITION.

The tangent at any point to a circle is at right angles to the radius drawn to the point of contact.

Then in the limit when HK becomes the tangent at A, we have L DAHL DAK.

Ex. 475.-Prove the same theorem by using III. 3, and taking the limit.

When using this method, the student should be careful to draw two diagrams, one illustrating what happens before, the other what happens at the coincidence of the two points which approach one another.

PROPOSITION.

Each angle contained by a tangent and a chord drawn from the point of contact is equal to the angle in the alternate segment of the circle.

[III. 33.

Let EBB'F be the secant of ABB'D through B, B', and BD any other chd. through B.

A

B

LFB'D+LEB'D=2 rt. s.

=LBAD+LEB'D.

... LFB'D= L BAD.

Now let B' move up to and coincide with B.

Then in the limit, when EBF becomes the tangent at B,

LFBD=4 BAD.

[III. 22.

[See diagram of III. 33.

From a given General Theorem, the student may often deduce a special case by the method of limits, of the truth of which he may afterwards satisfy himself by a demonstration which does not depend on that method.

The circum-circles of the four triangles formed by four intersecting straight lines all pass through one point.

Let ADD', ACC', OCD, OC'D' be four As formed by the four intersecting st. lines OC, OD, CD, C'D'.

Then the circum-Os of As ADD', ACC' meet in some other pt. B.

[See Ex. 366 (iii. ).

Join BD, BA, BC.

LABD=LOD'A in same segt. ABD'D,

and ABC ext. LAC'O of cyclic quadl. ABCC'.

... LCBD=LS OD'A, AC'O.

... Ls CBD, COD=LS OD'A, OC'A, COD,

= two rt. Ls.

... the circum- of OCD passes through B.
Similarly the circum- OC'D' passes through B.
Now, let C' move up to and coincide with C,

and.. D' move up to and coincide with D.

Then in the limit when OC, OD become the tangents at C and D the circum- of OCD passes through B.

Enunciate generally.
Again :-

If two circles intersect, and through one of the common points two straight lines be drawn and terminated each way by the circumference, they subtend equal angles at the other common point. Let ABC, ABD be the two Os, CAD, C'AD' the two st. lines through A. Join AB, DD', CC; produce DD', CC' to meet in O. (See p. 248.) Then it has been shown that

Similarly

Ls CBD, COD =2 rt. s.

Ls C'BD', C'OD=2 rt. 4 s.
.. 4 CBD= ▲ C'BD.

Now let C' and D each move up to and coincide with A.

Then in the limit when CA, D'A become tangents at A we have

If two circles intersect, and through one of the common points

a chord be drawn to each circle touching the other, these chords subtend equal angles at the other common point.

[See Ex. 369.

Ex. 476.-Prove the first Corollary to III. 36 without assuming III. 36, and then deduce III. 36 by the method of limits. Prove each rectangle = diffce. between sq. on ED,

and sq. on radius.

The Method of Limits can also be applied to the contact of circles with one another.

Ex. 477.-Deduce III. II and III. 12 from the theorem that

The straight line through the centres of two intersecting circles bisects their common chord at right angles.

Ex. 478.-In the figure on p. 248.

LCBD=LS AD'D, ACO.

Hence show by the Method of Limits that when the two circles touch at A, DD', CC' are parallel.