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16.
PROBLEM VII.

Having
two points A and B given in poſition, and likewiſe a right line EF

given in poſition, it is required to find the center of a circle, which ſhall paſs

through the given points and touch the given line.

Case
Iſt. When the points A and B are joined, ſuppoſe AB to be parallel to

EF: then biſecting AB in D, and through D drawing DC perpendicular to it,

DC will alſo be perpendicular to EF: draw a circle therefore which will paſs

through the three points A, B, andC, (by Euc. IV. 5.) and it will be the circle

required: (by a Corollary from Euc. III. 1. and another from Euc. III. 16.)

Case
2d. Suppoſe AB not parallel to EF, but being produced meets it in

E: then from EF take the line EC ſuch, that its ſquare may be equal to the

rectangle BEA, and through the points A, B, C, deſcribe a circle, and it will be

the circle required by Euc. III. 37.

This
is Vieta’s Solution. But Mr. Thomas Simpſon having conſtructed this,

and ſome of the following, both in the Collection of Problems at the end of his

Algebra, and alſo among thoſe at the end of his Elements of Geometry, I ſhall

add one of his Conſtructions.

Let
A and B be the points given, and CD the given line: drawing AB and

biſecting it in F, through E let EF be drawn perpendicular to AB and meeting

CD in F: and from any point H in EF draw HG perpendicular to CD, and

having drawn BF, to the ſame apply HI = HG, and parallel thereto draw BK

meeting EF in K: then with center K and radius BK let a circle be deſcribed,

and the thing is done: join KA, and draw KL perpendicular to CD, then be-

cauſe of the parallel lines, HG: HI: : KL: KB; whence as HG and HI are

equal, KL and KB are likewiſe equal. But it is evident from the Conſtruction

that KA = KB, therefore KB = KL = KA.

Because
two equal lines HI and Hi may be applied from H to BF each

equal to HG, the Problem will therefore admit of two Solutions, as the Figure

ſhews: except in the caſe when one of the given points, A for inſtance, is given

in the line CD, for then the Problem becomes more ſimple, and admits but of

one conſtruction, as the center of the circle required muſt be in the line EF con-

tinued, as alſo in the perpendicular raiſed from A to CD, and therefore in their

common interſection: and this is the limit of poſſibility; for ſhould the line CD

paſs between the given points, the Problem is impoſſible.

N. B.
Tho
’ Vieta does not take notice that this Problem is capable of two

anſwers, yet this is as evident from his conſtruction, as from Mr. Simpſon’s, for

EC (the mean proportional between E B and EA) may be ſet off upon the given

line EF either way from the given point E.