18(6)
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.
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.)
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.
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.
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.
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.
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.
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.
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