15(3)
Case 5th.
Let one of the given circles include the other, and let be required
that the circle to be deſcribed be touched outwardly by one of the given circles,
and inwardly by the other.
that the circle to be deſcribed be touched outwardly by one of the given circles,
and inwardly by the other.
Limitation.
Then it’s Diameter muſt not be given greater than the greater
ſegment of the right line, joining the centers of the given circles, which is in-
tercepted between the two concave circumferences of the ſaid circles, nor leſs
than the leſſer ſegment; viz. not greater than CD, nor leſs than MN.
ſegment of the right line, joining the centers of the given circles, which is in-
tercepted between the two concave circumferences of the ſaid circles, nor leſs
than the leſſer ſegment; viz. not greater than CD, nor leſs than MN.
Case 6th.
Let the two given circles cut each other, and let it be required
that the circle to be deſcribed, and to be touched by them both, ſhall alſo be
included in each of them.
that the circle to be deſcribed, and to be touched by them both, ſhall alſo be
included in each of them.
Limitation.
Then it’s Diameter muſt not be given greater than the ſeg-
ment of the right line, joining the centers of the given circles, intercepted by
their concave circumferences, which lies in the ſpace common to both the given
circles; viz. not greater than CD.
ment of the right line, joining the centers of the given circles, intercepted by
their concave circumferences, which lies in the ſpace common to both the given
circles; viz. not greater than CD.
There may be alſo three other Caſes of this Problem, when the given circles
cut each other; but becauſe they are ſimilar to the 1ſt, 2d, and 4th Caſes already
propoſed, and ſubject to juſt the ſame Limitations; except that which is ſimilar
to the 1ſt, which is ſubject to no Limitation at all, they are here omitted; as are
likewiſe thoſe Caſes where the given circles touch each other; becauſe they are
the ſame as the preceding, and ſolved in the ſame manner.
cut each other; but becauſe they are ſimilar to the 1ſt, 2d, and 4th Caſes already
propoſed, and ſubject to juſt the ſame Limitations; except that which is ſimilar
to the 1ſt, which is ſubject to no Limitation at all, they are here omitted; as are
likewiſe thoſe Caſes where the given circles touch each other; becauſe they are
the ſame as the preceding, and ſolved in the ſame manner.
The GENERAL Solution.
Join the given centers A and B, and where the Caſe requires, let AB be pro-
duced to meet the given circumferences in C and D: and let CI and DH be
taken equal to the given line Z: and let two circles be deſcribed; one with cen-
ter A and diſtance AI, and the other with center B and diſtance BH: and theſe
two circles will neceſſarily cut or touch each other by the Limitations given. Let
the point of concourſe be E: from E draw the right line EAF cutting the circle
whoſe center is A in F; as alſo EBG cutting the circle whoſe center is B in G:
then with center E and diſtance EF deſcribe a circle FK, this will be the circle
required: becauſe AF and AC are equal as alſo AI and AE; therefore FE
and CI are alſo equal: but CI was made equal to Z, therefore FE is equal to
Z. Again, becauſe BD and BG are equal, as alſo BH and BE, therefore DH
and EG are alſo equal: but DH was made equal to Z, therefore EG is equal to
Z. Hence it appears that the circle FK, paſſing through F will alſo paſs thro’
G, and likewiſe that it will alſo touch the given circles in F and G, becauſe EAF
and EBG are right lines paſſing through the centers.
duced to meet the given circumferences in C and D: and let CI and DH be
taken equal to the given line Z: and let two circles be deſcribed; one with cen-
ter A and diſtance AI, and the other with center B and diſtance BH: and theſe
two circles will neceſſarily cut or touch each other by the Limitations given. Let
the point of concourſe be E: from E draw the right line EAF cutting the circle
whoſe center is A in F; as alſo EBG cutting the circle whoſe center is B in G:
then with center E and diſtance EF deſcribe a circle FK, this will be the circle
required: becauſe AF and AC are equal as alſo AI and AE; therefore FE
and CI are alſo equal: but CI was made equal to Z, therefore FE is equal to
Z. Again, becauſe BD and BG are equal, as alſo BH and BE, therefore DH
and EG are alſo equal: but DH was made equal to Z, therefore EG is equal to
Z. Hence it appears that the circle FK, paſſing through F will alſo paſs thro’
G, and likewiſe that it will alſo touch the given circles in F and G, becauſe EAF
and EBG are right lines paſſing through the centers.
PROBLEM IV.
Having a given point A, and a given right line BC, it is required to draw a
circle, whoſe Radius ſhall be equal to a given line Z, which ſhall paſs through
the given point, and alſo touch the given line.
circle, whoſe Radius ſhall be equal to a given line Z, which ſhall paſs through
the given point, and alſo touch the given line.
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