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greater of the given circles the line EH = the difference of the Radii of the given
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circles, and with A and B as Foci and EH Tranverſe Axis, let two oppoſite
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Hyperbolas be deſcribed KEK and LHL: then I ſay that KEK will be the
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locus of the centers of all the circles which can be drawn ſo as to be touched
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outwardly by both the given circles; and LHL will be the locus of the centers
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of all the circles which can be drawn ſo as to be touched inwardly by both the
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given circles. For taking any point K in the Hyperbola KEK, and drawing KA
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and KB, let theſe lines cut the given circumferences in F and G reſpectively: and make KQ = KA: then from the nature of the curve QB = EH, but by con-
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ſtruction EH = BG - AF, therefore QB = BG - AF and hence QG = AF, and
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then taking equals from equals KG = KF, which is a demonſtration of the 1ſt
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Caſe.
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<span class="caps" style="font-size: 75%">Then</span>
with regard to the 2d Caſe, taking any point L in the Hyperbola
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LHL, and drawing LB and LA and producing them to meet the concave cir-
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cumferences in M and N, let alſo LR be taken equal to LB; then from the pro-
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perty of the curve AR = EH, but EH (by conſtruction) = BM - NA; there-
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fore AR = BM - NA, and NR = BM, and then adding equals to equals LN =
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LM, which is a demonſtration of the 2d Caſe.
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<span class="caps" style="font-size: 75%">Case</span>
3d. Suppoſe it be required that the circles to be deſcribed be touched
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outwardly by one of the given circles, and inwardly by the other.
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Then drawing AB, let it cut the convex circumferences in C and D, and the
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concave ones in P and O, biſect PD in E, and from E towards B ſet off EH =
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the ſum of the given radii. Then with A and B foci and EH tranſverſe axis,
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let two oppoſite hyperbolas be deſcribed KEK and LHL: and KEK will be the
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locus of the centers of the circles which are touched inwardly by the circle A
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and outwardly by the circle B; and LHL will be the locus of the centers of
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thoſe circles which are touched inwardly by B and outwardly by A. The demon-
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ſtration mutatis mutandis is the ſame as before.
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<span class="caps" style="font-size: 75%">Case</span>
4th. Suppoſe the given circle A to include B, and it be required that
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the circles to be deſcribed be touched outwardly by them both.
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Let AB cut the circumferences in C and D, P and O: and biſecting CD in
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I, and ſetting off from I towards P, IL = the ſum of the ſemidiameters of the
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given circles, and with A and B foci, and IL tranſverſe axis, deſcribing an
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ellipſe LKI, it will be the locus of the centers of the circles required. For
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taking any point K in the ellipſe, and drawing AK and BK, let AK be con-
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tinued to meet one of the given circumferences in G, and let BK meet the other
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F. Then from the property of the curve AK + BK = IL = AG + BF (by con-
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