G E O Ivl 
a’ifo be above it, fmce tlicfe pot-nts miifl be in the cii- 
cles L and N, and inlinitcly near to M ; and if L is be¬ 
low IMK, N, for tlic fame rcafon, mull be below it ; 
and Ivl is common to the circle and tlie curv^e; rhere- 
fbre tite curve touches the circle IMK at M, and is per- 
piendiciilar to the radius PM. 
Prop. 11. Prohltm.-—To draw a tangent to a cycloidal 
curve at any given poitu. —On tlie given point as a centre, 
deferibe a circle equal to the defci ibing circle of tl;e 
curve, and from the interfeftion of this circle, with the 
line deferibed by the centre of the generating circle, let 
fail a perpendicular on the bafts; the point thus found 
will be the point of coiitadb, and the tangent will be 
perpendictilar to the right line joining tliis point of 
contact and the given point, by the firft projiofition. 
It will be obvious, from infpeCtion, which of the two 
interfeClions of the circle to be deferibed, with the 
traCt of tlie centre, is to be taken as the place of tluit 
centre corre(ponding to the given point. 
pR OP. Ill, Problem. —Tofind the length of an cpitroclwid .— 
Let C be tJie cen¬ 
tre of the bafis VP, 
K that of the rota¬ 
ting circle PR, and 
of the defci ibing 
circle GL, P the 
point of contact, 
and M the deferib- 
ing point. Then 
joining MXC, and 
fiippoling V X to 
be an element rc- 
prefenting the mo¬ 
tion of the point P 
in either the balis 
or the generating 
circle, draw the arc 
MN on tlie centre 
C, and join CVN, 
then NM will re- 
prefent the motion 
of the point M as 
far as it is produced 
by the revolution round the centre C : take MO to VX 
as GK to PK, then MO will be the motion of M ariliug 
from tlie revolution round K, and NO will be the ele- 
ifieiit of the curve produced by the joint motion. Let 
CH be parallel to PM, then CX or CP : CM :: VX : MN, 
and PK : MK :: CP ; HM :; VX : MO, therefore CM : 
HM :: MN : MO, and tliefe lines being perpendicular to 
CM, HM, the triangle NMO is fimilar to CMH, and 
MN ; NO CM ; CH, hence CP : CH :: VX : NO. 
Take PY to CP as PK to CK, then CH ; CP :: PM ; 
PY NO : VX.' On L deferibe the circle PFB, and 
draw IMLF : let F'D be perpendicular to P-RB, take 
DE to DF, as PL, and E will be always in the elliplis 
EEP: let AE and AF be tangents to tlie ellipfis and 
circle at E and F; then the increment of the arc BF 
will be to MO as PL to GL, and to VX as PL to PR. 
Join GM, and parallel to it draw PI ; tlien PJL is a 
right angle, and ILP— AED, and IM : IL ;: PG : PL :: 
DE : DF, by conltruction; therefore the figure IPML 
is fimilar to DAEF, as PL to PM fo is AF to AE, and 
fo is tlie increment of the arc BF to that of BE ; but the 
increment of BF is to VX as PL to PR, therefore the 
increment of BE is to VX as PM to PR. Now, it was 
proved that NO : VX :: PM : PY ; therefore the incre¬ 
ment of BE is to NO as PY to PR, or as CP to 
aCK: and the whole elliptic arc BF) is to the whole 
as the radius of the bafis to twice tlie dillanCe of 
the centres. 
Corollary i. The fluxion of every cycloidal arc is pro¬ 
portional to the diftance of the delcribing point from the 
point of contat.t- 
CoroUary 2. In the epicycloid the ellipfis coincides 
with its axis B P, and tlie arc BE witli B D, which is 
double the verfed fine of half the arc G M, in the de- 
V«J.. Vni.No. 515. 
E R Y. . 4 u 
feribing or genciwting circle; ih 'refore the length (d 
the curve is to this verfed line as four times the dillanc-* 
of t!ic centres to the I'adius of t!;e ba.tis. 
Pro?. IV'. Problem.—To frnd the centre of ciirvatnre of an 
cpitrochoid. —Let PY be, as in tlie iaii j roi ofiiiun, to C.P 
as PK to CK, .and on 
tlie diameter PY dc- 
fcrlbe the circle PZ^', 
cutting PO ia Z : take 
OW a third propor¬ 
tional to OZ and OP, 
and '.V will be the 
ceulrc of curvature. 
For, let QP—V'X be 
the fpace deferibed 
by P, while NO is de- 
Icribed b\' O; it is obvious from Prop, r, that tlie in- 
terfedtion of NQ and OF mull be the centre of curva¬ 
ture. Let Qr be perpendicular to FA parallel to QNj 
then, by Prop. .1, NO : VX or QP ::,PO : PY, but by 
fiiiiihir triangles QP : CF PY : PZ ; tlicretore NO ; 
Qr :: PO : PZ, and by divifion, NO : aO :: OP : OZ, 
and by liniilar triangles OW : OF or OP. 
Corollary. Y.’lien Z coiiicide.s tvith O or M, OW is i.n- 
fiiiite ; therefore whenever PZY interfetts the defenb- 
ing circle, the epitiochold will liav.e a point ofcoarraty 
flexure at tlie f.ime diflaiice from C as this intcrfectioi:. 
and the circle P Z Y is given when tlie bafis and gene¬ 
rating circle are given, whatever the magnitude ot tiic 
delcril-'ing circle may he. 
Prop. V. Problem.—Tojindlheevolutecf art epicycloid .—> 
In tlie epicycloid SM, 
the point M being in 
the circumference of 
PMR, PZ will be to 
PM in the coiillaiit 
ratio of PY to PR, 
and MZ to PM as 
RY to PR ; arid PM 
to MW in tlie fame 
ratio ; iience PM : 
PW :: RY : PY :: 
CR : CP, therefore 
the poiiitW is always 
in a circle PWs of 
wiiicli the radius is 
to PK in that proportion, and which touciics SP in P. 
On tlie centre C deferibe a circle AH© toucliing PWs 
in H ; then, fince CR : CP :: PR : PH, we iiave by di¬ 
vifion CR : CP :: CP ; P3, and the circle PW3 being 
to AH© as PMR to SP, the arc PM being equal to SP, 
the fimilar arc PW will be equal to AH, and, taking AH 0 
rr PWs, H© will be always equal to hW, and W in a 
curve ©WS fimilar to SM, of which it is the evolute. 
Prop. VI. Problem.—To find the area of an epitrcchoid ,— 
On the centre C de¬ 
feribe a circle touch¬ 
ing the epitroclioid in 
S, take GH to GC as 
PR to PC, and let 
the circle GCiH de- 
Icribe on the balis SG 
the epicycloid S< 1 ). 
Then taking GM al¬ 
ways to GrJ) as G L to 
GH, M will be in ilie 
epitroclioid SM ; for 
the angular motion of 
the chord G(!i, is tiie 
lame as that of GXf 
in the primary epitro- 
choid. LetSfibethe 
evolute of S< 1 >, and 
GWs its generating 
circle. On diameters 
equal to hG, sL, .and 
HH, defciibe three circles, AD, AE, and AF, touching; 
5 U , ' lliv 
