10750An ESSAY
Point R, wherein it is cut by the Line m R per-
pendicular to m p. Aſſume A S = e, OQ = x,
and M P = y. Then in the ſimilar Triangles O q m,
and m R D, we have,
O q (x): q m (e + y): : m R (d x) R D: ({edx + ydx/x})
The ſimilar Triangles O p q and p T D, give,
oq (x): q p (e): : p P (d x): P T ({edx/x})
P R is equal to y + dy, and if P T ({edx/x}) be added
to it, and then from the Aggregate be taken R D
({edx + ydx/x}) we ſhall have T D = y + dy - {ydx/x}
pendicular to m p. Aſſume A S = e, OQ = x,
and M P = y. Then in the ſimilar Triangles O q m,
and m R D, we have,
O q (x): q m (e + y): : m R (d x) R D: ({edx + ydx/x})
The ſimilar Triangles O p q and p T D, give,
oq (x): q p (e): : p P (d x): P T ({edx/x})
P R is equal to y + dy, and if P T ({edx/x}) be added
to it, and then from the Aggregate be taken R D
({edx + ydx/x}) we ſhall have T D = y + dy - {ydx/x}
Now to find the Points of Interſection of the two
Circles, whoſe Radii are T D and P M, and Centers
diſtant from each other, by the Space T P, the Square
of T D leſs the Square of P M muſt be divided 1165. Twice P T, and then half of P T muſt be taken there-
from, which may be here neglected, becauſe it is infinite-
ly ſmall in compariſon of the reſt; and we ſhall have
{xydy/dex} - {yy/e} for the Part of the Line PM included be-
tween P and the Point wherein this Line is cut by a
Line joyning the two Points of Interſection of the two
Circles.
Circles, whoſe Radii are T D and P M, and Centers
diſtant from each other, by the Space T P, the Square
of T D leſs the Square of P M muſt be divided 1165. Twice P T, and then half of P T muſt be taken there-
from, which may be here neglected, becauſe it is infinite-
ly ſmall in compariſon of the reſt; and we ſhall have
{xydy/dex} - {yy/e} for the Part of the Line PM included be-
tween P and the Point wherein this Line is cut by a
Line joyning the two Points of Interſection of the two
Circles.
Now before what I have here demonſtrated be ap-
ply’d to the Problem, we muſt obſerve, that if from
the Point M, a Line be drawn thro’ the Center B,
the Triangles M P C and m R M will be ſimilar;
for the Angle m M P is the Exterior Angle of the
Triangle m R M, and the Angle m M C is a right
one. And conſequently
m R (dx): R M (d y): : M P (y): P C ({ydx/dx}).
ply’d to the Problem, we muſt obſerve, that if from
the Point M, a Line be drawn thro’ the Center B,
the Triangles M P C and m R M will be ſimilar;
for the Angle m M P is the Exterior Angle of the
Triangle m R M, and the Angle m M C is a right
one. And conſequently
m R (dx): R M (d y): : M P (y): P C ({ydx/dx}).
Now if S A = s a in the 32d and 33d Figures
22Fig. 32,
33. be repreſented by e in this Computation; as
22Fig. 32,
33. be repreſented by e in this Computation; as
zoom in
zoom out
zoom area
full page
page width
set mark
remove mark
get reference
digilib