A sequel to the first six books of the Elements of Euclid, containing an easy introduction to modern geometry, with numerous examples. By John Casey.

BOOK VI.

109

chord which subtends a right Z at a fixed point P;
AE, BE tangents at A and B, then E is the pole of
AB: it is required to find the locus of E. Let 0 be
the centre of X. Join OE, intersecting AB in I; then,
denoting the radius of X by r, we have OI2 + AI2 = r2;
but AI = IP, since the Z APB is right;.0. OI2 + IP2
= r2;.. in the A OIP there are given the base OP in
magnitude and position, and the sum of the squares of
OI, IP in magnitude. Hence the locus of the point I
is a ( (Prop. 2, Cor., Book II.). Let this be the 0
INR. Again, since the Z OAE is right, and AI is L
to OE, we have OI. OE = OA2 = r2. Hence the point
E is the inverse of the point I with respect to the (
X; and since the locus of I is a 0, the locus of E will
be a circle (see Prop. 1).
Prop. 14.-If two circles, whose radii are R, r, and
distance between their centres 8, be such that a quadrElateral inscribed in one is circumscribed about the other;
then
1 1 1
(R + a)2 + (R - 8)2 = r2'
Dem.-Produce AP, BP (see last fig.) to meet the
O X again in the points C and D; then, since the
chords AD, DC, CB subtend right s at P, the poles
of these chords, viz., the points H, G, F, will be points
on the locus of E; then, denoting that locus by Y, we
see that the quadrilateral EFGH is inscribed in Y and
circumscribed about X. Let Q be the centre of Y; then
radius of Y = R, and OQ = 8. Now, since N is a point
on the locus of I (see Dem. of last Prop.), ON2 + PN2
= r2; but PN = OR;.-. ON2 + OR2 = r2. Again, let
OQ produced meet Y in the points L and M; then
L and M are the inverses of the points N and R with
respect to X. Hence
ON. OL = r2; that is ON.(R + ) = r2
r2
therefore ON = + i.
1~+ T '