1131 
(DA) asa, Fha, (1 + A) b=}, + Ab, (1 + Wee, 4-ae, 
From this we find for the images 
vd, Vs 2 
tn Vy Yan eg tee 
A pencil of circles is therefore represented by a straight line. 
Its intersections with G are the images of the point-circles of the 
pencil. The point at infinity of the line represents the axis of the 
pencil. 
A tangent at G is the image of a pencil of circles of which the 
limiting points have coincided; any two points of a tangent are 
therefore the images of two touching circles. 
This may be confirmed as follows. Let d be the distance of the 
centres of two circles with radii rand 7'; we have then d= r + r' or 
V(a—a’)? + (bb =V ae Hb — ce 4 Va? — bd. 
After some reduction we find for the images 
z+ 2'\? 
: )=e Hayter + y? — 2), 
which relation expresses that the images lie on a tangent of (@. 
Wid 2! 
wv 
~§ 3. A net of circles is represented by C, + 4C,-+ uC, = 
From this it ensues for the images 
(l4+ 2+ u)e=—2a, + de, + ua, etc. consequently 
| 
2 
A a A 
Re ees OSE Ss geen e 
A net of circles ús therefore represented by a plane. 
Plane sections of G have circles as horizontal projections. For 
the section of a+ y? =z with e= ar + By 4 y has as projection 
the figure represented by «# + 4° — aa — By — y= 0. 
The point-circles of a net of circles lie therefore on a circle; this 
proposition is reversible. 
The net that corresponds to z == ar + 8y + y, has as equation 
X*? + Y? — 2aX — 2bY + (aa + Bb + y)=0, 
where a and h are variable parameters. If we write for this 
X?+ Y?+a(a—2X)+b0(@—2Y)4+y=—0, 
it appears that all circles have in the point Gea, 4) equal power 
4 (a? +3) Hy; this point is the centre of fhe circle that 
contains the point-circles of the net. 
To a tangent plane of G corresponds a net of circles that’ pass 
through a fixed point. For, to 27,7 + 2y,y=z-+ 2, corresponds a 
