36 s Dr, R. Hey's Propositions containing 
perpendicular, as at H or K, the circle is found from the pro- 
portion, DK to KV as DH to HV. 
Cor. 6. Fig. 7. If two circles, SQT and HFK, have the 
same vertex V to a given base LR ; they are not concentric, 
and one includes the other and does not touch it. Thus : if E 
be the centre of SQT, and G of HFK, and EV be greater 
than GV ; SQT includes HFK, and does not touch it. For, let 
VDbe the perpendicular on LR. Draw VQ parallel to LR, 
cutting SQT in Q and HFK in F. Draw DQ, DF : which are 
tangents. Then, if any point X were the common centre of 
the circles, and XF and XQ were drawn, both XFD and XQD 
would be right angles : which is impossible. E and G are * 
in DV produced : which line will cut SQT first in a point S, 
then in a point T ; and in like manner cut HFK, in H and K. 
Then DK is to KV as DH to HV, and DT to TV as DS to 
SV. But DK is less than DT ; whence the ratio of DK to KV 
is greater than of DT to TV. Therefore that of DH to HV 
is greater than of DS to SV ; whence SV is greater than HV. 
A fortiori the whole of SQT falls without HFK. 
Prop. VII. Fig. 7. Problem. Two circles, not concentric, 
being given, of which one includes the other and does not 
touch it ; to find a base, having a vertex common to the two 
circles. 
Solution, Let E be the centre of the outer circle SQT, and 
G that of the inner HFK. Draw EG indefinite ; cutting the 
circles in H and S beyond G, in K and T beyond E. Find a 
right line A equal -f to divide it X 
* Cor. 1 to prop. I. f See 1 Eucl. prop. 47 ; and VI Eucl. prop. 17 and 1 1. 
X Cut it perpendicularly by a chord equal to 2GH, in a circle on the diameter A. 
