1915-16.] The “ Geometria Organica ” of Colin Maclaurin. 109 
Draw PT parallel to 0 1 0 2 cutting 0 2 Q in T. Describe the semicircle 
OjROa. Then PO^Q is a rectangle, and PO^T is a parallelogram. 
Hence 
QR = Oj_P — 0 2 T, 
so that 
TQ = 0 2 R. 
Now T traces out a line V. 
Thus, to get the locus of Q, take T any point on V and let the circle 
determine the chord 0 2 R on 0 2 T. Produce 0 2 T to Q so that TQ = 0 2 R. 
Cor. 2. The same results obtain if on 0 1 0 2 is described a segment of a 
circle instead of a semicircle. 
Cor. 3 gives another method of generating the curve. 
Fig. 22. 
On 0 1 0 2 describe the semicircle 0 1 R0 2 . Draw 0 2 D at right angles to l. 
Then 0 2 RDPQ lie on a circle whose diameter is 0 2 P ; and PR0 2 Q is a 
rectangle. Also RDQ = R0 2 Q = 
Hence rotate two right angles RDQ and R0 2 Q round the two fixed 
points D and 0 2 , and let R trace out the semicircle, when Q generates 
the cubic. Cf. § 39. 
Cor. 4 contains a generalisation of Cor. 3, as Cor. 2 is of Cor. 1. 
Let 
OjPQ = a, 0 2 QP = p. 
Describe a circle round 0 2 QP cutting l in D, and O x P again in R. D 
is a fixed point, and 0 1 R0 2 = 7r — /3, so that R generates a segment of a 
circle on 0 1 0 2 . 
