The Astrological Symbols were also employed by the Alchemists to indicate the seven metals then known. PLATE III.--SQUARING THE CIRCLE. Mr. James Smith, of Liverpool, the most laborious among recent workers in this field of enquiry, claiming to have propounded several simple and exact methods, offers the following as sufficiently demonstrative:-- I construct my diagrams in the following way:--I draw two straight lines at right angles, making O the right angle. From the point O, in the direction OA, I mark off four equal parts together equal to OA, and from O, in the direction of OB, I mark off three of such equal parts together, equal to OB, and join AB. It is obvious, or rather self-evident, that AOB is a right-angled triangle, of which the sides that contain the right angle are in the ratio of 4 to 3, by construction. With A as centre and AB as interval, I describe the circle X, produce AO and BO to meet and terminate in the circumference of the circle at the points G and C, and join AC, CG, and BG, producing the quadrilateral ACGB. I bisect AG at F, and with O as centre, and OF as interval, describe the circle Z. The line OF is the line that joins the middle points of the diagonals in the quadrilateral ACGB; and it follows that, {AG^2 + CB^2 + 4(OF^2)} = {AC^2 + CG^2 + BG^2 + AB^2.} When AO = 4, we get the following equation:-- {5^2 + 6^2 + (4 × 1'5^2)} = {5^2 + sqrt(10^2) + sqrt(10^2) + 5^2,} or, {25 + 36 + 9} = {25 + 10 + 10 + 25} = 70. From the points B and C, I draw straight lines at right angles to AB and AC, and therefore tangential to the circle X, to meet AG produced at D, and join BD and CD, producing the quadrilateral ACDB. I bisect AD at E, and with O as centre, and EO as interval describe the circle XY, and with E as centre, and EA or ED as interval describe the circle Y. Now, to square the circle, or, in other words, to get exactly equal in superficial area to the circle X, I will show how to find it. From the point G draw a straight line--say G _m_--perpendicular to ED, making G _m_ equal GD. Produce GA to a point _n_, making G _n_ equal to 2AG - GD, and join _n m_. The square on _n m_ will be the required square. (I have indicated this square by dotted lines.) For example:--If AO = 4, then AG = 5, and GD = 1'25; therefore {2 AG - GD} = {10 - 1'25} = 8'75 = Gn: and Gm = 1'25; therefore, Gn^2 + Gm^2 = 3-1/8 (AB^2); that is, {8'75^2 + 1'25^2} = 3-1/8 (5^2), or, {76'5625 + 1'5625} = {3'125 × 25}; and this equation=Area of the Circle X; and area of the square on _n m_ :: and it follows, that the area of every circle, is equal to the area of a square on the hypotenuse of a right-angled triangle, of which the sides that contain the right angle are in the ratio of 7 to 1, and the sum of these two sides equal to the diameter of the circle. In many ways I have proved this fact, by practical or constructive geometry. PLATE IV.