COR. I. If the given point be not on the Oce, then that straight line, which if produced would pass through the centre, is the shortest. For if the given point A be within the O, then CA, AS are together and .:. > CD; .. AS is > AD; CS, and if the given point A be without the O, .. AS is > AD. COR. 2. To every one of the straight lines, on one side of that through the centre, there can be drawn one, and only one, equal to it on the other side. For the figure may be turned about the line through the centre till the Oce on one side of that line shall fall on the former position of the Oce on the other side; and .. to all the straight lines AR, AS &c. on one side of the line through the centre there will be equal straight lines on the opposite side. Also there cannot be two equal straight lines on the . same side. COR. 3. If the distance between the centres of two Os be > the sum or < the difference of their radii, the s will not meet. If the distance be = the sum or difference, the s will meet in one point only. If the distance be < the sum and > the difference the OS will meet in two points. But if If the Os are equal, the construction is easy. not, let A be the centre of the smaller and B of the greater. With centre B and radius = the difference between the radii of the given Os describe a o PQ. From A draw a tangent AP to o PQ. Join BP and produce it to T; draw AS || to BT; join ST. Then ST shall be the tangent required. ... AS is and I to PT, .. AT is a , (1. 23) = and . TPA is a right 4, .. AT is a rectangle; (1. 27) 4s at S and Tare rights; ... ST touches the Os RS, TV. (III. 6) We may also draw a common tangent to RS, TV thus: with centre B describe a HK whose radius is = the sum of the radii of the given circles. From A draw a tangent AH to HK. Join BH cutting TV in T; draw AR to BH; join RT. Then RT shall be a common tangent to RS, TV. For AR is and || to HT, .. HR is a □, = and AHT is a right ; (III. 6) ... HR is a rectangle, and ... ▲ s at T, R are right <s; .. RT touches the given s. (III. 6) Proofs of the two following theorems depending on the first three Books are here given; but more elegant demonstrations will be found in Book VI. THEOREM (d). If two chords of a circle intersect, the rectangles contained by their segments are equal. Let the two chords PQ, RS of the PRQ intersect in X; then shall rectangle (PX, XQ) be rectangle (RX, XS). = If PQ, RS intersect in the centre of the -O: then PX, XQ, RX, XS are all equal; and rectangle (PX, XQ) is = rectangle (RX, XS). If one of them, as PQ, pass through the centre C, and R S P cut the other RS which does not pass through the centre at right angles, and accordingly bisect RS, = join CR. Then PX is the sum and XQ= the difference of CX and CQ; .. rectangle (PX, XQ) + square on CX= square on CQ, (II. 2) and ... = square on CR, and... .= squares on CX,XR; (1. 35) .. rectangle (PX, XQ) = square on XR, i. e. = rectangle (RX, XS). If one of them, as PQ, pass through the centre and cut the other RS which does not pass through the centre, but not at right angles, from Clet fall CH1 to RS, and join CR. Then rectangle (PX, XQ) + square on CX= square on CQ; (II. 2) .. rectangle (PX, XQ) + squares on CH, HX = square on CR; and squares on CH, HR; (1. 35) .. rectangle (PX, XQ) + square on HX = square on HR; but rectangle (RX, XS) + square on HX=square on HR, ... rectangle (PX, PQ) = rectangle (RX, XS). Lastly, if neither of them pass through the centre, R B (II. 2) P through X draw the diameter ACXB. Then rectangle (PX, XQ) is = rectangle (AX, XB), and... rectangle RX, XS. = |