Exercises.
1. The shortest chord which can be drawn through a given point within a circle is the perpendicular to the diameter which passes through that point.
2. Through a given point, within or without a given circle, draw a chord of length equal to that of a given chord.
3. Through one of the points of intersection of two circles draw a secant—1. the sum of whose segments intercepted by the circles shall be a maximum; 2. which shall be of any length less than that of the maximum.
4. Three circles touch each other externally at A, B, C; the chords AB, AC of two of them are produced to meet the third again in the points D and E; prove that DE is a diameter of the third circle, and parallel to the line joining the centres of the others.
PROP. XVI.—Theorem.
1. The perpendicular (BI) to the diameter (AB) of a circle at its extremity (B) touches the circle at that point. 2. Any other line (BH) through the same point cuts the circle.
Dem.—1. Take any point I, and join it to the centre C. Then because the angle CBI is a right angle, CI2 is equal to CB2 + BI2 [I. xlvii.]; therefore CI2 is greater than CB2. Hence CI is greater than CB, and the point I [note on I., Def. xxxii.] is without the circle. In like manner, every other point in BI, except B, is without the circle. Hence, since BI meets the circle at B, but does not cut it, it must touch it.
2. To prove that BH, which is not perpendicular to AB, cuts the circle. Draw CG perpendicular to HB. Now BC2 is equal to CG2 + GB2. Therefore BC2 is greater than CG2, and BC is greater than CG. Hence [note on I., Def. xxxii.] the point G must be within the circle, and consequently the line BG produced must meet the circle again, and must therefore cut it.