228 Vector and complex-number methods Ch. 11
It will also meet the nine-point circle at the point
z 22 =z′ 16 −r 1
|z 15 −z′ 16 |(z 15 −z′ 16 ).The mid-point of these two points is the centreZ′ 16 of the nine-point circle and so
[Z 21 ,Z 22 ]is a diameter of this nine-point circle.
Similarly the half-line[Z 15 ,Z 16 ′ has pointsZ=Z 15 −t(Z 16 ′ −Z 15 )and meets the
incircle at the point
z=z 15 −r 2
|z′ 16 −z 15 |(z′ 16 −z 15 ).Then clearly this is the pointZ 21 again and the corresponding point
z 24 =z 15 +r 2
|z′ 16 −z 15 |(z′ 16 −z 15 )will be a diametrically opposite point on the incircle. Then the nine-point circle and
the incircle have a common tangent at the pointZ 21.
It is clear from elementary geometry that, if two circles touch at just one point, the
centres of the two triangle and the point of tangency must be collinear. The following
is a skeleton picture of a possible situation.
Figure 11.18.Now we turn to the supposition thatr 2 ≥r 1. We first suppose thatr 2 >r 1. We treat
the incircle as above and take the point
z 24 =z 15 +r 2
|z′ 16 −z 15 |(z′ 16 −z 15 ).ThenZ 24 is a point on the incircle and the point
z 25 =z 15 −r 2
|z′ 16 −z 15 |(z′ 16 −z 15 ).is such that[z 24 ,z 25 ]is a diameter of the incircle.