Apollonius Theorem
**A**_{k}(A,B) is a line if k is 1, and a circle if k is not 1.- If
**A**_{k}(A,B) is the circle**C**with centre C and radius r, then- A, B and C are collinear, with A and B on the same side of C, and
- CA.CB = r
^{2}.
- If A, B, and the circle
**C**satisfy (1) and (2), then**C**, is in**A**(A,B).
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ProofWe choose the perpendicular bisector of AB as the y-axis, and the line AB as the x-axis, so if A is (a,0), then B is (-a,0).
Then P(x,y) lies on
If k = 1, this gives the C : x^{2} + y^{2} + 2ax(k^{2}+1)/(k^{2}-1) + a^{2} = 0.
For brevity, put K = (k ^{2}+y^{2} = a^{2}(K^{2}-1),^{2} = a^{2}(K^{2}-1). Thus, C lies on the x-axis, i.e. on AB, and, as |K| > 1, lies outside [A,B]. Also CA.CB = (a-aK)(-a-aK)=a ^{2}(K^{2}-1) = r^{2}.This proves the second part.
Finally, suppose that A, B and the circle |

apollonian families |