OK. The reason I recommended using jonah2.0's method after your post #8 is that you had used the formula [MATH]r^2\left(k^2+1\right)=\left(kp-q+n\right)^2[/MATH], rather than jonah2.0's bisector line method. I know they are essentially the same, however your method gave you 2 possible solutions [MATH]\mathrm{I}[/MATH], [MATH]\mathrm{II}[/MATH] (in your work).
Anyway, if we work with that, in [MATH]\mathrm{II}[/MATH] you made a slip -40 instead of +40, so you should get:
[MATH]8p-16q+88=0[/MATH][MATH]p-2q+11=0[/MATH]and this happens to be the solution you want to work with.
(The centre [MATH](p,q)[/MATH] lies on the bisector line. In your solution [MATH]\mathrm{I}\text{, } 6p+3q-4=0[/MATH] represents the bisector line [MATH]6x+3y-4=0[/MATH]. In your solution [MATH]\mathrm{II}\text{, } p-2q+11=0[/MATH] represents the bisector line [MATH] x-2y+11=0[/MATH]. From the picture in post #2, clearly [MATH]x-2y+11=0[/MATH] is the correct bisector line we want - looking at the 2 straight lines and the position of the point (1,6).)
Anyway, now you have [MATH]p=2q-11[/MATH] *, so the centre is [MATH](2q-11,q)[/MATH] and:
[MATH]\left(\left(2q-11\right)-1\right)^2+\left(q-6\right)^2=\frac{\left(2\left(2q-11\right)-q+8\right)^2}{5}[/MATH] ** by the second part of jonah2.0's method,
which easily gives two values for [MATH]q[/MATH], then * gives corresponding values for [MATH]p[/MATH] and finally either side of equation ** gives [MATH]r^2[/MATH].
brilliant.org
