Найдите точку, равноудаленную от точек A (6,7), B (3,3), C (1,-5) и вычислите это...

Решите задачу:

$A(6;\ 7);\ B(3;\ 3)\\\\k:\\(x-6)^2+(y-7)^2=(x-3)^2+(y-3)^2\\x^2-12x+36+y^2-14y+49=x^2-6x+9+y^2-6y+9\\-12x-14y+85=-6x-6y+18\\-14y+6y=-6x+12x+18-85\\-8y=6x-67\ \ \ \ /:(-8)\\\\\underline{y=-\frac{3}{4}x+\frac{67}{8}}$

$B(3;\ 3);\ C(1;-5)\\\\l:\\(x-3)^2+(y-3)^2=(x-1)^2+(y+5)^2\\x^2-6x+9+y^2-6y+9=x^2-2x+1+y^2+10y+25\\-6x-6y+18=-2x+10y+26\\-6y-10y=-2x+6x+26-18\\-16y=4x+8\ \ \ \ /:(-16)\\\\\underline{y=-\frac{1}{4}x-\frac{1}{2}}$

$-\left\{\begin{array}{ccc}y=-\frac{3}{4}x+\frac{67}{8}\\y=-\frac{1}{4}x-\frac{1}{2}\end{array}\right\\--------------\\.\ \ \ \ \ 0=-\frac{2}{4}x+\frac{67}{8}+\frac{4}{8}\\\\.\ \ \ \ \frac{1}{2}x=\frac{71}{8}\ \ \ \ /\cdot2\\\\.\ \ \ \ x=\frac{71}{4}\\\\y=-\frac{1}{4}\cdot\frac{71}{4}-\frac{1}{2}=-\frac{71}{16}-\frac{8}{16}=-\frac{79}{16}\\\\O:\left(\frac{71}{4};-\frac{79}{16}\right)$