write the equation of the circle of the triangle with vertices $$A = (5 ,\ -4 ),\ B = (6 ,\ -1 ),\ C = ( 2,\ 3)$$ examine the relative position of this district and its image in axial symmetry about the line $$3x + 4y + 26 = 0$$
I think about this: (drawing paper)
Substitute $(A,B,C)$ into $(x-a)^2 + (y-b)^2 = R^2$ : $$ (5-a)^2 + (-4-b)^2 = R^2 \\ (6-a)^2 + (-1-b)^2 = R^2 \\ (2-a)^2 + (3-b)^2 = R^2 $$ $$ a^2 - 10 a + 25 + b^2 + 8 b + 16 = R^2 \\ a^2 - 12 a + 36 + b^2 + 2 b + 1 = R^2 \\ a^2 - 4 a + 4 + b^2 - 6 b + 9 = R^2 $$ Subtract the second equation from the first one and the third equation from the second one: $$ 2 a - 11 + 6 b + 15 = 0 \\ - 8 a + 32 + 6 b - 3 = 0 $$ $$ 2 a + 6 b = - 4 \\ - 8 a + 8 b = - 24 $$ Add the second equation to $4 \times$ the first one .. and do the rest: $$ 32 b = - 40 \quad \Longrightarrow \quad b = - 5/4 \\ 2\cdot 2 a = - 8 + 15 = 7 \quad \Longrightarrow \quad a = 7/4 \\ R^2 = (2-a)^2 + (3-b)^2 = 145/8 $$ So the circle is: $$ (x-7/4)^2 + (y+5/4)^2 = \left(\sqrt{145/8}\right)^2 $$
Still don't know what you mean by "district" ..