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(D) The intersection of the perpendicular bisectors is the circumcenter $O$. Solving $x-3y-5=0$ and $x+5y-9=0$ gives $O(\frac{13}{2}, \frac{1}{2})$.Si

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$4\sqrt{2}$

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(D) The intersection of the perpendicular bisectors is the circumcenter $O$. Solving $x-3y-5=0$ and $x+5y-9=0$ gives $O(\frac{13}{2}, \frac{1}{2})$.Since $O$ is the circumcenter,$OA=OB=OC$.$OA^2 = (\frac{13}{2}-1)^2 + (\frac{1}{2}-2)^2 = (\frac{11}{2})^2 + (-\frac{3}{2})^2 = \frac{121+9}{4} = \frac{130}{4} = \frac{65}{2}$.Since $O$ is the circumcenter,$OB^2 = OA^2 = \frac{65}{2}$.Let $B=(x_1, y_1)$. Since $B$ lies on the perpendicular bisector of $AB$,$OB=OA$. Also,$B$ is the reflection of $A$ across the line $x-3y-5=0$.The line $AB$ is perpendicular to $x-3y-5=0$,so its equation is $3x+y+k=0$. Since $A(1,2)$ lies on it,$3(1)+2+k=0 \Rightarrow k=-5$. So $AB: 3x+y-5=0$.The intersection of $AB$ and its perpendicular bisector is the midpoint $P$. Solving $x-3y-5=0$ and $3x+y-5=0$ gives $P(2,-1)$.Since $P$ is the midpoint of $AB$,$2 = \frac{1+x_1}{2} \Rightarrow x_1=3$ and $-1 = \frac{2+y_1}{2} \Rightarrow y_1=-4$. Thus $B(3,-4)$.Since $O$ is the circumcenter,$OC=OA$. $C$ is the reflection of $B$ across the line $x+5y-9=0$.The line $BC$ is perpendicular to $x+5y-9=0$,so its equation is $5x-y+k'=0$. Since $B(3,-4)$ lies on it,$5(3)-(-4)+k'=0 \Rightarrow k'=-19$. So $BC: 5x-y-19=0$.The intersection of $BC$ and its perpendicular bisector is the midpoint $Q$. Solving $x+5y-9=0$ and $5x-y-19=0$ gives $Q(4,1)$.Since $Q$ is the midpoint of $BC$,$4 = \frac{3+x_2}{2} \Rightarrow x_2=5$ and $1 = \frac{-4+y_2}{2} \Rightarrow y_2=6$. Thus $C(5,6)$.The length $AC = \sqrt{(5-1)^2 + (6-2)^2} = \sqrt{4^2+4^2} = \sqrt{32} = 4\sqrt{2}$.

Let $ABC$ be a triangle and $A=(1,2)$. If $x-3y-5=0$ and $x+5y-9=0$ are the perpendicular bisectors of the sides $AB$ and $BC$ respectively,then the length of the side $AC$ is

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