${{\left( x-6 \right)}^{2}}+{{\left( y-6 \right)}^{2}}=9,{{x}^{2}}+{{y}^{2}}-2x-2y+2-{{a}^{2}}=0$ 

${{C}_{1}}:{{x}^{2}}+{{y}^{2}}-2x-2y+2-{{a}^{2}}=0$ 

${{O}_{1}}\left( 1,1 \right),{{R}_{1}}=\frac{1}{2}\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -2 \right)}^{2}}-4\left( 1-{{a}^{2}} \right)}=\left| a \right|$ 

${{C}_{2}}:{{\left( x-6 \right)}^{2}}+{{\left( y-6 \right)}^{2}}=9$ 

${{O}_{2}}\left( 6,6 \right),{{R}_{2}}=3$ 

${{O}_{1}}{{O}_{2}}=\sqrt{{{\left( 6-1 \right)}^{2}}+{{\left( 6-1 \right)}^{2}}}=\sqrt{50}=5\sqrt{2}$

مطابق شكل هر كدام از دو چهار ضلعی ايجاد شده، يك مربع است و در نتيجه طول قطر آن، $\sqrt{2}$ برابر طول ضلع آن است. داريم:

\[\begin{matrix}    {{O}_{1}}M=\sqrt{2}{{R}_{1}}  \\    {{O}_{2}}M=\sqrt{2}{{R}_{2}}  \\ \end{matrix}\left. \begin{matrix}    {}  \\    {}  \\ \end{matrix} \right\}\Rightarrow {{O}_{1}}M+{{O}_{2}}M=\sqrt{2}\left( {{R}_{1}}+{{R}_{2}} \right)\] 

$\Rightarrow \left| {{O}_{1}}{{O}_{2}} \right|=\sqrt{2}\left( {{R}_{1}}+{{R}_{2}} \right)\Rightarrow 5\sqrt{2}=\sqrt{2}\left( \left| a \right|+3 \right)\Rightarrow \left| a \right|+3=5\Rightarrow \left| a \right|=2\xrightarrow{a \gt 0}a=2$