Help algebra

If $x$ and $y$ satisfy $(x-3)^2 + (y-7)^2 = 64$,
what is the minimum possible value of $x^2 + y^2$?

$\text{Let the origin $O=(0,0)$ } \\ \text{Let the center of the circle $C=(3,7)$ } \\ \text{Let the radius of the circle $r=\sqrt{64}=8$ } \\ \text{Let $x^2+y^2=r_{\text{min}}^2$ }$

$\text{1. Distance between origin and center of the circle:}\\ \begin{array}{|rcll|} \hline \overline{OC} &=& \sqrt{(0 - 3)^2 + (0 - 7)^2} \\ \overline{OC} &=& \sqrt{9+49} \\ \mathbf{\overline{OC}} &=& \mathbf{\sqrt{58}} \\ \hline \end{array}$

$\text{2. } x^2+y^2 = r^2_{\text{min}} \\ \begin{array}{|rcll|} \hline r_{\text{min}} &=& r-\overline{OC} \\ r_{\text{min}} &=& 8-\sqrt{58} \\ \hline x^2+y^2 &=& r_{\text{min}}^2 \\ x^2+y^2 &=& \left(8-\sqrt{58}\right)^2 \\ x^2+y^2 &=& 64-16\sqrt{58} + 58 \\ \mathbf{ x^2+y^2 } &=& \mathbf{ 122 -16\sqrt{58} } \\ \text{The minimum value of }~\mathbf{ x^2+y^2 } &=& \mathbf{0.1476303062 } \\ \hline \end{array}$