The real numbers x and y satisfy (x - 3)^2 + (y - 4)^2 = 18. Find the minimum value of x^2 + y^2.
The real numbers x and y satisfy \((x - 3)^2 + (y - 4)^2 = 18\).
Find the minimum value of \(x^2 + y^2\).
\(\text{Let the origin $O=(0,0)$ } \\ \text{Let the center of the circle $C=(3,4)$ } \\ \text{Let the radius of the circle $r=\sqrt{18}=3\sqrt{2}$ } \\ \text{Let $x^2+y^2=r_{\text{min}}^2$ }\)
1. Distance between origin and center of the circle:
\(\begin{array}{|rcll|} \hline \overline{OC} &=& \sqrt{(0 - 3)^2 + (0 - 4)^2} \\ \overline{OC} &=& \sqrt{9+16} \\ \overline{OC} &=& \sqrt{25} \\ \mathbf{\overline{OC}} &=& \mathbf{5} \\ \hline \end{array}\)
2. \(x^2+y^2 = r^2_{\text{min}}\)
\(\begin{array}{|rcll|} \hline r_{\text{min}} &=& \overline{OC} - r \\ r_{\text{min}} &=& 5 -3\sqrt{2} \\ \hline x^2+y^2 &=& r_{\text{min}}^2 \\ x^2+y^2 &=& \left(5 -3\sqrt{2}\right)^2 \\ \mathbf{ x^2+y^2 } &=& \mathbf{ 43 -30\sqrt{2} } \\ \text{The minimum value of }~\mathbf{ x^2+y^2 } &=& \mathbf{ 0.5735931288 } \\ \hline \end{array}\)