If x and y satisfy (x-3)^2 + (y- 7)^2 = 64, what is the minimum possible value of x^2 + y^2?
+526 \((x-3)^2+(y-7)^2=64\) \(...(1)\)
⇒ \(x^2-6x+9+y^2-14y+49=64\)
⇒ \(x^2+y^2-6x-14y=6\)
∵ \(x\) and \(y\) satisfy equation (1), \((x-3)^2\) and \((y-7)^2\) are perfect squares
⇒ \((x-3)^2≥0\) and \((y-7)^2≥0\)
⇒ \(x≥3\) \(y≥7\)
When \(x=3\), ⇒\(y=15\)
When \(y=7\), ⇒\(x=11\)
1st Case : \(x^2+y^2=9+225\)
\(=234\)
2nd Case : \(x^2+y^2=49+121\)
\(=170\)
Thus the minimum possible value of x2 + y2 is 170.
N.B. I'm not too sure about the final answer, feel free to suggest corrections if any.
Thank you!
~Amy
+26367 If \(x\) and \(y\) satisfy \((x-3)^2 + (y-7)^2 = 64\),
what is the minimum possible value of \(x^2 + y^2\)?
\(\text{Let the origin $O=(0,0)$ } \\ \text{Let the center of the circle $C=(3,7)$ } \\ \text{Let the radius of the circle $r=\sqrt{64}=8$ } \\ \text{Let $x^2+y^2=r_{\text{min}}^2$ }\)
\(\text{1. Distance between origin and center of the circle:}\\ \begin{array}{|rcll|} \hline \overline{OC} &=& \sqrt{(0 - 3)^2 + (0 - 7)^2} \\ \overline{OC} &=& \sqrt{9+49} \\ \mathbf{\overline{OC}} &=& \mathbf{\sqrt{58}} \\ \hline \end{array} \)
\(\text{2. } x^2+y^2 = r^2_{\text{min}} \\ \begin{array}{|rcll|} \hline r_{\text{min}} &=& r-\overline{OC} \\ r_{\text{min}} &=& 8-\sqrt{58} \\ \hline x^2+y^2 &=& r_{\text{min}}^2 \\ x^2+y^2 &=& \left(8-\sqrt{58}\right)^2 \\ x^2+y^2 &=& 64-16\sqrt{58} + 58 \\ \mathbf{ x^2+y^2 } &=& \mathbf{ 122 -16\sqrt{58} } \\ \text{The minimum value of }~\mathbf{ x^2+y^2 } &=& \mathbf{0.1476303062 } \\ \hline \end{array}\)
