SOLUTION
Write out the equation of the line and the circle, we have
[tex]\begin{gathered} 5y-4x=k\rightarrow5y-4x-k=0 \\ \text{and } \\ (x-3)^2+(y+2)^2=41 \end{gathered}[/tex]We want to find the possible values of for which the line does not intersect the circle this implies that the distance of the from the center of the circle to the line is greater than the radius of the circles.
From the equation of the circle, we have the radius to be
[tex]\begin{gathered} (x-3)^2+(y+2)^2=41 \\ \text{Hence } \\ \text{center}=(3,-2),\text{raduis}=\sqrt[]{41} \end{gathered}[/tex]The distance is given by
[tex]\begin{gathered} d=\frac{|Ax_1+By_1+c|}{\sqrt[]{A^2+B^2}} \\ \text{Where } \\ c=k,(x_1,y_1)=(3,-2),A=-4,B=5 \end{gathered}[/tex]Then
[tex]\begin{gathered} d=\frac{|-4(3)+5(-2)+(-k)|}{\sqrt[]{(-4)^2+5^2}} \\ \text{Then } \\ d=\frac{|-12-10-k|}{\sqrt[]{41}} \end{gathered}[/tex]if d >r, we have
[tex]\begin{gathered} |-22-k|>\sqrt[]{41} \\ \text{When } \\ -22-k>0\Rightarrow k\le22 \end{gathered}[/tex]when
[tex]\begin{gathered} |-22-k|=-22-k>41\Rightarrow-k>41+22\Rightarrow-k>63 \\ \text{Then } \\ k<-63 \end{gathered}[/tex]So,
[tex]K<-63\text{ or K>-22}[/tex]because
[tex]\begin{gathered} -22-k<0\Rightarrow k>-22 \\ |-22-k|=22+k>41\Rightarrow k>19 \end{gathered}[/tex]Hence
We have
[tex]k<-63\text{ or k>19 are the possible values of k}[/tex]Hence
The possible vales of k are
K<-63 or K>19
