Question 1008349
Complete the square.
{{{x^2-6x+y^2+8y=k-5}}}
{{{(x^2-6x+9)+(y^2+8y+16)-9-16=k-5}}}
{{{(x-3)^2+(y+4)^2=k+20}}}
If the right hand side is greater than zero, then it is the square of the radius of the circle centered at (3,-4).
{{{k+20>0}}}
{{{k>-20}}}
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If the right hand side is equal to zero, then it is a single point, namely the center (3,-4).
{{{k+20=0}}}
{{{k=-20}}}
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If the right hand side is less than zero, then the solution is the empty set.
{{{k<-20}}}