Answer
$(x+2)^2+(y+3)^2 =30$
This is the standard form of a circle.
Work Step by Step
Given: $x^2+y^2+4x+6y-17=0$
$(x^2+4x)+(y^2+6y)-17=0$
This gives: $(x^2+4x+4-4)+(y^2+6y+9-9)-17=0$
and
$(x^2+4x+4)+(y^2+6y+9)-30=0$
Thus, we have
$(x+2)^2+(y+3)^2 =30$
This is the standard form of a circle.