## Algebra 2 (1st Edition)

$(x+2)^2+(y+3)^2 =30$ This is the standard form of a circle.
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.