Answer
$(x-2)^2+(y+3)^2=16$,
$ x^2+y^2-4x+6y-2=0$,
See graph.
Work Step by Step
Step 1. Based on the given data, we can write the standard form as $(x-2)^2+(y+3)^2=16$,
Step 2. The general form is $x^2+y^2-4x+6y+4+9-15=0\longrightarrow x^2+y^2-4x+6y-2=0$,
Step 3. See graph.
