[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
Let's write the equation in standard form ~
[tex]\qquad \sf \dashrightarrow \: {x}^{2} + {y}^{2} - 2y - 15 = 0[/tex]
[tex]\qquad \sf \dashrightarrow \:(x ){ }^{2} + {y}^{2} - 2y = 15[/tex]
[tex]\qquad \sf \dashrightarrow \:(x ){ }^{2} + {y}^{2} - 2y + 1 = 15 + 1[/tex]
[tex]\qquad \sf \dashrightarrow \:(x ){ }^{2} +( {y}^{} - 1 ) {}^{2} = 16[/tex]
Now, it's the required form of circle ~