Part A:
[tex] x^{2} + y^{2} -10x+6y=15 \\ \\
x^{2} -2(x)(5)+ y^{2}+2(y)(3)=15 \\ \\
x^{2} -2(x)(5)+(5)^{2}+ y^{2}+2(y)(3)+(3)^{2}=15+ (5)^{2}+(3)^{2} \\ \\
(x-5)^{2}+(y+3)^{2}=49 [/tex]
Part B:
Since the signs and the coefficient of squared terms are the same in the given equation, the given equation represents a circle. The center of the circle is (5, -3) and the radius of the circle is 7.
