Quadrilateral ABCD has vertices A(2,3), B(7,10), C(9,4), and D(4,-3). Prove that ABCD is a parallelogram but not a rhombus. (include conclusion statement)

How is the equation of this circle written in standard form? x2 + y2 - 6x + 14y = 142 a) (x - 3)2 + (y + 7)2 = 200 b) (x+ 3)2 + (y - 7)2 = 200 (x - 6)2 + (y + 14)2 = 142 d) (x+6)2 + (y- 14)2 = 142