Let D be a region bounded by a simple, closed, positively oriented curve C. Prove: (20 #0, ±√2) j z ³ 2 z dz = c (2-20)3 8 6πTiZO 0 ZOED zo&D

By expanding (xh)2+(yk)2=r2, we obtain x22hx+h22ky+k2r2=0. When we compare this result to the form x2+y2+Dx+Ey+F=0, we see that D=2h,E=2k, and F=h2+k2r2. Therefore, the center and the length of a radius of a circle can be found by using h=D2,k=E2 and r=h2+k2F. Use these relationship to find the center and the length of the radius of each of the following circles. x2+y2+4x14y+49=0