We have the circle S S=x2+y2+2gx+4y+1=0 and the second circle be S1, S1=x2+y2−2x−3=0 ∵ Centre (1,0) and radius =2 Now, common chord of circles S and S1 is ‌S−S1=0 ‌(2g+2)x+4y+4=0 If a circle bisects the circumference of another circle, then common chord is a diameter of the second circle. ‌2g+2+4=0⇒g=−3 ∵‌‌S=x2+y2−6x+4y+1=0 Centre (3,−2) Radius =√9+4−1=√12