If the two circles x2+y2+2gx+c=0  and x2+y2−2fy−c=0  have equal radius then locus of (g,f)  is

Given circles are x2+y2+2gx+c=0 C1(−g,0), r1=g2−c  andx2+y2−2fy−c=0C2(0,f), r2=g2−cSince r1=r2⇒g2−c=f2+cSquaring on Both Sides⇒g2−c=f2−c∴  Locus of (g,f)  isx2−c=y2+c⇒x2−y2=2c