Let equation of circle be x2+y2+2gx+2fy+c=0 Given equation of circle is x2+y2−4x+8=0 The centres of above circles are (−g,−f) and (2,0). Condition of orthogonality is ‌2(g1g2+f1f2)‌=c1+c2 ‌∴‌2(g×(−2)+(f)×0)‌=c+8 ⇒‌−4g‌=c+8 . . . (i) Also, the assume circle touch the line x+1=0. ∴ The perpendicular drawn from centre to the line is equal to radius. ∴‌‌
−g+1
√12
=√g2+f2−c ⇒−g+1=√g2+f2−c On squaring both sides, we get ‌g2+1−2g=g2+f2−c ‌⇒‌‌c=f2+2g−1 ‌ Putting the value of c in Eq. (i), we get −4g=f2+2g−1+8 ⇒f2+2g+4g+7‌=0 ⇒‌f2+6g+7=0 ∴ Locus of centre of circle is y2+6x+7=0.