Let the equation of the required circle be x2+y2+2gx+2fy+c=0 and centre is (−g,−f) Given circles, C1:x2+y2−2x−4y−4=0 Here, g1=−1,f1=−2,c1=−4 C2:x2+y2−10x+12y+52=0 Here, g2=−5,f2=6,c2=52 Condition of two circles cuts orthogonally 2g1g2+2f1f2=c1+c2 For C1:2(g)(−1)+2(f)(−2)=C−4 or 2g+4f=−c+4...(i) For C2:‌‌2(g)(−5)+2(f)(6)=c+52 10g−12f=−c−52....(ii) Subtracting on Eqs. (i) and (ii), we get −8g+16f=56 or g−2f=−7 or g=2f−7...(iii) Multiplying Eq. (i) by 5 and then subtracting Eq. (ii), we get  10g+20f=−5c+20 +10g−12f=−c−52 − ‌‌‌+ ‌‌‌‌‌‌‌+ + ────────────── 32f=−4c+72 or ‌‌8f=−c+18 or c=−8f+18 Hence, radius r=√g2+f2−c =√(2f−7)2+f2(−8f+18) =√5f2−20f+31
=√5(f2−4f)+31=√5(f−2)2+11
For minimum, value f−2=0 ⇒‌‌f=2 Putting the value of ' f in Bq. (iii), we get g=2×2−7⇒g=−3 Therefore, coordinate of centre is (+3,−2).