Let the equation of circle whose centre is (h,k)= (x−h)2+(y−k)2=r2 ⇒x2+h2−2xh+y2+k2−2ky=r2 ⇒x2+y2−2xh−2ky=r2−h2−k2 ⇒x2+y2−2xh−2ky+c=0 (Let h2+k2−r2=c) Thus also pass origin then, c=0 ∴x2+y2−2xh−2ky=0 ....(i) (i) is orthogonal to x2+y2+2x+6y+12=0 and x2+y2+4x−6y+9=0 then −4h+6k=9.....(ii) and −2h−6k=12 ...(iii) Adding Eqs. (ii) and (iii), we get −6h=21 h=−