Given, equation of circles x2+y2−10x−10y+41=0⋅⋅⋅⋅⋅⋅⋅(i) x2+y2−22x−10y+137=0⋅⋅⋅⋅⋅⋅⋅(ii) Centre (C1) and radius (r1) of Eq. (i) are C1=(5,5) and r1=3 and centre (C2) and radius (r2) of Eq. (ii) are C2=(11,5) and r2=3
We observe that, C1C2= distance between centres =(11−5)2+(5−5)2=6=r1+r2 ∴ Circles Eqs. (i) and (ii) touch each other externally. Hence, circles have only one meeting point.