JEE Main 18 March 2021 Shift 1 Solved Paper

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=3We observe that,
C1C2= distance between centres
∴ Circles Eqs. (i) and (ii) touch each other externally.
Hence, circles have only one meeting point.