BITSAT 2021 Paper

The centre of the circle is on the perpendicular bisector of the line joining (−1,2) and (1,2), which is the y - axis. The ordinate of the centre is given by
(5−y)2=1+(y−2)2
⇒ y=

10

3

Hence, eq. of the circle is :
x2+(y−

10

3

)2=(

5

3

)2
⇒ 9x2+9y2−60y+75=0