The equation of the circle with centre $(0,2)$ and radius 2 is $x^2+y^2-m y=0$. The value of $m$ is
1
2
4
3
Solution
$(x-0)^2+(y-2)^2=(2)^2$
or $x^2+y^2-4 y+4=4$ or $x^2+y^2-4 y=0$
Asked in: BITSAT 2023 (Memory Based Paper 2)
Asked in: BITSAT 2023 (Memory Based Paper 2)