Find the point on $y$-axis which is at a distance of 4 units from a point P(4,2).

The correct option is C

$(0,2)$

Any point on the $y$-axis is of the form $(0,a)$.

Now, distance from P(4,2) to $(0,a)$ is given by $\sqrt{(4-0{)}^2+(2-a{)}^2}$

According to the question,

$\sqrt{4^2+(2-a{)}^2}=4$

$⟹\sqrt{16+4+a^2-4a}=4$

$⟹\sqrt{20-4a+a^2}=4$

$⟹20-4a+a^2=16$

$⟹a^2-4a+4=0$

$⟹(a-2{)}^2=0$

$⟹a-2=0$

$⟹a=2$

Therefore the required point is $(0,2)$.