Find the point on y-axis which is at a distance of 4 units from a point P(4,2).
(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 √(4−0)2+(2−a)2
According to the question,
√42+(2−a)2=4
⟹√16+4+a2−4a=4
⟹√20−4a+a2=4
⟹20−4a+a2=16
⟹a2−4a+4=0
⟹(a−2)2=0
⟹a−2=0
⟹a=2
Therefore the required point is (0,2).