A circle touches the y-axis at (0, 4) and passes through the point (-2, 0). Then, the radius of the circle is

A circle touches the y-axis at (0, 4). Therefore, the centre of the circle is on the horizontal line whose y coordinate = 4. Thus, the coordinates of the center of the circle will be (a,4).

Now, the circle also passes through the point (-2,0)

The radius of the circle remains constant. 

$$\left(a-0\right)^2+\left(4-4\right)^2=\left(a-\left(-2\right)\right)^2+\left(4-0\right)^2$$

$$a^2=a^2+4a+4+16$$

=> $$a=-5$$

Therefore, the radius of the circle = $$\sqrt{\left(4-4\right)^2+\left(0-\left(-5\right)\right)^2}=5\ cm$$