Let the centre of the required circle be (h, k) with radius r.

Since the circle touches the line x = 0, the perpendicular distance from the centre to this line equals r: \(|h| = r\).

Since the circle touches the line y = 0, similarly \(|k| = r\).

Since the circle touches the line x = 4, the distance from the centre to this line equals r: \(|h-4| = r\).

Taking h and r positive and the centre lying between the lines x = 0 and x = 4, h = r gives \(|r-4| = r \Rightarrow 4-r = r \Rightarrow r = 2\), so h = 2.

From \(|k| = r = 2\), k can be either +2 or −2, since nothing in the given conditions fixes whether the circle lies above or below the x-axis.

Hence two circles satisfy all three tangency conditions: \((x-2)^2+(y-2)^2=4\) and \((x-2)^2+(y+2)^2=4\), which together are written compactly as \((x-2)^2+(y\pm2)^2=4\).