A moving circular disc of radius 2 centimetres touches two fixed discs from outside…

Step 1. The two distances. The first fixed disc has centre $(0, 0)$ and radius $1$, so the moving centre is $2 + 1 = 3$ centimetres from it. The second has centre $(6, 0)$ and radius $3$, so the moving centre is $2 + 3 = 5$ centimetres from it.

Step 2. Two equations. $x^2 + y^2 = 9$ and $(x - 6)^2 + y^2 = 25$.

Step 3. Subtract. $x^2 - (x - 6)^2 = 9 - 25$, which is $12x - 36 = -16$, so $x = \dfrac{5}{3} \approx 1.67$.

Step 4. The height. $y^2 = 9 - \dfrac{25}{9} = \dfrac{56}{9}$, so $y = \dfrac{2\sqrt{14}}{3} \approx 2.49$ centimetres in the upper half.