Given the circle A
As shown :
DC is tangent to the circle
AD is the radius of the circle
so, AD will be perpendicular to the radius
Let AD = r
so,
[tex]\begin{gathered} AD=r \\ DC=4 \\ AC=AB+BC=r+2 \end{gathered}[/tex]The triangle ADC is a right triangle
Using the Pythagorean theorem:
[tex]\begin{gathered} AC^2=AD^2+DC^2 \\ (r+2)^2=r^2+4^2 \end{gathered}[/tex]solve the equation to find r:
[tex]\begin{gathered} r^2+4r+4=r^2+16 \\ 4r=16-4 \\ 4r=12 \\ \\ r=\frac{12}{4}=3 \end{gathered}[/tex]So, the answer will be:
The radius of the circle A = 3