AC = radius of circle with centre A + radius of circle with centre C = x +5

Since angle bac is 90 degrees, bc becomes the hypotenuese. Using the pythagorus theorum,

AB^2 + AC^2 = BC^2

(x+2)^2 + (x+5)^2 = (5+2)^2

x^2 + 4 + 4x + x^2 + 25 + 10x = 49

2x^2 + 14x + 29 = 49

2x^2 + 14 x = 20

2(x^2 + 7x) = 2(10)

x^2 + 7x = 10

x^2 + 7x – 10 = 0

Now we know that roots of ax^2 + bx + c = 0 equation are 1/2a (-b + sqrt (b^2-4ac) and 1/2a (-b – sqrt (b^2-4ac)

Root 1 is 1/2a (-b + sqrt (b^2-4ac) = 1/2 { – 7 + sqrt [ 7^2 – 4(1)(-10)]}

= 1/2 [-7 + sqrt (49 +40)]

= 1/2 (-7 + sqrt 89)

= 1/2 (-7 + 9.43)

= 2.43 / 2 = 1.215

Root 2 is 1/2a [-b – sqrt (b^2-4ac)]= 1/2 [-7 – sqrt (49 + 40)]

= 1/2 (-7 – 9.43)

= 1/2 (-16.43) = -8.215

Therefore we get x = 1.215 and x = -8.215

But since x implies the radius of the circle with centre A, x cannot be equal to -8.215 ( radius cannot be negative)

Therefore the radius of the circle with centre A = 1.215 cm

= 12.15 mm