OA=OC, then ∠OAC=∠CAB=∠OCA=30°
∴∠OCE=∠ACD-∠OCA=75°-30°=45°
∵OE⊥AD, then △OCE is isosceles Right triangle
∴OE=√2/2OC=√2
10, (1)Connect OM, make OE⊥CD at E
∵BC is the circle tangent, then OM⊥BC
∴∠OMC=∠MCE=∠OEC=90°
that is, OMCE is a rectangle
∵∠MCD=∠ ECO=45°
∴OMCE is a square
∴OM=OE, then DC is the tangent to the circle
(2)∵OMCE is a square
∴OM=CM=r
∵AC=√2(Pythagorean theorem: AC?=2AB?)
then OC=AC-OA=√2-r
∴Collinear theorem: r squared + r squared = (√2-r) squared
r squared + 2√2r-2 = 0
r = (-2√2±4)/2
r = 2-√2 (negative values are rounded off)