locus

Sorry to say that your diagram is not good.
Point O (origin) should lie on the line L1 because L1 : y=x.

(a)

Since, L1: y=x.
angle AOC = 45degree
angle ADC = 180 - angle A - angle AOC = 180 - 90 - 45 = 45 degree (given angle A = 90)

So, Triangle OAD has two equal base angles.
So, side OA = AD.
AC bisects angle A (since AB and AE tangents to circle))
AC is common edge.

So, triangle ACO is equivalent to triangle ACD


(b)
angle OBC = 90 because line OB is tangent to circle.
angle BOC = 45 because slope of L1 = 1
angle BCO = 180 - 90 - 45 = 45 degree

So, triangle OBC has two equal base angles.
So, side OB = BC
So, BC = 1 (given OB = 1)

Radius = BC = 1

(c)

All the interior angle of ABCE are right angles. (AB, AE is tangent to circle C and Angle A = 90)

Since, ABCE is a rectangle. Opposite sides are equal. Now, BC = CE (radius of circle)
So, ABCE is also a square!
So, AB = BC = CE = AE

So, AB = 1--------------(*). so, OA = OB + AB = 1 + 1 = 2.
So, A = (2cos45, 2 sin45) = (sq. root 2, sq. root 2)

Slope of L2 = -1 / slope of L1 = -1/1 = -1
L2 passes throught A.
The equation of L2 is :

y - sq root 2 = -1 ( x - sq root 2)
x + y - 2* sq root 2 = 0

(d)

AB = 1 (from above (*) )

(e)

B = (1* cos 45, 1*sin45) = (1 / sq root 2, 1 / sq root 2)

To show: OBEC is a parallelogram.
Angle OBC = 45
Angle EDC = 180 - angle A - angle OBC = 180 - 90 -45 = 45
Angle ECD = 180 - angle CED - angle EDC = 180 - 90 - 45 = 45

So, OB is parallel to EC.
OB = BC (based angles equal)
EC = ED (based angles equal)
OB = BC = EC
Triangle OBC is equivalent to triangle ECD

X-ordinate of point E = X-ordinate of point B + radius of circle
Y-ordinate of point E = Y-ordinate of point B

So, E = ( 1+ (1/sq root 2), 1/sq root 2).

2006-10-17 18:58:19 補充：
(a)So, triangle ACO is equivalent to triangle ACD -- should be conguent not equivalent(e)Triangle OBC is equivalent to triangle ECD -- should be conguent not equivalent