( 1)

As shown in the figure:

L 1: MX-Y = 0, passing the fixed point (0,0), and the slope KL1= m.

L2: x+my-m-2 = 0, and the slope kl2 =-1/m.

= & gtm(y- 1)+x-2=0

Let y- 1 = 0 and x-2 = 0.

Y = 1，x = 2。

∴l2 crossing point (2 1)

∵kl 1？ kl2=- 1

∴ The straight line l 1 is perpendicular to the straight line l2.

∴ The intersection of the straight line l 1 and the straight line l2 must be on a circle with (0,0) and (2, 1) as the diameter endpoints.

And the center of the circle is (1, 1/2) and the radius is r = 1/2 √ (2? + 1? )=√5/2

∴ The equation of the circle is (x- 1)? +(y- 1/2)？ =5/4

x？ +y？ -2x-y=0

(2)

From (1):

P 1(0，0)，P2(2 1)

When point P moves on a fixed circle, the bottom of △PP 1P2 is a fixed value 2r.

When the height of the triangle is the largest, the area of △PP 1P2 is the largest.

So s △ PP 1p2max = 1/2? 2r？ r=5/4

And the intersection of l 1 and l2 is p (? (m+2)/(m？ + 1)，[m(m+2)]/(m？ + 1)? )

The angle between OP and P 1P2 is 45.

∴|OP|=√2？ r=√ 10/2

That is [(m+2)/(m? + 1)]? +[? [m(m+2)]/(m？ + 1)? ]? =5/2

Solution: m = 3 or m =- 1/3.

Therefore, when m = 3 or m =- 1/3, the area of △PP 1P2 reaches the maximum value of 5/4.

?