The center of the circle is on x-y- 1=0, which is known from symmetry.

Circle x 2+kx+y 2 = 0

(x+k/2)^2+y^2=k^2/4

Central coordinate (-k/2,0)

When x=-k/2,

y=(- 1+k/2)=0，k=-2

Center of the circle (1, 0), r= 1

(x-1) 2+y 2 =1is a circular equation.

According to the distance formula from point to line, the distance from the center of the circle to AB is (3√2)/2.

So the height of the triangle Pab with AB as the base is 3√2/2+ 1.

S=(3√2/2+ 1)*2√2/2

=3+√2