Construct a circle with a radius of (2) at the origin

The line l will be tangent to this circle

The equation of this circle is

x^2 + y^2 = 4 (1)

The slope of a tangent line to this circle at any point will be -x /y

So

-x/ y = 1

-x = y (2)

Sub (2) into (1)

x^2 + (-x)^2 = 4

2x^2 = 4

x^2 = 2

x = sqrt2

y =-sqrt 2

The equation of the line is

y = 1 (x - sqrt 2) - sqrt 2

y = x - 2sqrt 2

y= x -sqrt 8

When x = 0, y= - sqrt 8

When y = 0 x = sqrt 8

So points (0.-sqrt 8) and (sqrt 8,0) are on the line

The base of the triangle is sqrt [ (sqrt 8)^2 + (-sqrt 8)^2 ] ]= sqrt (16) = 4

Its area is (1/2) base * height = ( 1/2)(4) (2) = 4