The line L passing through the point P (- 2 √ 2,0) intersects the circle O: X * 2 + y * 2 = 4 with two points a and B, and finds the maximum OAB area of the triangle and the L equation
∵S△AOB=(1/2)OA*OB*sin∠AOB,OA=OB=2
When ∠ AOB = 90 °, the maximum value of s △ AOB = 2
∵ here △ AOB is an isosceles right triangle,
The distance between point O (0,0) and line AB d = (1 / 2) AB = (1 / 2) √ (2 ^ 2 + 2 ^ 2) = √ 2
Let the equation of line AB be y = K (x + 2 √ 2), that is, kx-y + 2 √ 2K = 0
∵d=|0-0+2√2k|/√(k^2+1)
∴|2√2k|/√(k^2+1)=√2
That is 8K ^ 2 = 2K ^ 2 + 2, K ^ 2 = 1 / 3
∴k=±√3/3
So the equation of line AB is y = (± √ 3 / 3) (x + 2 √ 2)
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