Let R be the region of the disc $ x^2+y^2\leq1 $ in the first quadrant. Then the area of the largest possible circle contained in R

Question

Let R be the region of the disc $ x^2+y^2\leq1 $ in the first quadrant. Then the area of the largest possible circle contained in R is $\pi.(3-2\sqrt 2)$.

My Attempt :

If I draw a tangent of the circle $ x^2+y^2=1 $at the point $(1/\sqrt 2 , 1/\sqrt 2)$. Then we will get a isosceles triangle whose sides are the tangent , X axis and Y axis . Now the circle inscribed in this triangle will be the biggest among the all circles which completely lie in the triangle. So the area of this circle will be the required area.

Can anyone please tell me if I have gone wrong anywhere?

Answer 1

One strategy to prove this is the optimum would be as follows:

1. Assume the center of the circle lies on the line $x=y$, for each point on the line compute the area of the circle and show that your solution is the maximum

2. Assume the optimum does not lie on the line $x=y$, then shifting the center towards this line will allow a bigger radius to circle to fit into the region. Hence the point away from this line are not optimal.