Square ABCD encloses a circle of area 154 at its center

The answer is definitely not C since A is enough to answer the question. Still can't wrap my head around if B is enough or not though.
If we can find the length of the diagonal of the square, the rest should be possible to calculate. We already know the radius of the circle. So 2 times the radius plus two times the shortest distance should add up to the diagonal of the square. And since the diagonal splits the square into two isosceles similar triangles, A should be enough.

Area of circle = 154

So \({\pi}r^2=154\)
So \(\frac{22}{7}r^2=154\)
So \(22*r^2=154*7\)
So \(r^2=7*7\)
So \(r=7\)

So we can find r

Statement 1: Shortest distance from any of the vertices is \(10\sqrt{2}-7\)

So diagonal of square would be \(2*{(10\sqrt{2}-7)}+7+7\)
Which is \(2*10\sqrt{2}\) = \(20\sqrt{2}\)
At this point you have to realize that traingle formed by the diagonal is an isocelese one.
If you remember the proportions of an isosceles triangle \(1:1:\sqrt{2}\)
Hence the side of the square = \(20\)
So Area of Square = \(400\)
So Area of the remaining space = \((400-154)=246\)

So A is sufficient

Still wrapping my head around B though

Statement B: The length of tangent drawn from any of the four vertices A, B, C, or D of the square ABCD to the circle is √151.

Length of tangent is given. Tangent forms an angle of 90 degrees with the radius of the circle. Length of tangent is \(\sqrt{151}\)
Let half the diagonal of square be = \(x\)
Length of Radius = \(7\)
So \(7^2+{(\sqrt{151})}^2=x^2\)
So \(x^2=200\)
So \(x=10\sqrt{2}\)
So length of diagonal = \(20\sqrt{2}\)
Again, it forms an isosceles triangle so the side of square = \(20\)

Answer should be D

There might be something wrong with my calculations since I do not know if the tangent extends all the way to the edge of the square or just the circle but to sum it up, both statements should be individually sufficient. What's the original answer?

Just on a side note, if you face a question like this in reality on the GMAT, i wouldn't be doing all these calculations. I just did the calculations to come up with a proof. Once I can spot that the statement is sufficient, there is no need to prove it in the actual exam since you just have to pick the right option. I was just a bit skeptic about the question (and wrongly so) so solved it to prove things to myself. But more importantly, the answer to this question could be spotted in less than a minute on the actual exam.