Well, here's my solution - it's late over here, though, so there may be an error! I imagine there are a few ways to get an answer here.
-call the centre of the circle O, and the centre of the square C
-draw the diameter of the circle passing through A, C and O. This will meet the left side of the square at a point we'll call B
-call the upper left hand corner of the square D
Now:
-AC is half of one side of the square; its length is 2.
-CB is also one half of one side of the square, so is also 2.
-let x equal the length of OC, the short distance between the center of the square and the center of the circle;
-the distance OB is then 2-x, since CB is 2;
-now, the radius is the distance OA, which is just OC + CA, and must therefore be 2+x in length
Finally, we have a right triangle ODB. OD is the hypotenuse of this triangle, and it is also a radius; its length is 2+x. The legs of this triangle are OB, which has length 2-x, and DB, which is half of one side of a square, so has length 2.
That is, ODB is a right triangle with side lengths 2-x, 2, 2+x. We can use algebra to find x, or we can see that if x = 1/2, we have sides that are exactly half those in a 3-4-5 triangle. Either way, the radius is 2+1/2 = 5/2.
Would be a bit easier to explain with a diagram! Am curious to see your solution, since I don't see an error in mine on a second pass.
_______
edit - and I'd add, nice work, aroon, with your estimation approach - that's a very clever way to get a maximum possible value for the radius.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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. Sorry about that. However, you can still try to find the correct solution. Just know that it's not listed in the answer choices.
