Bunuel wrote:

If x and y are both positive, is x greater than 4?

(1) x > y

(2) In the coordinate plane, the point (x, y) lies outside a circle of radius 5 centered at the origin.

Kudos for a correct solution.

Given: x > 0, y > 0

x > 4?

Statement 1:

x > y

This statement is insufficient to prove whether x > 4 because there are innumerable pairs of positive numbers where x > y and x < 4 such as (2, 1) and also x > y and x > 4 such as (5, 3), etc.

Statement 2:

Given: the point (x, y) lies outside a circle of radius 5 centered at the origin.

The point (x, y) lies on a circle of radius \(\sqrt{x^2 + y^2}\) centered at the origin.

Since the point (x, y) lies outside the circle of radius 5, this implies that the radius of the circle with the point (x, y) is greater than 5.

So, \(\sqrt{x^2 + y^2} > 5\)

\(x^2 + y ^2 > 25\)

But, there are many pairs of numbers which satisfy this inequality without giving a definite answer on whether x > 4 such as (2, 5), (4, 4), (5, 1), etc.

Since the individual statements are insufficient individually, on combining them:

x > y and \(x^2 + y ^2 > 25\)

Since, x > y and x and y are positive

\(x^2 > y^2\)

\(x^2 + x^2 > y^2 + x^2 > 25\)

\(2x^2 > 25\)

x > \(5/\sqrt{2}\)

x > 3.5

But, combining the statements hasn't concluded if x > 4.

Hence, the statements are sufficient neither individually nor combined.

Therefore, the answer is

E.