If x^2 +y^2 is less than 9, is x^2 less than x ? (1) y^2 is

I would like to say (E)

I prefer to solve it with the XY Plan.

x^2 < x ?
<=> x^2 - x < 0 ?
<=> x*(x-1) < 0 ?
<=> 0 < x < 1 ?

x^2 +y^2 < 9
<=> x^2 + y^2 < 3^2 >>> In the XY plan, this represents a surface delimited by the circle with 0(0,0) as center and 3 as radius.

So, 0 < x < 1 creates a part of this circle's surface.

From 1
y^2 > 8
<=> y > 2*qrt(2) or y < -2*qrt(2)

Creating 2 horizontal lines (Fig 1), one at -2*qrt(2) and one at 2*qrt(2), we observe especially for negative values of x that we have a surface out of 0 < x < 1.
Also, we have a part of the surface created by these 2 lines that are in the 0 < x < 1.

INSUFF.

From 2
x=3*y
<=> y = 1/3*x >>> it's a line passing by 0(0,0) (Fig 2).

All negative values of x will create negative values of y. In this case, we are not in the section 0 < x < 1.

Also, we have a part of the line that passes throught 0 < x < 1.

INSUFF.

Both (1) & (2)
Here, looking the 2 conditions and so 2 graphs, there is no common area within the circle.... So, it's pretty weird and we cannot conclude...

I tend to say (E).