Hello everyone.
We are going to proceed to case #2 when Ix >Iy but Ixy is negative.
We look to the expression of tan(2θ)=  (-Ixy/(Ix-Iy)/2).
Our Ixy this Ixy value is negative so negative/negative will be positive and the denominator.
This (Ix-Iy)/2  is positive value so we have a positive value of tan(2θ) which means
that the angle θ will become positive i.e. the x'-axis will be enclosed between axes x and y while 
y' or sometimes call u' and v', will be between the positive y-direction and the negative
x' direction.
Again we will draw two.
perpendicular  axes, the first axis is for the maximum and minimum values of the moment of inertia and pointing
to the right, while perpendicular axis for Ixy is pointing downwards.
We have two points points, point A and point B, 
If we start by point A this time we will go and
get the Iy value and since Ix value  always with the direction  Iy
So if we draw  Iy value it will be accompanied by the opposite sign of Ixy.
So when Ixy is negative we are going to draw +ve Ixy
the reverse of Ixy.
This time will go downwards, while  for point B, it has the coordinate of Ix  same value of Ix
value, sine which is negative.
So we are going to come upwards because as we agree that Ixy positive value is pointing downwards.
We have these two points A and B will intersect with the horizontal X. We get the central point O, the
angle enclosed between OB and the horizontal X will be 2θp and  the other angle also is 2θp,
the intersection of  AB with the x-axis will give us the center of the radius and the radius as before. 
The radius value is (Ix-Iy)/2
This will give us this horizontal value we are going to square it and we square,  plus the square of
Ixy at the end, the value we are going to take SQRT of  that value will give us the radius.
This is the center of the circle.
We will draw by the value of the radius then we can we will get this point at that point if we call
this when as b' and that point is A' , so B'
represents Ix' maximum while the A' will represent the I x' minimum.
And again this (2 θp) which, in this regard is a positive value.
This is a central angle and this value is equal to that value.
So this is (1/2*2*θ) will be the θ and this is the direction we are looking for, for the
angle.
This is representing the x' axis, represented by the line A'B, and perpendicular to that axis
will give us  y'-direction as we can see, always remember for the x'-direction you join the point
A' that is having I-minimum with point B which you (Ix,-Ixy)
in that case this will give us the x'- direction.
Thanks a lot and we'll continue to see case number #3.
Goodbye and take care.
