When k is the value, the intersection of the line x + 2Y + K + 1 = 0 and 2x + y + 2K = 0 is in the fourth quadrant

When k is the value, the intersection of the line x + 2Y + K + 1 = 0 and 2x + y + 2K = 0 is in the fourth quadrant

X + 2Y + K + 1 = 0, x = 1-k-2y is deduced
Substituting 2x + y + 2K = 0 gives y = (2-k) / 3
Substituting y = (2-k) / 3 into x = 1-k-2y
Because the intersection is in the fourth quadrant, so X

When k is the value, the intersection of the line x-2y-2k = 0 and 2x-3y-k = 0 is outside the circle x ^ 2 + y ^ 2 = 9

The intersection point of x-2y-2k = 0 and 2x-3y-k = 0 is (- 4K, - 3K)
The intersection point outside the circle x ^ 2 + y ^ 2 = 9 should satisfy
（-4k)^2+(-3k)^2>9
That is 16K ^ 2 + 9K ^ 2 > 9
25k^2>9
k^2>9/25
k> 3 / 5 or K