Given the circle C: (x-1) 2 + (y + 2) 2 = 9, the straight line L: 2ax-y + 2a-1 = 0, we prove that no matter what the real number a is, the straight line l always intersects the circle

Given the circle C: (x-1) 2 + (y + 2) 2 = 9, the straight line L: 2ax-y + 2a-1 = 0, we prove that no matter what the real number a is, the straight line l always intersects the circle

Circle C: (x-1) 2 + (y + 2) 2 = 9,
The straight line L: 2ax-y + 2a-1 = 0, y + 2 = 2aX + 2A + 1 substituting (x-1) 2 + (y + 2) 2 = 9:
（x-1)^2+(2ax+2a+1)^2=9
(4a^2+1)x^2+2(4a^2+2a-1)x+(4a^2+4a-7)=0
Discriminant △ = 4 (4a ^ 2 + 2a-1) ^ 2-4 (4a ^ 2 + 1) (4a ^ 2 + 4a-7)
=4[(16a^4+16a^3-4a^2-4a+1) - (16a^4+16a^3-24a^2+4a-7)]
=4(20a^2-8a+8）
=16(5a^2-2a+2)
=16 [(radical 5A - 1 / radical 5) ^ 2-1 / 5 + 2]
=16 [[(radical 5A - 1 / radical 5) ^ 2 + 9 / 5] ≥ 144 / 5
The line l always intersects the circle no matter what the real number a is

If the equation x ^ 2 + y ^ 2 + KX + 2Y + 1 / 4K ^ 2 + k = 0 represents a circle, then the value range of real number k is

x^2+y^2+kx+2y+1\4k^2+k
=(x+k/2)^2+(y+1)^2+k-1=0
That is, (x + K / 2) ^ 2 + (y + 1) ^ 2 = 1-k
The equation (x + K / 2) ^ 2 + (y + 1) ^ 2 = 1-k denotes a circle
So 1-k > = 0
So K

When k is a different real number, what is the characteristic of the geometry represented by the equation KX + y + 3K + 1 = 0?
A: All pass through the first quadrant B: form a closed circle C: represent all the lines in the rectangular coordinate plane D: intersect at a point

Constant crossing (- 3, - 1) point

Let y = KX ^ 2 + (3K + 2) x + 1, for any real number k, if x

When x