If the circle x2 + Y2 + 2x-4y + 1 = 0 is symmetric about the straight line 2aX by + 2 = 0 (a, B ∈ R), then the value range of AB is () A. (−∞，1 4] B. [1 4，+∞) C. (−1 4，0) D. (0，1 4)

If the chord length of the straight line 2aX by + 2 = 0 (a > 0, b > 0) cut by the circle x2 + Y2 + 2x-4y + 1 = 0 is 4, then the maximum value of AB is () A. 1 Four B. 1 Two C. 2 D. 4

The equation of a circle is transformed into a standard equation and the result is: (x + 1) 2 + (Y-2) 2 = 4, so the coordinates of the center of the circle are (- 1,2), r = 2, the chord length intercepted by the line by the circle is 4, and the diameter of the circle is also 4

If the straight line 2aX by + 2 = 0 (a > 0, b > 0) always bisects the circumference of circle x2 + Y2 + 2x-4y + 1 = 0, then the maximum value of AB is () A. 4 B. 2 C. 1 Four D. 1 Two

The circle x2 + Y2 + 2x-4y + 1 = 0, i.e. (x + 1) 2 + (Y-2) 2 = 4, which means that the center of the circle is (- 1, 2), and the radius is equal to 2,
The center of the circle (- 1,2) is on the straight line 2aX by + 2 = 0 (a > 0, b > 0),
∴-2a-2b+2=0．
And then from a + 2 = 1
ab，∴1≥4ab，ab≤1
4，
So the maximum value of AB is 1
4，
Therefore, C

If the line ax by + 2 = 0 (a > 0. B > 0). The chord length cut by the circle x ^ 2 + y ^ 2 + 2x-4y + 1 = 0 is 4. Then the minimum value of a ^ 2 + 4B ^ 2-AB is

x²+y²+2x-4y+1=0
(x+1)²+(y-2)²=4
The center of the circle is (- 1,2) radius = 2
Chord length =4 so chord pitch =0
The line ax by + 2 = 0 passes through the center of the circle
-a-2b+2=0
a+2b=2
(a+2b)²=4>=8ab
ab=4ab-ab
=3ab
=3/2
Minimum = 3 / 2

If the chord length of the line ax by + 2 = 0 (a > 0, b > 0) cut by the circle x2 + Y2 + 2x-4y + 1 = 0 is 4, then 1 A+1 The minimum value of B is () A. 1 Four B. Two C. 3 2+ Two D. 3 2+2 Two

The circle x2 + Y2 + 2x-4y + 1 = 0, i.e. (x + 1) 2 + (Y-2) 2 = 4, represents a circle with m (- 1, 2) as its center and 2 as its radius,
The center of the circle is on the straight line ax by + 2 = 0 (a > 0, b > 0), so - 1a-2b + 2 = 0,
That is, a + 2B = 2, ν 1
A+1
b=a+2b
Two
a+a+2b
Two
B=1
2+b
A+a
2b+1≥3
2+2
One
2=3
2+
2，
If and only if B
a＝a
When 2B, the equal sign is established,
Therefore, C

straight line 3x + y − 2 = 0, and the chord length is () A. 1 B. 2 Three C. 2 Two D. 2

The radius of the circle is 2, and the distance from the center of the circle (0, 0) to the straight line is d = | - 2|
3+1=1，
The chord length is 2
r2−d2=2
4−1=2
3，
Therefore, B

Find the equation of straight line which passes through P (1, - 2) and intersects the circle x ^ 2 + y ^ 2 = 4, and cuts the chord length to 2 times the root sign 3

Circle equation:
Y (2) x = 0,2
The point (1, - 2) is outside the circle
Let the line passing through the point P intersect the circle as y + 2 = K (x-1), that is, kx-y-k-2 = 0
According to Pythagorean theorem, a right triangle is formed by radius, half chord length and chord center distance
The chord center distance is obtained as D 2 = 4 - (√ 3) 2 = 1, d = 1
The distance from the center of the circle to the straight line is 1
that
｜k+2｜/√(1+k²)=1
k²+4k+4=1+k²
4k=-3
k=-3/4
In this case, the straight line is - 3x / 4-y + 3 / 4-2 = 0, that is, 3x + 4Y + 5 = 0
When K does not exist, that is, the chord length cut by the line x = 1 is also 2 √ 3
At this time, the chord center distance is 1, the radius is 2, and the chord length is 2 √ 3
So there are two straight lines: x = 1 or 3x + 4Y + 5 = 0

If the common chord length of circle x2 + y2 = 4 and circle x2 + Y2 + 2ay-6 = 0 (a > 0) is 2 3, then a is equal to () A. 1 B. Two C. Three D. 2

Given that the radius of x2 + Y2 + 2ay-6 = 0 is 6 + A2, the coordinates of circle center are (0, - a), the radius of circle x2 + y2 = 4 is 2, the coordinates of center of circle are (0, 0) ∵ the length of common chord of circle x2 + y2 = 4 and circle x2 + Y2 + 2ay-6 = 0 (a > 0) is 23, then the distance from the center of circle (0, 0) to the common chord is 1

If the length of the common chord of the circle x? + y? = 4 and the circle x? + y? + 2ay-6 = 0 (a > 0) is twice the root sign 3, then a = () Note: there is an operation symbol different from the one on the Internet. Please don't paste it directly I'm sorry that the title is wrong. The correct one is as follows: if the length of the common chord of the circle x? + y? = 4 and the circle x? + y? + 2ay + 6 = 0 (a > 0), then a = ()

In the first circle, according to Pythagorean theorem, it is easy to calculate that the distance between the common chord and the center of the circle (0,0) is √ (2 ^ 2-3) = 1, so - 1 / a = 1, a > 0, and a = 1
The result of + 6 or - 6 is the same, because it just reverses the circle along the X axis and the distance does not change

If the common chord length of circle x ^ 2 + y ^ 2 = 4 and circle x ^ 2 + y ^ 2 + 2ay-6 = 0 (a > 0) is 2 Radix 3, then a=

X ^ 2 + y ^ 2 = 4 and circle x ^ 2 + y ^ 2 + 2ay-6 = 0
If the equation of two circles is different, we can get that the line where the common chord of two circles is located is ay = 3, y = 3 / A
The straight line is cut by the circle x ^ 2 + y ^ 2 = 4, resulting in a chord length of 2 and a sign of 3
According to the chord length formula: (L / 2) ^ 2 = R ^ 2-D ^ 2, so d ^ 2 = R ^ 2 - (L / 2) ^ 2 = 4-3 = 1
D is the distance from the center of the circle x ^ 2 + y ^ 2 = 4 to the straight line y = 3 / A
So d = 3 / | a | 1
A = 3 or a = - 3

If the circle x2 + Y2 + 2x-4y + 1 = 0 is symmetric about the straight line 2aX by + 2 = 0 (a, B ∈ R), then the value range of AB is () A. (−∞，1 4] B. [1 4，+∞) C. (−1 4，0) D. (0，1 4)