If the line X-my + 1 = 0 is tangent to the circle x ^ 2 + y ^ 2-2x = 0, what is the value of M

If the line X-my + 1 = 0 is tangent to the circle x ^ 2 + y ^ 2-2x = 0, what is the value of M

Circle x ^ 2 + y ^ 2-2x = 0
That is, (x-1) ^ 2 + y ^ 2 = 1
Center C (1,0), radius r = 1
The line X-my + 1 = 0 is tangent to the circle x ^ 2 + y ^ 2-2x = 0
Then the distance between the straight line and the center of the circle C is equal to radius 1
That is | 1 + 1 | / √ (1 + m ^ 2) = 1
∴1+m^2=4
m^2=3 ,m=±√3

If the distance between point (4, a) and line 4x-3y-1 = 0 is not greater than 3, then the value range of a is ()
A. [0，10]B. [13，313]C. （0，10）D. （-∞，0]∪[10，+∞）

The distance d from point (4, a) to line 4x-3y-1 = 0 = | 16 − 3a − 1 | 42 + (− 3) 2 = | 15 − 3a | 5 ∵ D is not greater than 3, | 15 − 3a | 5 ≤ 3, which is reduced to | 15-3a | ≤ 15, - 15 ≤ 15-3a ≤ 15, 0 ≤ a ≤ 10, so a is selected

Given that real numbers x and y satisfy x ^ 2 + y ^ 2 = 1, find the value range of (y + 2) / (x + 1)

Method 1
Let (y + 2) / (x + 1) = t, then y = t (x + 1) - 2, substitute into the known equation, and sort out the quadratic equation of one variable about X, so the equation discriminant is greater than or equal to 0. After sorting out, t > = 3 / 4, which is the value range of (y + 2) / (x + 1)
Method 2
k=(y+2)/(x+1)
So K is the slope of the line passing through (- 1, - 2)
x. Y satisfies x ^ 2 + y ^ 2 = 1
So it is to find that the line passing the point (- 1, - 2) and the unit circle have common points, and the point is the value range of the slope
Obviously, there is a maximum value when tangent
y+2=kx+k
kx-y+k-2=0
The distance from the tangent circle center to the straight line is equal to the radius
So | 0-0 + K-2 | / √ (k ^ 2 + 1) = 1
|k-2|=√(k^2+1)
k^2-4k+4=k^2+1
k=3/4
Another tangent is x = - 1, because the radius from (0,0) to x = - 1
Then K does not exist, that is, infinite
So k > = 3 / 4
(y+2)/(x+1)>=3/4

If real numbers x and y satisfy 4 ^ x + 4 ^ y = 2 ^ (x + 1) + 2 ^ (y + 1), then the value range of S = 2 ^ x + 2 ^ y

Let 4 ^ x = (2 ^ x) ^ 2, 4 ^ y = (2 ^ y) ^ 2 hold, let 2 ^ x = a, 2 ^ y = B, then there is a ^ 2 + B ^ 2 = 2A + 2B, and the inequality a ^ 2 + B ^ 2 ≥ [(a + b) ^ 2] / 2 hold, that is, 2 (a + b) = a ^ 2 + B ^ 2 ≥ [(a + b) ^ 2] / 2, then (a + b) ^ 2-4 (a + b) ≤ 0, the solution is a + B ∈ [0,4], and s ∈ (0,4) is obtained from 2 ^ x ＞ 0,2 ^ y ＞ 0