Given that the real numbers x and y satisfy the equation x2 + y2-4x + 1 = 0, then the value range of YX + 1 is () A. [-1，1]B. [-22，22]C. [-3，3]D. [0，2]

The equation x2 + y2-4x + 1 = 0 is simplified to (X-2) 2 + y2 = 3, the equation represents a circle with a point (2,0) as its center and a circle with a radius of R = 3, YX + 1 represents two points (x, y), and the slope of (- 1,0) is k = YX + 1, that is, kx-y + k = 0. When the line is tangent to the circle, K takes the maximum and minimum value. In this case, the distance from the center of the circle to the line d = R, that is, d = | 2K + K | 1 + K2 = r = 3 | k = 22. The value range of YX + 1 is [− 22,22] B

If the real number x.y satisfies the elliptic equation x2 / 16 + Y2 / 9 = 1, then the value range of X + y + 10 is?

From the elliptic equation, we can set the parameter equation x = 4cosa, y = 3sina
∴x+y+10=4cosa+3sina+10
=5sin (a + b) + 10, where b = arctan 4 / 3
∵ -5≤5sin(a+b)≤5
∴ 5≤x+y+10≤15

X is a real number, and y = 4x + 3 of x2 + 1

The title is y = (4x + 3) / (x ^ 2 + 1)!
Multiply both sides by x ^ 2 + 1 to get y (x ^ 2 + 1) = 4x-3
It is concluded that: YX ^ 2-4x + (Y-3) = 0, a quadratic equation of one variable about X, because x is a real number, it has a solution
So: △ = B ^ 2-4ac 0
So: 16-4y (Y-3) 0
So: (y-4) (y + 1) 0
So: Y-1 or y-4

Given that the real numbers x and y satisfy the equation x2 + y2-4x + 1 = 0, then the value range of YX + 1 is () A. [-1，1]B. [-22，22]C. [-3，3]D. [0，2]