If the straight line x + y = K and the curve y = radical 1-x ^ 2 have exactly one common point, then the value range of K is? RT see that it is not the previous one, please do not copy and paste. Thank you

If the straight line x + y = K and the curve y = radical 1-x ^ 2 have exactly one common point, then the value range of K is? RT see that it is not the previous one, please do not copy and paste. Thank you

Substituting y = K-X into y ^ 2 = 1-x ^ 2
2Y ^ 2-2ky + K ^ 2-1 = 0
Because there is only one common point, then there is only one solution
Therefore (- 2K) ^ 2-8 (k ^ 2-1) = 0
k=±√2
That's what you want

If the line y = x + B and the curve X= If 1 − Y2 has a common point, then the value range of B is______ .

The line y = x + B is a straight line with slope 1 and intercept B;
Curve X=
The deformation of 1 − Y2 is x2 + y2 = 1 and X ≥ 0
It is obviously a right semicircle with a center of (0,0) and a radius of 1
According to the meaning of the title, the straight line y = x + B and the curve X=
1 − Y2 has and has a common point
If we make their figures, we can easily get the value range of B is: - 1 < B ≤ 1 or B=-
2．
So the answer is: - 1 < B ≤ 1 or B=-
2．

If the straight line y = x + K and the curve X = (under the root sign) 1-y ^ 2 have exactly one common point, then the value range of K is? I figured it was [- 1,1] u {- 2},

First of all, the curve is the right half of the unit circle, and includes the points (0,1) and (0, - 1). Then it moves down to (- 1,1], and there is a tangent point below. If the distance between the center of the circle and the straight line is 1, then k = radical 2. In summary, the range of K is (- 1,1] u {- radical 2}

If the quadratic power of the line y = x + K and the curve X = radical 1-y have exactly one common point, then the value range of K is A. k = positive and negative root 2 b. K greater than or equal to 2 or K less than or equal to negative root 2 C. K greater than negative root 2 less than root 2 d. K equal to negative root 2 or K greater than negative 1 less than or equal to 1

X = √ (1-y 2) denotes a half circle with the center at the origin and radius 1 on the right side of the y-axis. Using the graph, we can get: - 1

The curve y = 1+ If 4 − x2 and the line L: y = K (X-2) + 4 have two different intersections, then the value range of real number k is () A. (5 12，+∞) B. (1 3，3 4] C. (0，5 12) D. (5 12，3 4]

Draw a graph according to the meaning of the title, as shown in the figure: the straight line L passes through a (2, 4), B (- 2, 1), and the curve y = 1 + 4 − x2. The image is a semicircle with (0, 1) as the center and 2 as the radius. When the line L is tangent to the semicircle and C is the tangent point, the distance from the center of the circle to the line L is d = R, that is, | 3 − 2K | K2 + 1 = 2

If the line y = x + m and the root of the curve (1-y ^ 2) = x have two different intersections, find the value range of real number M

The curve is a semicircle with a positive half axis X and a radius of 1. You should be able to see this (if you can't see it, you don't have to do this problem, go back and eat the textbook thoroughly)
If the slope of the line is 1, you start to move down from the point where the line passes (- 10) until it is tangent to the circle
Draw your own picture. If you can do it, you should be able to do it

If the straight line y = x + m and the curve If 1 − y2 = x has two different intersections, then the value range of real number m is () A. （- 2， 2） B. （- 2，-1] C. （- 2，1] D. [1， 2）

From the meaning of the title: curve
1 − y2 = x represents the right semicircle of the circle, as shown in the figure
When the line y = x + m is tangent to the circle x2 + y2 = 1, then M = ±
2，
If the straight line y = x + m and the curve
When 1 − y2 = x tangent, then M=-
2．
By translating the straight line y = x, if the line y = x + m and the curve
If 1 − y2 = x has two different intersections, then the value range of real number m is: (-
2，-1]．
Therefore, B

Given that the straight line y = x + m and the curve y = radical (1-x ^ 2) have two focal points, we can find the value range of real number M The focus equation I worked out 2x ^ 2 + 2mx + m ^ 2-1 = 0 Then Δ 〉 0 is the - 1 of - radical 2 = 0=

It's easy for you to figure it out
The curve is a semicircle in the coordinate system. Find out the position relationship between the curve and the straight line. When the straight line is tangent to the semicircle, M = root 2. When the straight line passes through (- 1,0), M = 1. These are two special positions. The position of the straight line between them (including the case of M = 1) is in line with the meaning of the question, Sometimes when you don't pay attention to special circumstances, it's easy to make mistakes. It's time-saving to use image method

The curve y = 1+ When there are two intersections between 4 − X2 (| x | 2) and the line y = K (X-2) + 4, the value range of real number k is () A. (5 12，3 4] B. （5 12，+∞） C. (1 3，3 4) D. (0，5 12)

The curve y = 1 + 4 − X2 (| that is, X2 + (Y-1) 2 = 4, (Y ≥ 1), represents that the part of the circle with a (0, 1) as the center and 2 as the radius is located above the straight line y = 1 (including the intersection points c and D of the circle and the line y = 1), which is a semicircle, as shown in the figure: the straight line y = K (X-2) + 4 crosses the fixed point B (2, 4)

There are two intersections between the straight line y = KX + 2K + 4 and the curve y = radical (4-x * 2). Find the range of K

Note that the former is a straight line passing through the (- 2,4) point, and the latter is the upper semicircle. Therefore, after drawing the graph, the range of K can be solved from two equations. One is the line passing through (- 2,4) and (2,0), that is, k = - 1. The other is the slope of the line tangent to the circle