If the line y = KX + 4 + 2k and the curve y= If 4-x2 has two intersections, then the value range of K is______ .

If the line y = KX + 4 + 2k and the curve y= If 4-x2 has two intersections, then the value range of K is______ .

The curve y = 4-x2, that is, X2 + y2 = 4, (Y ≥ 0) represents a semicircle above the x-axis with (0, 0) as the center and 2 as the radius, as shown in the figure: the straight line y = KX + 4 + 2K, that is, y = k (x + 2) + 4 is the straight line with constant crossing point (- 2,4) slope K. combining with the graph, we can get the following results: K AB = 4-4 = - 1, ∵ | 4 + 2k| 1 + K2 = 2, k = - 3

If the line kx-y-2 = 0 and the curve If 1 − (Y − 1) 2 = x − 1 has two different intersections, then the value range of real number k is () A. (4 3，2] B. (4 3，4] C. [−2，−4 3)∪(4 3，2] D. (4 3，+∞)

The straight line kx-y-2 = 0 is transformed into y = kx-2, so it must pass through the point (0, - 2), while the curve 1 − (Y − 1) 2 = x − 1 can be deformed into (x-1) 2 + (Y-1) 2 = 1 (x ≥ 1) ﹥ the curve takes (1,1) as the center of the circle, the circle with radius of 1 is located at the right part of the line x = 1

Given that there are two intersections between the straight line y = KX + 2 and the curve y = radical 4x-x ^ 2, then the value range of K is

Substituting y = KX + 2 into y = 2x-x? Gives x? + (K-2) x + 2 = 0
Two intersections, so the discriminant is greater than 0
That is (K-2) 2 - 8 > 0
So K, 2 + 2 root sign 2

If the curve y = radical (1-x2) always has an intersection with the straight line y = x + B, then the value range of B is Thank you. Thank you. I'd like to have the right answer X2 is the square of X.... Thank you Forget the problem is not finished: complete problem: if the curve y = root (1-x2) and the straight line y = x + B always have an intersection point, then what is the value range of B? If there is an intersection point, the value range of B is? If there are two intersections, what is the value range of B?

B> Radical 6 / 2 or B < - radical 6 / 2

If the curve y = radical (1-x ^ 2) always has an intersection point with the straight line y = x + B, what is the value range of rigidity B?

The idea of combining trees
The image with y = √ (1-x?) is a semicircle above the x-axis
When the line y = x + B passes through point (1,0), the value of B is the minimum, and B = - 1
When the straight line y = x + B is tangent to the semicircle (the tangent point is in the third quadrant, and the tangent point coordinates are (- √ 2 / 2, √ 2 / 2)), the value of B is the largest,
B = √ 2
So the value range of B is [- 1, √ 2]

Equation | x − 4−y2|+|y+ If the curve represented by 4 − x2 | = 0 intersects with the straight line y = x + B, then the value range of real number B is______ .

∵ by the equation | x −
4−y2|+|y+
4−x2|＝0，
It is possible to obtain | x −
4 − Y2 | = 0 and | y+
4−x2|＝0，
ν x2 + y2 = 4 and X ≥ 0, y ≤ 0, which means that the circle with the origin as the center and 2 as the radius is located in the fourth quadrant,
Including the intersection with the axis, as shown in the figure:
When the line coincides with AB, there are two intersections between the curve and the straight line,
When the line coincides with L, the curve is tangent to the line and has only one intersection point,
The intercept of AB on the y-axis is - 2, and it is easy to know that the intercept of the straight line L on the y-axis is - 2
2, and ab ‖ straight line L, so the value range of real number B is [− 2
2，−2]，
So the answer is [− 2
2，−2]．

If there are two different intersections between curve y = 1 + root sign 4-x ^ 2 and straight line y = K (X-2) + 4, find the value range of real number K It's the simplification of that formula 4-x ^ 2 under y = 1 + radical Add square This is where I have a problem First, y ^ 2 = 1 + 4-x ^ 2 Then x ^ 2 + y ^ 2 = 5 But I made Y-1 = root 4-x ^ 2 X ^ 2 + (Y-1) ^ 2 = 4

Y ^ 2 = 1 + radical 4-x ^ 2
It should be (Y-1) ^ 2 = 4-x ^ 2

Curve C: x = big root (4-y ^ 2) (- 2 < = y < = 2) and the straight line y = K (x-1) + 3 have only one intersection point to find the range of real number K

x^2+y^2=4
And x > = 0
So it's a half circle, to the right of the y-axis
The intersection point with y is a (0,2), B (0, - 2)
Y = K (x-1) + 3 over C (1,3)
In the drawing, the straight line has an intersection point between AC and BC, and the tangent is also a
AC slope = (3-2) / (2-1) = 1
BC slope = (3 + 2) / (2-1) = 5
1<=k<=5
kx-y+3-k=0
Center (0,0), radius = 2
The distance from the center of the circle to the tangent point is equal to the radius
So | 0-0 + 3-K | / root sign (k ^ 2 + 1) = 2
k^2-6k+9=4k^2+4
3k^2+6k-5=0
As can be seen from the graph, the tangent K < 0
So k = (- 3-2 √ 6) / 3
1<=k<=5,k=(-3-2√6)/3

Given that the straight line y = x + m and the curve y = radical 1-x ^ 2 have two intersections, find the value range of real number M

Y = radical 1-x ^ 2 x ^ 2 + y ^ 2 = 1 (Y > = 0) the graph is a semicircle with the origin as the center and 1 as the radius above the X axis
Y = x + m is a set of parallel lines with slope 1
Two intersections,
One

If there is a range of X-1 and x 2, find the intersection point of X-2 and x 2

If the straight line y = K (x-1) + 2 and the curve y = √ (1-x 2) have two intersections, find the range of K value
The definition domain of curve y = √ (1-x 2) is - 1 ≤ x ≤ 1, and the range is [0,1]; its image is a parabola passing through three points a (- 1,0), B (0,1), C (1,0); the straight line y = K (x-1) + 2 crosses the fixed point P (1,2);
Let the slope of the tangent line of the curve drawn by p be K ₁, and the slope of the straight line connecting PA be K Ψ
When k ₁