There is a common point between the square of the straight line y x b and the curve x followed by th

Solution: Because they only have one common point.

So there is only one solution to the system of equations they are made of.

Substituting y=x+b into y=1 x squared under the root sign yields:

x+b 1 x squared under the root number.

Then square the two sides to get:

x²+2bx+b²=1-x²

2x²+2bx+b²-1=0

Explain that this equation has two equal real roots.

4b²-8(b²-1)

4b²+84(b²-2)=0

Solution: b 2

And at this time x=-2b 4= 2 2, which is in line with 1 x 0 so: b 2

Simultaneous y=x+b and curve y=1 x

x+b= 1 x, squared on both sides gives x +2bx+b = 1-x , 2x +2bx+b -1=0, since the curve has a common point, so =4b -8b +8=0, b= 2

Solution: 1 x 2 under the y root number to obtain x 2 y 2 1, we can see that the curve takes the origin as the upper semicircle with a radius of 1 and the slope of y x b is 1, and the intercept is a straight line with b. There is exactly one common point, i.e., there is only one semicircle, and you can draw it to get b root number 2 and 1 or b root number 2

If the negative root number 2 is greater than or equal to the positive root number 2, the curve can be drawn as a circle with a radius of 1, and then the root finding formula is used to lead derta [triangular symbol] to get 0 to get the knot b equal to the negative root number 2 and the positive root number 2.

You can use the image method, the former is a straight line with a slope of 1, the latter is the center of the circle in the center of the right semicircle of the coordinate center intuitively, the intercept should be -1 and there is a pending order is the tangent line below, the image can obtain its intercept < 0, the following calculation is calculated by y=x+b, x=y-b brings in the curve x = (1-y 2)y-b = (1-y) after the square of both sides.

2y 2-2by+b 2-1 =0 Because there is only one common point, =(2b) 2 - 8(b 2-1) = 0 gives b = 2 (non-compliant) or - 2

Hence -1< b 1 or b = - 2< p>

Substituting y=x+b into the curve equation.

The problem is converted into a quadratic equation with one and only one solution.

Note: The curve "y=3 - the square of 4x-x under the root number" is not a perfect circle, but a semicircle with the mouth facing up, which is a fan-shaped arc with a fan angle of 180 degrees, that is, the part of the circle below the straight line y=3 is a fan-shaped semicircular arc.

When the straight line y=x+b is cut down with the circle, you are doing it right, at this time b=1 2 root number 2 and the straight line y=x+b is cut up with the circle, it has exceeded the y=3 straight line above, which is meaningless, at this time you should consider the upper junction of the line and the arc p(0,3).

Bringing p into the line y=x+b, we get: b= 3

You get: b (1-2 root numbers 2,3).

Now you can see what you're doing wrong, right?

Here's how I do it.

Curve y=3- (4x-x) is transformed into.

x²-4x+y²-6y+9=0

The center of the circle is (2,3) and the radius is 2

Substituting y=x+b into the circular equation obtains.

x -4x+x +2bx+b -6x-6b+9=02x +2(b-5)x+b -6b+9=0 The line has an intersection with the circle.

=4（b-5)²-8(b²-6b+9)≥0b²-10b+25-2b²+12b-18≥0b²-2b-7≤0

b-1)²-8≤0

1-2√2≤b≤1+2√2

The answer is the same as yours.

y= (1-x 2) is a semicircle, above the x-axis, substituting y=x+b into the semicircle equation, x+b= (1-x 2), (1).

2x 2+2bx+b 2-1=0, because there are two intersections, the discriminant formula should be greater than 0,4b 2-8(b 2-1)>0,b 2<2, 2=0

To ensure that there are two intersection points, y=x+b should be on the left side of the line connecting the vertices and the left vertices on the semicircle, x<=-1,1>=x,b>1,1

Curve y=3 - root number.

Copy 4x-x 2

y=3- (4x-x 2)=3- [4-(x-2) raid 2] because: the root number is greater than or equal to 0 and less than or equal to 4, so the value of y between 1 and 3 is: (y-3) 2+(x-2) 2=4, that is, the curve is part of the circle, and the coordinates of the center of the circle are (2, 3).

It is easy to know that the curve represents the lower half of the circle.

If the line y=x+b has an intersection with the curve, then.

b 3 (there is exactly one intersection point when b = 3, no when it is larger) When the line and the semicircle are tangent, b is taken to the minimum.

This can be done in this way:

The distance between the line L1 tangent to the circle and the line L2 when b is taken as the maximum is 2 + 2 (the slope of the line is 1 and the radius of the circle is 2).

Let the intersection of the x-axis of the straight line L1 at the point D, and the perpendicular line from D to L2, the distance is obviously 2+ 2, so the distance between the two straight lines at the intersection point of the X-axis is 2+2 2, because the straight line L2 intersects the X-axis at (-3,0).

Therefore, the x-axis of l1 intersection is (2 2-1,0).

Thus the minimum value of b is 1-2 2

The value range of b is [1-2, 2, 3].

y=3 - root number (

du4x-x 2) can be turned into.

x-2)²+y-3)²=4

0 x 4 is one.

Zhi Yuan. There is an intersection with the straight line daoy=x+b.

It is necessary that the distance from the center of the circle (2,3) to the straight line is less than or equal to half the diameter of 2, which is d=|2-3+b|The root number (1+1) <2 is |b-1|< 2 root number 2

It's 1-2 root number 2

(x-2) 2+(y-3) 2=4 (1<=y<=3) is obtained from y=3- (4x-x 2), which represents the lower half of the circle with a radius of 2 and the center of the circle is (2,3), y=x+b represents a simple parallel line with a slope of 1, where b is the intercept of the line on the y-axis, and when the line is tangent to the circle, the distance from the center of the circle to the straight line is equal to the radius, i.e., |2+b-3|1+1)=2 , the solution is b= 1-2 2 (rounding off 1+2 2), and the value range of b is [1-2 2,3].