M is known to belong to R, straight line L MX M square 1 Y 4M and circle C X2 Y2 8X 4Y 16 0

1.The slope of the straight line is m (m squared + 1).

Because m belongs to r

If the upper and lower divisions are equal to m, the slope is 1 (m+1 m).

It can be obtained with basic inequalities.

When m>0 and m+1 m is greater than or equal to 2 (equal sign if and only if m=1 is taken), the slope is less than or equal to 1 2

When m<0 and m+1 m is less than or equal to -2 (if and only if m=-1 is equal sign), then the slope is greater than or equal to -1 2

When m=0, the slope is zero.

In summary, -1 2 is less than or equal to the slope less than or equal to 1 2

2.A circle can be formed into (x-4) squared + (y-2) squared = 4, with the center of the circle being (4,2) and the radius being 2

If the straight line l can divide the circle c into two arcs with a ratio of 1 2 arcs, then the central angles of the circle are 120° and 240°

Passing through the center of the circle is the perpendicular line of the straight line.

Use the formula to calculate the distance from the center of the circle to the straight line as half the radius.

The resulting equation is 3m to the fourth power + 5m square + 3 = 0 without solution. So no.

m2 represents the square of m).

1.Since the slope of the straight line is m (m 2+1)=1 (m+1 m);

Since m+1 m>=2; So 1 (m+1 m)<=1 2;

Therefore, the value range of the slope of the straight line is: (-1 2).

2.Assuming that the circle c can be divided into arcs at both ends with a ratio of 1 2, then, since the equation for the circle can be reduced to: (x-4) 2+(y+2) 2=4;

It can be seen that the radius of the circle is r=2, and the center of the circle is at the point o(4,-2);

Let the smaller central angle be a and the larger central angle be b, then the long arc is r*b=2b, and the short arc length is r*a=2a.

So (2a) (2b) = 1 2....1),a+b=360°..2), from (1) and (2), it can be solved: a=120°, b=240°;

Drawing the diagram in a Cartesian coordinate system, the distance from the center of the circle to the straight line can be solved from the smaller central angle a=120° and the radius of the circle r=2 as:

r*cos60°=1...3)；

And because the distance from the center of the circle to the straight line is:

4m+2m^2+2-4m|/[√m^2+(m^2+1)^2]..4);

From (3) and (4) we get: |4m+2m^2+2-4m|/=1;

Simplified as: 2(m 2+1) = [m 2+(m 2+1) 2];

Square on both sides of the equation: 4(m 2+1) 2=m 2+(m 2+1) 2;

Finally, it is simplified as: 3m 4+5m 2+3=0;

From the discriminant formula of the root, it can be seen that the equation has no solution;

So there is no m that makes the assumption true;

So the assumption is not valid;

So the straight line l cannot divide the circle c into arcs at both ends with a ratio of 1 2 to the arc length.

First, m=0 gives us a write rate of 0, and then there is an important inequality: k=m (m2-1), the numerator and denominator, and divide by m to calculate the slope between plus and minus.

I don't think it's possible to simplify the equation of the circle, and if the radius is 2, then the arc should be divided into 1 to 2, then the central angle of the circle should be changed to 120°, then the distance from the straight line to the original center is 1, and it is not valid to bring it in.

Problem solving ideas: (1) the straight line l: (2m+1)x+(m+1)y-7m-4=0, that is, m(2x+y-7)+(x+y-4)=0, then the letter obviously passes through the intersection point a of the straight line 2x+y-7=0 and the straight line x+y-4=0, and the coordinates of the intersection point a are obtained by solving it

2) Turning the equation of the circle c into a standard form, finding the coordinates and radius of the center of the circle c, to make the length of the line segment cut by the circle c to be the minimum, the distance d from the center c to the line l is the maximum, and the maximum of d is the length of the line segment of the ca At this time, Ca and the line L are perpendicular, the product of the slope is equal to -1, and the value of m is obtained by solving the equation

1) Straight line l: (2m+1)x+(m+1)y-7m-4=0, that is, m(2x+y-7)+(x+y-4)=0, obviously through the intersection point a of the straight line 2x+y-7=0 and the straight line x+y-4=0

Composed. 2x+y−7＝0

x+y 4 0 The coordinates of the intersection point a are (3,1), so the line l:(2m+1)x+(m+1)y-7m-4=0 passes through the fixed point a(3,1).

2) Circle c: x2+y2-2x-4y-20=0 i.e. (x-1)2+(y-2)2=25, which means a circle with c(1,2) as the center and 5 as the radius

Let the distance from the center of the circle c to the line l be d, so that the length of the line segment cut by the circle c is the minimum, d needs to be the maximum From the title, it can be seen that the maximum of d is the length of the line segment of ca

The formula for the distance between two points gives ca=

In this case, Ca and the straight line l are perpendicular and the product of the slope is equal to -1, [1 2 3 1] (

2m+1m+1)=-1, and the solution is m=-[3 4].

Comments: Test points for this question: A straight line that is constant past a fixed point; The positional relationship between a line and a circle

Test Center Comments: This question mainly examines the problem of straight lines passing through fixed points, the application of the position relationship between straight lines and circles, and the maximum distance d from the center of the circle C to the direct grandson imitation wheel line L is the length of the Ca line segment, which is the key to solving the problem

Solution: (1) The equation for the straight line l can be reduced to y=mm2+1x-4mm2+1, and the slope k=mm2+1

Because |m|12(m2+1), so |k|=|m|m2+1 12, therefore, the range of slope k is [-12,12] (2) cannot be known by (1 The equation for l is y=k(x-4), where |k|≤12；

The center of the circle c is c(4,-2) and the radius r=2;The distance from the center c of the circle to the line l d=21+k2

Composed. k|12, get d 45 1, that is, d r2, thus, if l intersects the circle c, then the central angle of the chord obtained by the truncated line l of the circle c is less than 2 3, so l cannot divide the circle c into two arcs with a ratio of arc length of 12

Center of the circle (4,-2) radius r=2 The distance from the center of the circle to the straight line d=|2m^2+2|Root number (m 4 + 3m 2 + 1) Half of the string t = root number [4-d 2].

sabc=d*t=8 5 d 2=4 5 or d 2=16 5 when d 2=4 5 m has no solution.

When doing the knowledge of d 2=16 5, m= 1 The equation for the pure cover of the straight line to bury l is x-2y-4=0 or x+2y-4=0

C:x2+y2-8x+4y+16=0,x-4) +y+2) =4 center c(4,-2) radius r=2

Let the distance from the center c of the circle to the line l be h, and the intersection point is m

The intersection of a straight line and a circle is pure return a, b

The small arc is known: the large arc=1:2 then ACB=120° ACM=60°CM=ca*cos ACM, i.e., h=r*cos60°H=2*(1 2)=1

By h=i4m-(m +1)(-2)-4mi [m + (m +1) ]1

4(m²+1)²=m²+(m²+1)²

3(m +1) = m 3(m +1) -m =0 3m +m+1)( 3m -m+1) = 03m +m+1=0 No solution.

or 3m -m+1=0 without a group of pants late solution.

Therefore, there is no m that belongs to r, and the condition is satisfied, that is, the circle cannot be divided into two arcs with a ratio of 1:2 arcs.

Hope it helps you o(o

The arc length of the straight line over the fixed point (old punch 4,0) is 1:2, and the corresponding ratio of the center angle is 1:2, that is, 120:

240 。If the perpendicular line passes through the center of the circle to make a straight line, then the distance from (4,-2) to the straight line is 1. According to the formula for the distance from a point to a straight line, the waiter can be solved.

1.The slope of the straight line is m (m squared + 1).

Because m belongs to r

If the upper and lower divisions are equal to m, the slope is 1 (m+1 m).

It can be obtained with basic inequalities.

When m>0 and m+1 m is greater than or equal to 2 (equal sign if and only if m=1 is taken), the slope is less than or equal to 1 2

When m<0 and m+1 m is less than or equal to -2 (if and only if m=-1 is equal sign), then the slope is greater than or equal to -1 2

When m=0, the slope is zero.

In summary, -1 2 is less than or equal to the slope less than or equal to 1 2

2.A circle can be formed into (x-4) squared + (y-2) squared = 4, with the center of the circle being (4,2) and the radius being 2

If the straight line l can divide the circle c into two arcs with a ratio of 1 2 arcs, then the central angles of the circle are 120° and 240°

Passing through the center of the circle is the perpendicular line of the straight line.

Use the formula to calculate the distance from the center of the circle to the straight line as half the radius.

The resulting equation is 3m to the fourth power + 5m square + 3 = 0, so there is no solution, so it can't.

If the line l divides the circle c into two arcs with a ratio of 1 2 to the arc length, then the central angle of the pair is 120 degrees, so the distance from the center of the circle to the straight line should be 1 2

x^2+y^2-8x+4y+16=0

x-4)^2+(y+2)^2=6^2

The distance from the center of the circle (4,-2) to the line l is.

3 m does not exist.

So no.

Whether a straight line l can divide a circle c into two arcs with a ratio of 1 2 is equivalent to asking:

The distance from the center of the circle to the straight line l can be 1, and you can understand it by drawing the following diagram yourself. Very good solution, y=m(x+4) (m 2+1).

Let d = 1 solve the equation to see if m has a real solution.

1.It is known that m belongs to r, and the line l:mx-(m2+1)y=4m is obliquely trunched y=m (m 2+1)x+4m (m 2+1)k=m (m 2+1).

1)m=0 k=0

2) m>0 k=1 (m+1 m) m+1 m>=2 so 0=2 so -1 2<=k<0

Therefore, the value range of the slope of the straight line l is [-1 2,1 2] 2The center of the circle is (4,-2) and the radius is 2, and if the circle c can be divided into two arcs with a ratio of 1 2 to the arc length, then the distance from the center of the circle to the straight line is d = 3

From (1) we know that -1 2<=k<=1 2

So when k = 1 2, the center of the circle is closest to the straight line, and at this time m = 1, the closest distance = 4 5> 3

So the straight line l cannot divide the circle c into two arcs with a ratio of 1 2 to the arc length.

The slope ranges from minus infinity to positive one-half.

It can be divided because the center of the circle is (4,-2) and the radius is 2, then the circle will pass the point (4,0) and the straight line will pass the point (4,0).

1) y=m (m 2+1)x-4m (m 2+1), the slope is m (m 2+1)=1 (m+1 m), when m>0, m+1 m is greater than or equal to 2, so the slope 02) straight line x-4=(m 2+1) m*y, circle(x-4) 2+y 2+4y=0, the linear equation is substituted into the circle equation to obtain (m 4+3m 2+1) m 2*y 2+4y=0, because the y=0 equation has a solution, That is, the intersection of the line and the equation is a little bit on the x-axis.

In order for the straight line l to be able to divide the circle into two segments of the circle with a ratio of one-half of the arc length, that is, the angle between the two intersection points and the center of the circle is 120 degrees, that is, the slope is plus or minus 3 of the root number three, and the root of the 3rd of the dena is 1 2, so it can't.