When the straight line l passes through the point 2,0 , when the line l and the circle x 2 y 2 2x,

y^2+x^2-2x=0

x-1)^2+y^2=1

It is a circle with (1,0) as the center and 1 as the radius.

Let the straight line be y=kx+b

Crossing point (-2, 0) b = 2k

y=kx+2k, i.e. kx-y+2k=0, if there are two intersections, then the distance from the center of the circle to the straight line should be less than 1, and the distance formula d=|k+2k|Root number (k 2+1) < 1 gives k 2<1 8

Then the value of k (-root number 2 4, root number 2 4).

y^2+x^2-2x=0

x-1)^2+y^2=1

3 = hypotenuse, 1 = right angled edge, solve right triangle.

Another right-angled side length = 2 2

tanθ=2√2/1=2√2=k1

k2=-2√2

2√2

y 2+x 2-2x=0 can be obtained by matching (x-1) 2+y 2=1, the center of the circle is (1,0), and the radius is 1Let the straight line be y=kx+b through the point (-2,0)b=2k y=kx+2k, that is, kx-y+2k=0 have two points, that is, intersect, then the distance from the center of the circle to the straight line is less than the radius root number (k 2+1) <1 gets k 2<1 8, so one of the two solutions is the root number of the quarter, and the other is its complex number.

Straight hole age line L over the point (-2,0).

Let the straight line y=k(x+2).

This is kx-y+2k=0

When the straight line l and the round sun x 2 + y 2 = 2x have two intersections.

x-1)^2+y^2=1

Utilize that the distance from the center of the circle to the line is less than the radius.

d=|k+2k| /1+k^2)<1

3k<√(1+k^2)

9k^2<1+k^2

8k^2<1

k^2<1/8

2/4

Solution: Let the slope of the straight line l with the point (-2,0) and two intersections with the circle x 2+y 2 2x be k, then the straight line l:y k(x 2), substitute y k(x 2) into x 2+y 2 2x, and get (1+k 2) bi number shirt x 2-(4k 2-2) x 4k 2 0 4k 2-2) 2-4(1+k 2) 4k 2 4-32k 2 0,- 2 4 k 2 4 k (repentance - 2 4, Bi Qi 2 4).

The square pattern of the straight line with the slope of the point (-2, 0) is k=k(x+2), and it is substituted by x2+y2=2x.

x 2+k 2(x 2+4x+4)=2x, and (k 2+1) x 2+(4k 2-2)x+4k 2=0,

There are two intersections between a straight line and a circle, == there are unequal trembling bands with solid roots, ==4=(2k 2-1) 2-4k 2(k 2+1).

4k^4-4k^2+1-4k^4-4k^2

1-8k 2>0, k 2< 1 8, so - 2 4

The center of the circle o(0,0), the point (-2,0) is the point a, and the tangent point is boa=2

The tangent is perpendicular to the radius.

radius = 1 = ob

The angle between the straight line and the x-axis is a

sina = 1 2 = >a = 30 degrees.

Linear efficiency = tana = root number 3 3, or - root number 3 3

With the easiest way:

Tangents, radii, and dim sum lines form a right-angled triangle

Tangent length = 3

Radius = 1 positive slope k = tana = r tangent length = 1 3

Negative slope k=-1 3

Let the equation for the straight line be:

y=k(x+2)

kx-y+2k=0

Because tangent, the distance from the center of the circle to the straight line = radius = 1

i.e. d=|2k|/√k²+1)=1

4k²=k²+1

3k = 1k = 3 3 so that is.

The slope is 3 3.

Let the slope of the straight line be k, then the equation for the straight line is: y=k(x+2)=kx+2k, and the equation substituted for the circle is split

x^-2x+(kx+2k)^=0

That is, (1+k)x+(4k-2)x+4k=0 has two intersections between the straight line and the circle, that is, the quadratic equation:

Discriminant = (4k -2) -4(1+k )*4k >0, i.e. k <1 8

2/4

The center of the circle o(0,0), the point (-2,0) is the point a, and the tangent point is boa=2 The tangent is perpendicular to the radius.

radius = 1 = ob

The angle between the straight line and the x-axis is a

sina = 1 2 = >a = 30 degrees.

Linear efficiency = tana = root number 3 3, or - root number 3 3

Solution: According to the meaning of the topic. Got: y 2

x^2-2x=0

x-1)^2

y 2=1 is a circle with (1,0) as the center and 1 as the radius.

Let the straight line be y=kx

b passes through the dot (-2,0) b = 2k

y=kx2k

That is. kx-y

2k=0If there are two intersections, then the distance from the center of the circle to the straight line should be less than 1 distance formula d=|k

2k|Root number (k 2

Get k 2<1 8

So. The value of k (-root 2 4, root 2 4).

The straight line l passes through the point (-2,0).

Let the straight line y=k(x+2).

This is kx-y+2k=0

When there are two intersections between a straight line l and a circle x 2+y 2=2x.

x-1)^2+y^2=1

Utilize that the distance from the center of the circle to the line is less than the radius.

d=|k+2k|

1+k^2)<1

3k<√(1+k^2)

9k^2<1+k^2

8k^2<1

k^2<1/8

The straight line l passes through the point (-2,0).

Let the straight line y=k(x+2).

This is kx-y+2k=0

When the line l and the circle x 2+y 2=2x have two intersections (x-1) 2+y 2=1

Utilize that the distance from the center of the circle to the line is less than the radius.

d=|k+2k| /√(1+k^2)<1

3k<√(1+k^2)

9k^2<1+k^2

8k^2<1

k^2<1/8

2/4

The straight line l passes through the point (-2,0).

Let the straight line y=k(x+2).

This is kx-y+2k=0

When the line l has two intersections with the circle x +y = 2x.

x-1)²+y²=1

Utilize that the distance from the center of the circle to the line is less than the radius.

d=|k+2k|

(1+k²)<1

3k<√(1+k²)

9k²<1+k²

8k²<1

k²<1/8

2/4

Circle x 2 + y 2 = 2x into the standard form (x-1) 2 + y 2 = 1 again by.

Point oblique. Write the equation of the straight line l, set the slope to kkx-y+2k=0, and use the distance from the center of the circle to the straight line is less than the radius to solve the range of k.