1. The point a (0,1) is known; the straight line L with slope k intersects the circle C: (X-2) ^ 2 + (Y-3) ^ 2 = 1 at two different points m and N. find out the value range of real number K 2) 3) if O is the origin of coordinate, and the vector om * vector on = 12. Find the value of K. in the afternoon, please see the question: l but point a

1. Radius (2,3) of circle center r = 1
Let L: y = KX + 1
The distance from the center of the circle to the straight line should be less than the radius
｜3k-2|

Given that the straight line L passes through the point (- 2,0), when the line L and the circle x2 + y2 = 2x have two intersections, the value range of its slope k is______ .

If the slope of the straight line L does not exist, then the straight line L is separated from the circle, which is inconsistent with the meaning of the title. Therefore, if the slope of the straight line L does not exist, then the straight line L is separated from the circle, which is inconsistent with the meaning of the topic ...

The straight line L passing through the point a (0,1) and the slope of K intersects the circle C: (X-2)? + (Y-3)? = 1 at M and n

(1) The equation has two unequal real roots, so [- 4 (K + 1)] ^ 2-4 * 7 (K + 1) x + 7 = 0, so [- 4 (K + 1)] ^ 2-4 * 7 (K + 1) 0, K < - 1 or K3 / 4, (2) because m, n are collinear, the angle between vector am and an is 0, am * an = | am|||||||||||||||||||||||||||||||

The line L passing through point a (0,1) and slope k intersects with circle C: (X-2) + (Y-3) = 1 at M and N. 1) find the range of real number K. 2) prove that the vector am × vector an is a fixed value 3) if O is the coordinate origin and the vector om * vector on = 12, find the value of K

(1) The equation has two unequal real roots, so [- 4 (K + 1)] ^ 2-4 * 7 (K + 1) > 0, the solution is K3 / 4, (2) because m, n are collinear, the angle of vector am, an = 0, am * an = | an | cos0 = | am | *

The straight line L passing through the point a (0,1) and the slope of K and the circle (X-2) ^ 2 + (Y-3) ^ 2 = 1, intersect at two points of Mn. It is proved that the vector am multiplied by the vector an is a constant value

The straight line L: y = KX + 1
Replace the circle C (X-2) ^ 2 + (Y-3) ^ 2 = 1
The result is: (X-2) ^ 2 + ((kx-2) ^ 2 = 1
That is, (1 + k? 2) x? - (4 + 4K) x + 7 = 0
Δ = 16 (1 + k) - 28 (1 + K 2) > 0 is required
Let m (x1, Y1), n (X2, Y2)
Then X1 + x2 = 4 (K + 1) / (K 2 + 1)
x1x2=7/(k²+1)
ν vector am. Vector an
=(x1+y1-1)●(x2,y2-1)
=x1x2+(y1-1)(y2-1)
=x1x2+kx1*kx2
=（1+k²)x1x2
=(1+k²)*7/(1+k²)
=7
I.e. vector am. Vector an = constant value 7
Method 2: geometric method
| AC|=2√2
Leading tangent line ad through a direction circle
|AD|²=|AC|²-r²=8-1=7
According to the cutting line theorem:
|AM||AN|=|AD|²=7
The angle between vector am and an is 0
| vector am. Vector an = | am | an | 7
Hope to help you, thank you!

Circle C (X-7) ^ 2 + (y + 1) ^ 2 = 50, if the straight line with slope of - 1 intersects with circle C at different two points m, n calculates the value range of vector am × vector an a (2,4)

Let x + C be the linear equation,
Because a line and a circle have two different intersections,
So the distance from the center of the circle to the straight line is less than the radius,
Namely | 7-1 + C | / √ 2

The straight line passing through the point m (m, 0) and the slope of - √ 3 / 3 and the circle x2 + y2 = 1 intersect at two points a, B, and vector am = 2, vector MB, calculate the value of M

Let the intersection be a (x1, Y1), B (X2, Y2)
Let the straight line passing through the point m (m, 0) be y = - √ 3 / 3 * (x-m)
If we bring in the circle equation, we get
X ^ 2 + (x-m) ^ 2 / 3 = 1
4X ^ 2-2mx + m ^ 2-3 = 0
x1+x2=m/2, x1x2=(m^2-3)/4; y1+y2=-(x1+x2-2m)/√3=√3/2*m
y1y2=(x1-m)(x2-m)/3=[x1x2-m(x1+x2)+m^2]/3=(m^2-1)/4
And vector am = (m-x1, - Y1), vector MB = (x2-m, Y2)
And vector am = 2 vector MB
ν there are m-x1 = 2 (x2-m), - Y1 = 2Y2
Combined with the equation obtained from the above-mentioned Veda theorem, it can be solved
x1=-2m, x2=5m/2
y1=√3*m, y2=-√3/2*m
m^2=1/7, m=±√(1/7)
The value of M is m = ± √ (1 / 7)

Given the circle C: (x-3) ^ 2 + (y-4) ^ 2 = 4, the straight line L1 crosses the fixed point a (1,0) If L1 and circle intersect at P and Q, the midpoint of PQ is m, and the intersection of L1 and L2: x + 2Y + 2 = 0 is n, what is the fixed value of am * an

Let the slope of the line L1 be K, then L1: y = K (x-1) L2: x + 2Y + 2 = 0, we can get the coordinates n of n [(2k-2) / (2k + 1), (- 3) K / (2k + 1)] let the coordinates of point m be (x0, K (x0-1)) from the coordinates C (3,4) of the center of circle C, we can get the slope of the line where cm is located K (CM) = [4-K (x0-1)] / (3-x0) and cm

Given the circle C: x ^ 2 + y ^ 2 + 2x-6y + 1 = 0, the straight line L: x + my = 3 1, if l is tangent to C, find the value of M 2 The circle C: x ^ 2 + y ^ 2 + 2x-6y + 1 = 0 and the straight line L: x + my = 3 are known If l is tangent to C, find the value of M 2 whether there is m value, so that l and C intersect with C at a and B, and the vector OA * vector ob = 0 (where o is the coordinate origin), if there is, find m, if not, please explain the reason

If the straight line L passes through the point (- 1,0), and l intersects the circle C: (x-1) 2 + y2 = 3 at two points a and B, then the probability of chord length | ab | ≥ 2 is___ .

The dot is (1, 0) and the radius is
3，
It is known that (- 1,0) outside the circle must have chord length | ab | 2, and the radius is
3，
Let the perpendicular foot of the line passing through the circle point perpendicular to ab be C, then we can get the distance between the circle point and ab is
2，
In the right triangle composed of (- 1,0) (1,0) and C, we can see that the straight line (- 1,0) is 45 ° to the X axis
When the line is tangent to the circle, the line passing through (- 1,0) is 60 ° to the X axis
So the probability is: 45 ° + 45 °
60°+60°=3
4．
So the answer is: 3
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