It is known that the equation of circle is x ^ 2 + y ^ 2 + KX + 2Y + K ^ 2 = 0. If the fixed point a (1,2) is outside the circle, the value range of K is obtained

It is known that the equation of circle is x ^ 2 + y ^ 2 + KX + 2Y + K ^ 2 = 0. If the fixed point a (1,2) is outside the circle, the value range of K is obtained

X ^ 2 + y ^ 2 + KX + 2Y + K ^ 2 = 0 (x + K / 2) ^ 2 + (K / 2) ^ 2 + y ^ 2 + 2Y + 1 + K ^ 2 - (k ^ 2) / 4-1 = 0 (x + K / 2) ^ 2 + (y + 1) ^ 2 + (3 / 4) k ^ 2-1 = 0 (x + K / 2) ^ 2 + (y + 1) ^ 2 = 1 - (3 / 4) k ^ 2 circle radius √ [1 - (3 / 4) k ^ 2], there should be 1 - (3 / 4) k ^ 2 ≥ 0, that is: K ^ 2 ≤ 4 / 3, the solution is: (2 √ 3) / 3 ≤ K ≤ (2 √ 3) / 3
Use a point to the center of the distance is greater than the radius on it! (x+k/2)^2+(y+1)^2=1-(3k^2)/4
x² + y² + kx + 2y + k² = 0
[x² + kx + (k/2)²] + [y² + 2y + 1] = - k² + (k/2)² + 1
(x + k/2)² + (y + 1)² = 1 - 3k²/4
Center C: (- K / 2, - 1) r
√[(1 + k/2)² + (2 + 1)²] > √(1 - 3k²/4)
==> k² < 4/3
==>- 2 / √ 3 < K < 2 / √ 3
First, the equation of the circle is transformed into a standard equation, which can be realized through the formula. In this way, (x + K / 2) ^ 2 + (y + 1) ^ 2 = 1-3k ^ 2 / 4 can be obtained,
So the one on the right side of the equation represents the square of the radius, so it's greater than 0, and we get the range of K
Then calculate the distance from the point P to the center of the circle to be greater than the radius. This can be achieved by the distance formula between two points. Combining these two aspects, find the intersection, ok... Expand
First, the equation of the circle is transformed into a standard equation, which can be realized through the formula. In this way, (x + K / 2) ^ 2 + (y + 1) ^ 2 = 1-3k ^ 2 / 4 can be obtained,
So the one on the right side of the equation represents the square of the radius, so it's greater than 0, and we get the range of K
Then calculate the distance between the point P and the center of the circle to be greater than the radius. This can be achieved through the distance formula between two points. Combining these two aspects, find the intersection and put it away
The value of 2cos π 2 − Tan π 4 + 34tan2 π 6 − sin π 6 + Cos2 π 6 + SIN3 π 2 is ()
A. 0B. 32C. −32D. 2
2cos π 2 − Tan π 4 + 34tan2 π 6 − sin π 6 + Cos2 π 6 + sin 3 π 2 = 2 × 0 − 1 + 34 × 13 − 12 + 34 − 1 = − 32, so select: C
Given the equation x ^ 2 + y ^ 2 + KX + 2Y + K ^ 2 = 0, what is the maximum area of a circle
X ^ 2 + y ^ 2 + KX + 2Y + K ^ 2 = 0
（x+k/2）²+（y+1）²=1-3k²/4
When the area of the circle is the largest, when k = 0
The area of the circle is equal to π * 1 ^ 2 = 3.14
Tan θ = 2, then Sin & sup2; θ + sin θ cos θ + 2cos & sup2; θ =?
tanθ=2
sinθ/cosθ=2
sinθ=2cosθ
sin²θ+cos²θ=1
5cos²θ=1
cos²θ=1/5
sin²θ+sinθcosθ+2cos²θ
=4cos²θ+2cos²θ+2cos²θ
=8cos²θ
=8/5
Given the equation x + y + KX + 2Y + k of a circle = 0, if there are two tangent lines of the circle passing through point (1,2), the value range of K is obtained
A:
x^2+y^2+kx+2y+k^2=0
(x + K / 2) ^ 2 + (y + 1) ^ 2 = 1-3k ^ 2 / 4 > 0, the solution is - 2 √ 3 / 30
To sum up, - 3 / 3
All the knowledge of triangles?
Beijing Normal University edition of junior high school students should master all the knowledge of triangle
1. Classification of triangles
The relationship between the sides of triangles is classified as follows
Triangles and isosceles are not included
Isosceles triangle includes isosceles triangle and equilateral triangle whose base and waist are not equal
Triangles are classified according to their angles as follows:
Triangles include right triangles (triangles with right angles) and oblique triangles
Oblique triangles include acute triangles (triangles with three acute angles) and obtuse triangles (triangles with one obtuse angle)
Connecting edges and angles, we have a special triangle: isosceles right triangle. It is a right triangle with two equal right sides
2. Trilateral relation theorem and inference of triangle
(1) Triangle trilateral relation theorem: the sum of two sides of a triangle is greater than the third side
Corollary: the difference between the two sides of a triangle is less than the third
3. The sum theorem of internal angles of triangles and its corollaries
Theorem of sum of internal angles of triangle: the sum of three internal angles of triangle is equal to 180 degrees
Inference:
① The two acute angles of a right triangle complement each other
② One exterior angle of a triangle is equal to the sum of two interior angles not adjacent to it
③ An outer angle of a triangle is greater than any inner angle not adjacent to it
Note: in the same triangle: equal angles to equal sides; equal sides to equal angles; big angles to big sides; big sides to big angles
4. Area of triangle
Area of triangle = × base × height
Congruent triangle
1. The concept of congruent triangle
Two triangles that are perfectly coincident are called congruent triangles
2. Determination of congruence of triangles
The theorem of triangle congruence
(1) Edge, corner and edge theorem: there are two triangles congruent (can be simply written as "edge, corner and edge" or "SAS") whose two sides and their angles are equal
(2) Angle, edge and angle theorem: there are two triangles congruent (which can be simply written as "angle, edge and angle" or "ASA") with two equal angles and their clip edges
(3) Edge theorem: two triangles with three corresponding equal sides are congruent
The judgement of congruent right triangle is as follows
For special right triangles, when they are congruent, there is HL theorem (hypotenuse, right edge theorem): two right triangles with hypotenuse and one right edge are congruent (which can be simply written as "hypotenuse, right edge" or "HL")
3. Congruent transformation
The graphic transformation that only changes the position of the figure without changing its shape and size is called congruent transformation
There are three kinds of congruent transformations
(1) Translation transformation: the transformation that the figure moves parallel along a line is called translation transformation
(2) Symmetrical Transformation: turn the figure 180 ° along a straight line. This transformation is called symmetrical transformation
(3) Rotation transformation: to rotate a figure around a certain point to another position by a certain angle. This transformation is called rotation transformation
an isosceles triangle
1. Properties of isosceles triangle
(1) The property theorem and inference of isosceles triangle
Theorem: the two base angles of isosceles triangle are equal
Inference 1: the bisector of the vertex angle of an isosceles triangle bisects the bottom and is perpendicular to the bottom. That is, the bisector of the vertex angle of an isosceles triangle, the middle line on the bottom and the height on the bottom coincide
Corollary 2: all the angles of an equilateral triangle are equal, and each angle is equal to 60 degrees
2. Median line in triangle
The line connecting the midpoint of a triangle is called the median line of the triangle
(1) There are three median lines in the triangle, and they form a new triangle
(2) You should be able to distinguish the middle line of a triangle from the middle line
The theorem of median line of triangle: the median line of triangle is parallel to the third side and equal to half of it
The function of the triangle median line theorem is as follows
Position relation: it can be proved that two lines are parallel
Quantity relation: it can prove the multiple relation of line segment
Common conclusion: any triangle has three median lines
Conclusion 1: three median lines make up a triangle whose circumference is half of the original triangle
Conclusion 2: three median lines divide the original triangle into four congruent triangles
Conclusion 3: three median lines divide the triangle into three parallelograms with equal area
Conclusion 4: a middle line of triangle and its intersecting median line are equally divided
Conclusion 5: the angle between any two median lines is equal to the vertex angle of the triangle
The line y = KX + 1 and the hyperbola 3x (square) - Y (square) = 1 intersect at two points a and B. what is the value of K when two points a and B are on the same branch of the hyperbola?
Suppose a (x1, Y1) B (X2, Y2)
A. B two points are on the same branch,
Then x1x2 > 0, y1y2 > 0
K = 0, does not meet the requirements
k≠0
Y = KX + 1 into hyperbola 3x (square) - Y (square) = 1
(3-K^2)X^2-2KX-2=0
Discrimination > 0
4k^2+8(3-k^2)>0,-√6
In △ ABC, if cot · B + C / 2 = 1, then what triangle is △ ABC
right triangle
Given that l: y = KX + 1 and hyperbola C: 3x square - y square = 1 intersect at two points a and B, is there a constant k such that a and B are symmetric with respect to y-2x = 0? If it exists, find K. if it does not exist, explain the reason
That's it. I'm sorry. With the help of a kind person~~~
Y = KX + 1 and 3x ^ 2-y ^ 2 = 1 eliminate y to get (3-K ^ 2) x ^ 2-2kx-2 = 0. ∵ there are two intersections between the straight line and the curve, the ∵ discriminant △ 0, the solution is: ∵ 6 ∵ K ∵ 6. (*) by Weida's theorem, X1 + x2 = 2K / (3-K ^ 2), Y1 + y2 = K (x1 + x2) + 2 = 6 / (3-K ^ 2), ∵ the midpoint coordinates of a and B (K / (3-K ^ 2), 3 / (3-K ^)
Let's multiply the slope of a by k-2x, and then we get the slope of a with respect to k-2x
Is the opposite side of a triangle tan or cot
It's Tan, cos, it's the adjacent side is bigger than the hypotenuse