Find the length of chord AB which is cut by circle C: x? + y? - 2x-4y = 0 for the straight line L: 3x-y-6 = 0. Use three different methods to calculate

Find the length of chord AB which is cut by circle C: x? + y? - 2x-4y = 0 for the straight line L: 3x-y-6 = 0. Use three different methods to calculate

(1) The coordinate of intersection point and the formula of distance between two points are solved;
(2) Calculate the distance and radius from the center of the circle to the straight line, and calculate the chord length;
(3) Chord length = │ x1-x2 │ √ (k ^ 2 + 1) = │ y1-y2 │√ [(1 / K ^ 2) + 1]

Bivariate linear equation: {3x = 2Y, 3x + 4Y = 36 {3x + 4Y = 10,4x + y = 9 {2x-y = 6, x + 2Y = - 2 {x + y = 420,30% x + 40% y = 160 × 80%

3x = 2x = 2Y, 3x + 4Y = 36x = 2Y = 2Y = 3Y + 4Y = 362y + 4Y = 36Y = 6y = 6x = 2 * 6 / 3 = 4 {3x + 4Y = 10,4x + y = 9x = (10-4y) / 3 generation into 4x + y = 94 (10-4y) / 3 + y = 940-16y + 3Y = 27y = 1x = (10-4 * 1) / 3 = 2 {2x- y = 6, x + 2Y = - 2 x = (6 + y) / 2, x + 2Y = (6 + y) / 2, x + 2Y = (6 + y) / 2, x + 2Y = (6 + y) / 2, x + 2Y + 2Y = (6 + y) / 2, x + 2- 2 (6 + y) / 2 + 2Y = - 26 + y + 4Y = - 4Y =

The circle x2 + y2-4x + 2Y + C = 0 and the straight line 3x-4y = 0 intersect at two points a and B. the center of the circle is p. if ∠ APB = 90 °, then the value of C is () A. 8 B. 2 Three C. -3 D. 3

The standard equation of circle is (X-2) 2 + (y + 1) 2 = 5-c, center C (2, - 1), radius R=
5−c，
∵∠APB=90°，
∴AP⊥BP，
The distance d from the center P of the circle to the line ab=
Two
2•
5−c，
That is, d = | 6 + 4|
5=
Two
2•
5−c，
C = - 3,
Therefore, C

We know that the straight line L: 3x-y-6 = 0, and the circle C: x ^ 2 + y ^ 2-2x-4y = 0. (1) find the distance between the center of the circle C and the line L; (2) find the length of the chord ab of the line L cut by the circle C

Circle C: (x-1) ^ 2 + (Y-2) ^ 2 = 5
The distance from center C (1,2) to straight line d = | 3-2-6 | / radical (9 + 1) = 5 / Radix 10 = (Radix 10) / 2
Because of: D ^ 2 + (AB / 2) ^ 2 = R ^ 2
10/4+AB^2/4=5
AB^2=10
So AB = radical 10

X = x y = x x in the plane of the right angle In the plane rectangular coordinate system xoy, if the line y = x = root 2 and X axis intersect at point a and the inverse scale function y = K / X in the first line intersect with point B, and the abscissa of B is root 2, then k =?

Because the image is in the first quadrant, so k > 0, because the abscissa of point B is root 2, so put x = root 2 into the line y = x + root 2, and get y = 2 root sign 2, that is B (root 2, 2 root sign 2), replace B coordinate into the inverse scale function y = K / x, and get k = 4

In the plane rectangular coordinate system xoy, the line L1 is known to pass through point a (- 2,0) and point B (0,2 / 3 root sign 3). The function analytic formula of line L2 is y = - radical 3 / 3x + 4 / 3 root sign * 3. L1 and L2 intersect with point P, and circle C is a moving circle, and its center is in a straight line

When L1 moves up, let the abscissa of the center of circle C be a. cross point C as cm vertical X axis, perpendicular foot as point M,
(2) When circle C is tangent to line L2, please prove that the distance from point P to line cm is equal to the radius r of circle C, and write the value of a when r = 3 root sign 2-2
(3) When the circle C and the straight line L2 are not separated, the radius of circle C is known to be r = 3, and the root sign is 2-2. Note that the area of the quadrilateral nmop is s (where the point n is the intersection of the line cm and L2.) is there a maximum value of S? If there is, find the maximum value and the value of a at this time; if not, please explain the reason

In the plane rectangular coordinate system, the coordinates of known points ABCD are (0,2), (- radical 3,0) (0, - 2) (root 3,0). Judge the shape of quadrilateral ABCD And say why

Let a (0,2), B (- radical 3,0), C (0, - 2), D (root 3,0), then a, C are symmetric about the origin, OA = OC; similarly, B, D are symmetric about the origin, OB = od.. because AC ⊥ BD, and OA = OB, OC = OD, the quadrilateral ABCD is a diamond

The number of points on the curve x2 + y2-4x-2y-11 = 0 to the straight line 3x + 4Y + 5 = 0 is () A. 1 B. 2 C. 3 D. 4

The formula of x2 + y2-4x-2y-11 = 0 shows: (X-2) 2 + (Y-1) 2 = 16, which is a circle with (2,1) as the center and 4 as the radius
The distance from the center of the circle to 3x + 4Y + 5 = 0 is d = 6 + 4 + 5
5＝3，
Therefore, if we make a line with a distance of 1 from the line 3x + 4Y + 5 = 0, we will find that there are two such lines (one above the line and one below the line). The upper line has an intersection with the circle, and the lower line has two intersections with the circle. Therefore, there are three points on the circle with a distance of 1 from the straight line
Therefore, C

(0.12x ^ 4Y? 3 - 0.9x? Y?) / 0.3x? Y? Is equal to?

(0.12x^4y³-0.9x²y³）÷0.3x²y²
=0.4x²y-3y

The minimum distance between the point on the circle x2 + y2 = 1 and the straight line 3x + 4y-25 = 0 is______ .

∵ the distance from the center of the circle (0, 0) to the straight line 3x + 4y-25 = 0 d = 25
5＝5
The minimum distance from the point on the circle x2 + y2 = 1 to the line 3x + 4y-25 = 0 is AC = 5-r = 5-1 = 4
So the answer is: 4