It is known that circle C: x2 + (Y-1) 2 = 5 and line L: mx-y + 1-m = 0. Let L and circle C intersect at two points a and B. if the inclination angle of line L is 120 degrees, the length of chord AB can be obtained

It is known that circle C: x2 + (Y-1) 2 = 5 and line L: mx-y + 1-m = 0. Let L and circle C intersect at two points a and B. if the inclination angle of line L is 120 degrees, the length of chord AB can be obtained

Provide ideas: the slope of the straight line is negative root sign 3, that is, the value of M ~ when the inclination angle is 120 °, then, substitute the value of m into the straight line, and then combine the equation of the straight line and the circle to solve two groups of solutions, which are two intersections ~ finally, use the two-point distance formula to find the two-point distance ~ which is the chord length ab~

Given the circle C: x2 + (Y-1) 2 = 5, the straight line L: mx-y + 1-m = 0. The proof: for m ∈ R, if the straight line L bisects the circle, find the value of M

Mx-y + 1-m = 0 through fixed point (1,1)
And (1,1) is inside the circle C
So for any R, there are two intersections between a line and a circle
When a line bisects a circle, it passes through the center (0,1)
So 0-1 + 1-m = 0
m=0

It is known that the circle C: x2 + (Y-1) 2 = 5, the line L: mx-y + 1-m = 0, and the line L intersects the circle C at two points a and B
(1) If the absolute value of AB = root 17, find the tilt angle of L;
(2) If point P (1,1) satisfies 2 vector AP = vector Pb, the equation of line L is obtained

(1) By substituting y = MX + 1-m into x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\35; 178; = 3M = ± √ 3  inclination angle = 60 ° or 120 ° (2

Given the circle C: (x-1) ^ 2 + y ^ 2 = R ^ 2 (r > 1), let m be the intersection of the circle C and the negative half axis of the x-axis. Let m be the chord Mn of the circle C through M, and let its midpoint p be exactly the same
When R ∈ (1, + ∞), the equation of locus e of point n is obtained
(2) A (x1,2), B (X2, Y2), C (x0, Y0) are different points on e, and ab is perpendicular to BC, so the value range of Y0 can be obtained
It is known that the vertex of the parabola C is at the origin, the focus is on the X axis, and the point P (2,4) on the parabola C is the moving chord PA of the parabola through P
, Pb, let the slopes be kPa, KPB respectively,
1. Solving parabolic equation
2. If kPa + KPB = 0, the slope of the straight line AB is determined and its value is obtained
3. If kPa + KPB = 1, the line AB is proved to pass through the fixed point and its coordinates are obtained

(1) Because m (1-r, 0), the midpoint is the Y axis, so n (R-1, y). The equation substituted into the circle is: n (R-1, under the positive and negative radicals (4r-4)). For the X and Y coordinates of N, the elimination parameter r is: e: y ^ 2 = 4x
(2) Determine a (1,2). By using the product of slopes as - 1 and two-point formula, we can get: Y0 = - (Y2 + 16 / (Y2 + 2)) (we can get rid of the coincidence point of B and C (that is, let Y0 = Y2 solve, there is no) and the coincidence point of a and B (that is, y2 = 2, Y0 = - 6)). By using the basic inequality, we can get the range of Y0: Y0 > = 10 or y

Go through the intersection of the circle O: x ^ 2 + y ^ 2 = 4 and the positive half axis of the y-axis, a make the tangent line L of the circle, M is any point on L, and then go through M to make all other lines of the circle, the tangent point is Q,
Then, when the point m moves on the straight line, the trajectory equation of the perpendicular center of the triangle MAQ is obtained
Steps to take
thank you

A(0,2).
Let the perpendicular center be h (x, y), q (x0, Y0)
Link AQ,
From the tangent property of plane geometry,
The triangle MAQ is isosceles triangle,
Point h is on OM, that is, the middle line of AQ at the bottom
kAQ=(y0-2)/x0,
kOM=y/x
∵AQ⊥OM
∴(y0-2)/x0= -x/y※
And x0 ^ 2 + Y0 ^ 2 = 4,
x=x0
Simplify
X ^ 2 + y ^ 2-4y = 0 is the trajectory equation

Given that there are two different intersections between the circle (x + 2) 2 + (y-m) 2 = 4 (m ∈ R) and the negative half axis of X axis, then the value range of real number m is

There are two different intersections between the circle and the negative half axis of the x-axis, let y = 0,
Then x ^ 2 + 4x + m ^ 2 = 0 has two unequal negative real roots
That is 4 ^ 2-4m ^ 2 ＞ 0 - 2 ＜ m < 2

Given that x ^ 2 + y ^ 2 = 1 / N ^ 2 and circle C (x-1) ^ 2 + y ^ 2 = 1, let the intersection of circle on and the positive half axis of Y axis be RN, the intersection of circle on and circle C above X axis be QN, and the intersection of line rnqn and X axis be at point PN. When n tends to infinity, point PN infinitely approaches a fixed point, and the abscissa of fixed point P is

Let m coordinate be (1-r, 0), because P is on the Y axis, and is the midpoint of M, N, then p coordinate is (0, y), n coordinate is (R-1, y). Substituting the coordinates of N point into the circular equation, we get y ^ 2 = 4 (R-1), y = under the root sign of plus or minus 2 * (R-1), so the coordinates of P point (0, plus or minus 2 times the root sign r-1.2) because P (0, 2), according to the answer in 1

If two circles intersect at points a (1,3) and B (m, - 1), and the centers of the two circles are on the straight line X-Y + C = 0, then the value of M + C is ()
A. -1B. 2C. 3D. 0

It can be seen from the meaning of the question: the straight line X-Y + C = 0 is the vertical bisector of the line AB, and the slope of the straight line X-Y + C = 0 is 1, then 3 − (− 1) 1 − M = - 1 ①, and M + 12-3 − 12 + C = 0 ②, M = 5 is obtained from the solution of ①, and M = 5 is replaced by the solution of ② to get C = - 2, then M + C = 5-2 = 3

If two circles intersect at points a (1,3) and B (m, - 1), and the centers of the two circles are on the straight line X-Y + C = 0, then M + C=______ ．

∵ if the centers of the two circles are on the straight line X-Y + C = 0, then the straight line X-Y + C = 0 is the vertical bisector of the line AB, that is KAB = - 1 = 3 + 11 − m, and the solution is m = 5, then the midpoint (3,1) of AB is on the straight line X-Y + C = 0, that is 3-1 + C = 0, and the solution is C = - 2 ∵ m + C = 3, so the answer is: 3

If two circles intersect at two points (1, 3) and (m, 1), and the centers of the two circles are on the straight line x − y + C2 = 0, then M + C = ()
A. -1B. 2C. 3D. 0

It is known that two circles intersect at two points (1,3) and (m, 1), and the centers of the two circles are on the straight line X-Y + C2 = 0, the slope of the common chord is: - 1, the common chord passing through point (1,3) is Y-3 = - 1 (x-1), so x + y-4 = 0, and because (m, 1) is on the common chord, so m + 1-4 = 0, the solution is m = 3; the midpoint of two points (1,3) and (3,1) is on the centripetal X-Y + C2 = 0, that is (2,2) On the connecting line X-Y + C2 = 0, so C = 0, so m + C = 3; so C