It is known that P is on the straight line L: x + Y-1 = 0 and Q is on the circle C: (X-2) 2 + (Y-2) 2 = 1. (1) make the tangent PM, PN and tangent point of circle C through P It is known that P is on the straight line L: x + Y-1 = 0 and Q is on the circle C: (X-2) 2 + (Y-2) 2 = 1. (1) through P, make the tangent PM, PN of circle C, and the tangent points are m, N, and find the minimum value of cos ∠ MPN. (2) through P, make the tangent PM, PN of circle C, and the tangent points are m, N, and find cos ∠ MPN ≤ 35, calculate the value range of abscissa of point P

To find the minimum value of cos ∠ MPN is to find the minimum value of cos ∠ CPN first
That is to find the maximum value of sin ∠ CPN first, that is to find the maximum value of CN / PC in the right triangle CPN
Obviously, CN = radius = constant, so we need to find the minimum value of PC first
It is to find the minimum value from the center of the circle C to the line L
It's a direct formula of distance
(2) It's the same with question (1)

It is known that the equation of circle C is (x-m) ^ 2 + (y + M-4) ^ 2 = 2. (1) find the trajectory equation of center C; (2) find the general equation of circle C when OC is minimum

1. C (m, 4-m) so the trajectory equation of center C is y = 4-x2, OC ^ 2 = m ^ 2 + (4-m) ^ 2 = 2m ^ 2-8m + 16 = 2 (m ^ 2-4m + 8) = 2 (m-2) ^ 2 + 8, so OC is the minimum when m = 2, so the general equation of circle C is (X-2) ^ 2 + (Y-2) ^ 2 = 2, do you understand? Do not understand Hi, I wish you study progress

Given that the equation of circle C is x square + y square = 4, two points a and B move on the circle, and the distance AB = 2 times root 3, find the trajectory equation of the midpoint m in ab

It is known that AB is the chord of circle C, ab = 2, root sign 3
Let m (x, y) be the midpoint of ab
Connecting center O and M OM
So om ⊥ AB, am = BM = radical 3
According to Pythagorean theorem, OM = radical (4-3) = 1
So the distance from m to the center of the circle (0,0) is 1
So the trajectory of M is
x^2+y^2=1

The moving circle P passes through point B and is circumscribed with the square of the circle + y square?
When the moving circle P passes through point B and is circumscribed with circle square + y square = 1, the trajectory equation of the center P of the moving circle P is obtained?

Let P = (x, y)
Root [(X-2) ^ 2 + y ^ 2] = root [(x + 2) ^ 2 + y ^ 2] - 1
X ^ 2-4x + 4 + y ^ 2 = x ^ 2 + 4x + 4 + y ^ 2 + 1-2 radical [(x + 2) ^ 2 + y ^ 2]
8x + 1 = 2 radical [(x + 2) ^ 2 + y ^ 2]
64x^2+16x+1=4x^2+16x+16+4y^2
60x ^ 2-4y ^ 2 = 15 is the trajectory equation of P

The moving circle P is circumscribed with the fixed circle a: x ^ 2 + (Y-3) ^ 2 = 9 and the fixed circle B: x ^ 2 + (y + 3) ^ 2 = 1. The trajectory equation of the center P is obtained

Let P (x, y) be circumscribed with fixed circle a: x ^ 2 + (Y-3) ^ 2 = 9 and fixed circle B: x ^ 2 + (y + 3) ^ 2 = 1, then the distance between P and other centers is constant R + 3 - (R + 1) = 2, which is obviously the locus of hyperbola. The intersection of hyperbola is (0,3) (0, - 3) according to the first definition of hyperbola and two fixed points F1, f on the plane

Passing through point a (2,1), the straight line and x-axis and y-axis intersect at two points B and C respectively, and the trajectory equation of the midpoint m of line BC is obtained

Let the line be y = K (X-2) + 1
Then the BC coordinates can be obtained as: B (2-1 / K, 0), C (0,1-2k)
Let the coordinates of the midpoint of BC be: m (x, y)
Then:
x=(2-1/k)/2=1-1/2k,
y=(1-2k)/2,k=1/2-y
therefore
x=1-1/2k=1-1/(1-2y)=-2y/(1-2y)
Therefore, the trajectory equation of the midpoint m of BC is as follows:
x(2y-1)=2y

Known: point a (5,0), point B is a moving point on the circle C: x ^ 2 + y ^ 2 = 9, find the trajectory equation of the midpoint P of line ab
Don't lose the answer. I know the answer,

Let P (x, y), B (x1, Y1) a (5,0)
∵ P is the midpoint of line ab
∴x1+5=2x ①
y1=2y ②
From ①: X1 = 2x-5 ③
∵ point B is the point on the circle C: x ^ 2 + y ^ 2 = 9
х take ③ and ② into the circle C equation
X1²+Y1²=9
(2x-5)²+4y²=9
(x-5/2)²+y²=(3/2)²

Through the origin, make a straight line intersection circle (X-8) ^ 2 + y ^ 2 = 1, and find the trajectory equation of the midpoint of line AB at two points a and B

Let the coordinates of the midpoint of AB be (x, y). 2. Let the linear equation be y = KX. 3. Combine the linear equation with the circular equation, and substitute y = KX into the circular equation to obtain the quadratic equation with respect to X. the abscissa of the midpoint of AB is x = (x1 + x2) / 2

The straight line passing through point P (0,2) and circle x + y = 2 intersect at two points a and B. let m be the midpoint of line AB, the trajectory equation of point m is obtained

Let a (x1, Y1), B (X2, Y2), m (x, y), the slope of the straight line be K. then there are X1 ^ 2-y1 ^ 2 = 1, X2 ^ 2-y2 ^ 2 = 1. The subtraction of the two formulas is: (y1-y2) / (x1-x2) = (x1, x2) / (Y1, Y2). Point m is the midpoint of line AB, so x1, X2 = 2x, Y1, y2 = 2Y, so k = 2x / (2Y) = x / y. point m is on the straight line, so y-0 = (x / y) (x

If two straight lines L1 and L2 passing through point P (1,1) and perpendicular to each other intersect with X and Y axes at two points a and B respectively, then the trajectory equation of point m in AB is______ ．

Let the coordinates of M be (x, y), then the coordinates of a and B are (2x, 0), (0, 2Y), connecting PM, ∵ L1 ⊥ L2, | PM | = | om |. And | PM | = (x − 1) 2 + (Y − 1) 2, | om | = x2 + Y2 (x − 1) 2 + (Y − 1) 2 = x2 + Y2, which is simplified to get x + Y-1 = 0, which is the trajectory