Let a (- 3,5), B (2, - 3) be on the line L: 3x-4y + 4 = 0, find a point P so that | PA | - | Pb | is the maximum, and find this value

Let a (- 3,5), B (2, - 3) be on the line L: 3x-4y + 4 = 0, find a point P so that | PA | - | Pb | is the maximum, and find this value

Let P (x, y) be a vector. First of all, let P (x, y) be a vector, then the vector PA's Mo (length) = root sign (- 3-x) ^ 2 + (5-y) ^ 2. The vector PB's Mo = root sign (2-x) ^ 2 + (- 3-y) ^ 2. Since it is the length and has the maximum value, the PA's length must be greater than the PB's length

Given two fixed points a (- 3,5), B (2,15), and the moving point P on the line 3x-4y + 4 = 0, then the minimum value of | PA | + | Pb | is ()
A. 513B. 362C. 155D. 5+102

Let point a (- 3, 5) be a symmetric point a '(m, n) with respect to line 3x-4y + 4 = 0, then 3 × m − 32 − 4 × n + 52 + 4 = 05 − n − 3 − m × 34 = − 1, and the solution is m = 3N = − 3, that is, a' (3, - 3). Connecting a ′ B and line intersecting at point P, then the minimum value of | PA | + | Pb | is | a ′ B | = (3 − 2) 2 + (− 3 − 15) 2 = 513

3. Introduce the tangent line of the circle from a point P (2.3) outside the circle (x-1) ^ 2 + (Y-1) ^ 2 = 1, and find the tangent equation

Center (1,1), radius r = 1
The distance from the center of the circle to the tangent is equal to the radius
If the tangent slope does not exist, then x = 2
The distance from the center of the circle to the tangent is equal to the radius
If the slope exists
y-3=k(x-2)
kx-y+3-2k=0
So the distance from the center of the circle to the tangent = | k-1 + 3-2k | / √ (K & sup2; + 1) = 1
(k-2)²=k²+1
k=3/4
So X-2 = 0 and 3x-4y + 6 = 0

The tangent equation of the circle passing through point P can be obtained from the tangent line of a point P (4,2) outside the circle x + y = 2
The second question is, if the tangent point is P1 P2, find the linear equation through the tangent point P1 P2

Let the tangent equation l of p be Y-2 = K (x-4), then the distance from the center of circle 0 to the straight line L is the radius, that is, the tangent equation l of | 4k-2 | / (1 + K ^ 2) = √ 2K = 1 or K = 1 / / 7p is Y-X + 2 = 0 or 7y-x-10 = 0. It is easy to know that the four points P1, P2, 0 and P are co circular, and the diameter is op, and the center point o '(2,1) is the center of the circle, that is, (X-2) ^ 2 +