The path equation of the midpoint m of the chord AB is obtained by the intersection of the secant of the leading circle of a point P (5,12) outside the circle x2 + y2 = 9 at two points ab

The path equation of the midpoint m of the chord AB is obtained by the intersection of the secant of the leading circle of a point P (5,12) outside the circle x2 + y2 = 9 at two points ab

Let the coordinates of the midpoint m be (x, y), then
∵ △ OAB is an isosceles triangle, and M is the middle point
∴OM⊥AB
A right triangle is a right triangle
From the formula of distance between two points:
OP=√(5²+12²)=13
In RT △ OPM, from the Pythagorean theorem, we obtain that:
OP²=OM²+PM²
13²=x²+y²+(x-5)²+(y-12)²
It is reduced to: X & # 178; + Y & # 178; - 5x-12y = 0
The locus of the midpoint m is the part in the circle

Senior one mathematics circle and equation, please answer in detail, thank you! (8 10:52:3)
1. The equation x ^ 2 + y ^ 2 + ax + 2ay + 2A ^ 2 + A-1 = 0 represents the element, then the value range of a is:?
2. The equation of the line parallel to the line 2x-y + 1 = 0 and tangent to the circle x ^ 2 + y ^ 2 = 5 is:?
3. If circle C1: (x-m) ^ 2 + (y + 2) ^ 2 = 9 and circle C2: (x + 1) ^ 2 + (y-m) ^ 2 = 4 are circumscribed, then the value of M:?

1.x^2+y^2+ax+2ay+2a^2+a-1=0
(x+a/2)^2+(y+a)^2=1-a-(3/4)a^2>=0
3a^2+4a-4

Mathematics of senior one (circle and line equation)
Given two points a (- 1,0), B (0,2), and point P is any point on the circle (x-1) ^ 2 + y ^ 2 = 1, what is the maximum and minimum value of △ PAB area?
I don't quite understand. Is it three and one

The bottom edge AB = √ 5, the height is the distance from P to AB, the line AB is 2x-y + 2 = 0, the box line is separated, so it passes through the center of the circle (1, 0) as the vertical line of AB, the intersection point of the vertical line and the circle is the farthest point and the nearest point respectively. The farthest point = the distance from the center of the circle to the straight line D + R, the nearest point is D-R R, r = 1 d = | 2-0 + 2 | / √ (2 ^ 2 + 1 ^ 2) = 4 √ 5 / 5, so the maximum height = 1 + 4 √ 5 / 5 = (5 + 4 √ 5) / 5, the minimum = (4 √ 5-5) / 5, so the maximum area = (5 + 4 √ 5) / 5 * √ 5 / 2 = (4 + √ 5) / 2, the minimum = (4 - √ 5) / 2

The circle C passes through the point a (2, - 1) and is tangent to the straight line x + y = 1, and the center of the circle is on the straight line y = - 2x. (1) find the equation of circle C; (2) find the equation of the straight line with a point B (2, − 52) in the circle

(1) Let the center of a circle (m, - 2m) be (x-m) 2 + (y + 2m) 2 = R2