C 1: x 2 + y 2-2x-6y-6 = 0, and the number of common tangents of circle C2: x2 + y2-4x + 2Y + 4 = 0

The radius of C1 circle center (1,3) is 4, and that of C2 circle (2, - 1) is 1

The equation of a circle passing through P (- 2,4) and taking the common chord of two circles x ^ 2 + y ^ 2-6x = 0 and x ^ 2 + y ^ 2 = 4 as a chord

Let x ^ 2 + y ^ 2-6x + k * (x ^ 2 + y ^ 2-4) = 0
Substituting the coordinates of point P (- 2,4), we get: (- 2) ^ 2 + 4 ^ 2-6 * (- 2) + k * (((- 2) ^ 2 + 4 ^ 2) - 4) = 0
The solution is: k = - 2
So the equation is: x ^ 2 + y ^ 2-6x-2 * (x ^ 2 + y ^ 2-4) = 0
After simplification, x ^ 2 + y ^ 2 + 6x-8 = 0
Or: (x + 3) ^ 2 + y ^ 2 = 17

Given that the circle x Λ 2 + y Λ 2-4x + 2Y = 0 and the circle x Λ 2 + y Λ 2-2y-4 = 0, the equation of the line where the common chord of two circles is located is solved

The difference between the two circular equations is the common string equation
4X-4Y-4=0,
That is, x-y-1 = 0

Find the equation of the line where the common chord of the intersecting circle C1: x ^ 2 + y ^ 2 + 4x + y + 1 = 0 and x ^ 2 + y ^ 2 + 2x + 2Y + 1 = 0

x²+y²+4x+y+1=0 (1)
x²+y²+2x+2y+1=0 (2)
(1)-(2)
2x-y=0
y=2x
The equation of the line where the common chord of two circles lies is y = 2x

Given the circle: x square + y square - 4x-6y + 12 = 0, point P (x, y) is any point on the circle, find the maximum value of X in y

Circle C: x ^ 2 + y ^ 2-4x-6y-12 = 0 the trajectory equation of the midpoint of a chord of length 8 in the circle C: x ^ 2 + y ^ 2-4x-6y-12 = 0? Please talk about the process and ideas

x^2+y^2-4x-6y-12=0
(x-2)^2+(y-3)^2=25,
The center of the circle is (2,3) and the radius is 5
A right triangle is formed by the center distance, radius and half of chord length,
Because half of the chord length is 4 and the radius is 5, the center distance is 3
That is, the distance from the center of the circle to the midpoint of the chord is 3, and the midpoint of the chord is (x, y),
Then there is (X-2) ^ 2 + (Y-3) ^ 2 = 9, which is the midpoint trajectory equation of the chord

Given the circle x ^ 2 + y ^ 2-4x = 0, a point a (2,2) passing through a leading circle string to find the trajectory equation of the focal point of the moving chord

Oh, circle: (X-2) ^ 2 + y ^ 2 = 4
Let chord intersection circle A, B (a, b)
Midpoint P (x, y)
Then: 2x = 2 + A, 2Y = B + 2
That is, a = 2x-2, B = 2y-2
Insert circle: (X-2) ^ 2 + (Y-1) ^ 2 = 1

Given that the equation of a circle is x ^ 2 + y ^ 2 -- 6y --- 8y = 0, let the longest chord and the shortest chord of the circle passing through the point (3,5) Given that the equation of a circle is x ^ 2 + y ^ 2 -- 6y --- 8y = 0, let the longest chord sum of the circle pass through the point (3,5) If the shortest chord is AC and BD, the area of the quadrilateral ABCD is () A 10 radical 6 B 20 radical 6 C30 radical 6 D40 radical 6 Please write down the idea and process of solving the problem. Thank you for your kindness. You must have a lot of fortune

First, the general equation is obtained
(x-3)^2+(y-4)^2=25
The longest string must be 10 in diameter
Center of the circle (3,4), so the distance between the centers of the circle = 1
The shortest chord = 2 root sign (25-1) = 4 root sign 6
S = 4 root number 6 * 10 / 2 = 20 root number 6, select b

It is known that the equations of the two circles are X^2+Y^2-2X-3=0 and X^2+Y^2+6Y-1=0 respectively

x^2+y^2-2x-3=0 (1)
x^2+y^2+6y-1=0 (2)
(1) (2) the equation of the line where the common chord is located is obtained
x+3y+1=0

The equation of common string of two circles x ^ 2 + y ^ 2-10x-10y = 0, x ^ 2 + y ^ 2 + 6x + 2y-40 = 0

X ^ 2 + y ^ 2-10x-10y = 0 = (X-5) ^ 2 + (Y-5) ^ 2 = 50, is the circle whose center is 5 √ 2 in (5,5). X ^ 2 + y ^ 2 + 6x + 2y-40 = 0 is the circle whose center is 5 √ 2 at (- 3, - 1) radius

C 1: x 2 + y 2-2x-6y-6 = 0, and the number of common tangents of circle C2: x2 + y2-4x + 2Y + 4 = 0