It is known that the circle C passes through point 10 and the center of the circle is on the positive half axis of the X axis. The straight line x + Y-1 = 0 is cut by the circle C, and the chord length is 2, and the circle standard is found equation

Given that the circle C passes through the point (1,0), and the center of the circle is on the positive half axis of the X axis, the straight line x + Y-1 = 0 is cut by the circle C, and the chord length is 2 √ 2. The standard equation of the circle is obtained by X + Y-1 = 0, and K1 = - 1. When y = 0, x = 1,  x + Y-1 = 0 passes through the (1,0) point, let x + Y-1 = 0 and the other intersection point of circle C is (x, y), so K1 = - 1 = (y-0) / (x-1)

Known straight line L: X- 3Y + 1 = 0, the center C of a circle is on the positive half axis of X axis, and the circle is tangent to the straight line L and Y axis (1) Find the equation of the circle; (2) If the straight line: MX + y + 1 2m = 0 intersects with circle C at two points a and B, and | ab|= 3. Find the value of M

(1) Let the center of a circle C (a, 0), a ﹥ 0, the radius is r,

If the circle C passes through the point (1,0), and the center of the circle is on the negative half axis of the X axis, the chord length of the straight line L: y = X-1 cut by the circle is twice the root sign 2, then the equation of circle C is The answer is: x ^ 2 + y ^ 2 + 6x-6 = 0?

It is known that if the center of the circle is C (a, 0), and the radius is r, then r = 1-A. in addition, the triangle formed by the center of the circle and the two intersections of y = X-1 and the circle is an isosceles right triangle, so r = 2 can be solved, so a = - 1
(x+1)^2+y^2=4

Find the equation of circle whose chord length is equal to 2 pieces of sign 7 which is tangent to X axis and whose center is cut by line y = x on line 3x-y = 0

Let the circle be (x-a) ^ 2 + (y-b) ^ 2 = C ^ 2
The center of the circle is on the line 3x-y = 0, so B = 3A
It is tangent to the X axis, that is, it has only one root with y = 0
(x-a) ^ 2 + (3a) ^ 2-C ^ 2 = 0
X ^ 2-2ax + (10a ^ 2-C ^ 2) = 0
△＝4a^2-4(10a^2-c^2)=0
c^2=9a^2
Circle equation (x-a) ^ 2 + (y-3a) ^ 2 = 9A ^ 2
The above equation is coupled with the line y = x again
By simplification, 2x ^ 2-8ax + A ^ 2 = 0 can be obtained
Because the length of a chord is two and seven
So the equation above must have two roots, x1, x2
We can get (x1-x2) ^ 2 + (y1-y2) ^ 2 = (2 radical 7) ^ 2
Here Y1 = x1, y2 = X2, we don't need to explain and continue to simplify
（x1＋x2）^2-4x1x2=0
It can be found that a ^ 2 = 1, so a = ± 1
So the equation of the circle is (x-1) ^ 2 + (Y-3) ^ 2 = 9
Or (x + 1) ^ 2 + (y + 3) ^ 2 = 9

The chord length of the circle whose center is (2, - 1) on the straight line x-y-1 = 0 is 2 times the root sign 2. Find the equation of the circle

The distance from the center of a circle to the straight line d = | 2 + 1-1 | / √ 2 = √ 2
The chord length is 2 √ 2, half of which forms a right triangle with D and radius R
So R ^ 2 = (√ 2) ^ 2 + (√ 2) ^ 2 = 4
So the equation of the circle is: (X-2) ^ 2 + (y + 1) ^ 2 = 4

Find tangent to the x-axis, the center of the circle C is on the line 3x-y = 0, and the chord length of the cut line X-Y = 0 is 2 The equation of a circle of 7

If the center of a circle (T, 3T) is tangent to the x-axis, the radius r = 3|t |
∵ distance from center of circle to straight line d = | t − 3T|
2=
2t，
ν from R2 = D2+（
7) 2, t = ± 1
The center of the circle is (1, 3) or (- 1, - 3), and the radius is equal to 3
The equation of circle C is (x + 1) 2 + (y + 3) 2 = 9 or (x-1) 2 + (Y-3) 2 = 9

Find tangent to the x-axis, the center of the circle C is on the line 3x-y = 0, and the chord length of the cut line X-Y = 0 is 2 The equation of a circle of 7

The Y-axis of a circular jade is tangent, the center of the circle is on x-3y = 0, and the length of the line segment cut on the straight line y = x is 2 root signs 7. The equation of the circle is solved

Tangent to Y-axis tangent to Y-axis distance is equal to radius (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2R = (a) ^ 2 + (y-b) C on the straight line x-3y = 0, a = 3B (X-3B) ^ 2 + (y-b) ^ 2 = 9b ^ 2 = 9b ^ 2 chord AB = 2 √ 7 midpoint is D, then ad = √ 7, AC = r = | 3b | CD = (9b ^ 2-7) C C to y = x distance = | 3b-b | / (1 + 1 + 1) = √ (1 + 1) = √ (1 + 1) = (1) = (1) = (1) = (1 + 1) = (1 + 1) = (1 + 1) = (1 9b ^ 2-7) B = 1, B = - 1 by (x-3) ^ 2 + (

Find the equation of the circle tangent to y axis, the chord length of which is 2 times the root sign 7 on the straight line y = x, and the center of the circle is on the line x = 3Y

The center of the circle is on x = 3Y
The center of the circle can be set as (3Y1, Y1)
It's tangent to the Y axis, so | 3Y1 | = radius
The distance from the center of a circle to the line y = x = (2 under the radical) Y1
Chord center distance = (2 under radical) Y1
2y1²+7=9y1²
y1=±1 x=±3
So the center of the circle is (3,1) and the radius is 3 or (- 3, - 1) and the radius is 3
The equation of a circle is (x-3) 2 + (Y-1) 2 = 9 or (x + 3) 2 + (y + 1) 2 = 9

The circle is tangent to the y-axis, the center of the circle is on the straight line x-3y = 0, and the chord length cut on the line y = x is twice the better root sign 7, and the equation of the circle is solved

Let the center coordinate a (a, b), radius r, a - 3B = 0, the distance from the center of a circle to the straight line y = x is | a - B | / 2 = | B | ^ 2 + 7 = R ^ 2B ^ 2 = 7, R ^ 2 + B ^ 2 = R ^ 2, a ^ 2 = 9 B ^ 210, B ^ 2 = R ^ 2 = 1 / 7, B ^ 2 = 7 / 69, a ^ 2 = 63 / 69, R ^ 2 = 1 / 69

It is known that the circle C passes through point 10 and the center of the circle is on the positive half axis of the X axis. The straight line x + Y-1 = 0 is cut by the circle C, and the chord length is 2, and the circle standard is found equation