(1) Find the equation of straight line passing through the intersection point of the line l1:7x-8y-1 = 0 and l2:2x + 17Y + 9 = 0 and perpendicular to the line 2x-y + 7 = 0 (2) The straight line L passes through the point P (5, 5) and intersects the circle C: x2 + y2 = 25, and the chord length is 4 5. Find the equation of L

(1) By the system of equations
2x+17y+9=0
7x-8y-1=0 ，
The solution
x=-11
Twenty-seven
y=-13
27, so the coordinates of the intersection point are (- 11
27，-13
27)．
Because the slope of the straight line is k = - 1
2，
So the linear equation is 27x + 54y + 37 = 0
(2) As shown in the figure, it is easy to know that the slope k of the straight line L exists. Let the equation of the straight line l be Y-5 = K (X-5)
The center of the circle C: x2 + y2 = 25 is (0, 0), the radius is r = 5, and the distance from the center of the circle to the straight line L is d = | 5-5k|
1+k2．
In RT △ AOC, D2 + ac2 = oa2,
(5-5k)2
1+k2+(2
5)2=25．
 2k2-5k + 2 = 0,  k = 2 or K = 1
2．
So the equation of line L is 2x-y-5 = 0 or x-2y + 5 = 0

Given that the radius of the circle is root 10, passing through point P (2,2), and the chord length cut on the straight line x = y is 4 times the root sign 2, what is the equation of the circle?

Let (x-a) ^ 2 + (y-b) ^ 2 = 10, (2-A) ^ 2 + (2-B) ^ 2 = 10, (a, b) to the straight line X-Y = 0, d = A-B / radical 2,
(4 root sign 2 / 2) ^ 2 = 8, d ^ 2 = root 2, A-B = 2, a = B + 2 or B = a + 2 into (2-A) ^ 2 + (2-B) ^ 2 = 10
The center of the circle (a, b) can be solved

1. 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 L: y = X-1 is cut by the circle, and the chord length is 2 * radical 2 1. If 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 L: y = X-1 is cut by the circle, and the chord length is 2 * root 2, then the circle C equation? 2. If the point P (- 3,0) is known, and the straight line with slope k intersects the part of the circle x ^ 2 + (Y-2) ^ 2 = 9 in the first quadrant, the value of K is taken

1、（x-m)^2+y^2=R^2,
d=√(R^2-2）,d=|m+0-1|/√2=|m-1|/√2,(1-m)^2=R^2,
m^2-2m-3=0,
m=3,R=2,(x-3)^2+y^2=4.
2. The intersection point between circle and Y axis (0,5) and X axis (±√ 5,0), K max = (5-0) / (0 + 3) = 5 / 3,0 ≤ K ≤ 5 / 3

The center of the circle C is on the positive half axis of the X axis (1,0). The radius of the circle C is 5. The circle C is cut by the line X-Y + 3 = 0 to obtain the chord length of 2 times and the root sign 17 (1) Let the line ax-y + 5 = 0 intersect the circle and a B to find the value range of A (2) Under the condition of (1), is there a such that a B is symmetric with respect to the line L passing through P (- 2,4)? If there is a, ask for the value of A

Circle C: (x-1) ^ 2 + y ^ 2 = 25,
(1) Distance from center C to line ax-y + 5 = 0 | a + 5 | / √ (a ^ 2 + 1)

The cross point of circle C (1, 0) is known, and the center of the circle is on the positive half axis of the X axis. The chord length of the straight line l:y=x-1 cut by circle C is 2 2, then the equation of the line passing through the center of the circle and perpendicular to the line L is______ .

Let x + y + M = 0, and let the coordinates of the center of the circle be (a, 0), then from the meaning of the question: (| a − 1 | 2) 2 + 2 = (a − 1) 2, the solution is a = 3 or - 1, and because the center of the circle is on the positive half axis of the X axis, so a = 3, so the coordinates of the center of the circle are (3, 0), ∵ the center of the circle (3, 0) is on the line

It is known that the circle C is tangent to the y-axis. The center point of the circle C is on the straight line x-3y = 0. The length of the line segment where the tangent line y = x is cut by circle C is two radical signs. 7. Solve the equation of circle C rttttttttttttttt

Tangent to y-axis
The distance to the y-axis is equal to the radius
(x-a)^2+(y-b)^2=r^2
r=|a|
The center point C is on the line x-3y = 0
a=3b
(x-3b)^2+(y-b)^2=9b^2
String AB = 2 √ 7
The midpoint is d
Then ad = √ 7, AC = r = | 3B|
CD=√(9b^2-7)
Distance from C to y = x = | 3b-b | / √ (1 + 1) = √ (9b ^ 2-7)
b=1,b=-1
(x-3)^2+(y-1)^2=9
(x+3)^2+(y+1)^2=9b^2

Given that a circle is tangent to y-axis, its center is on x-3y = 0, and the root sign 7 with chord length of 2 times cut on the straight line y = x, the standard equation of the circle is obtained There are two results in the answer to this question. I don't know how?

The answer is: (x + 1 / 4) ^ 2 + (y + 1 / 12) ^ 2 = 1 / 16
or
（X-1/4）^2+(Y-1/12)^2=1/16
Analysis ideas: 1. Tangent to Y-axis: R (radius) = X-axis coordinate of the center of the circle
2. If the center of a circle is on x-3y = 0, then x = 3Y, a = x = R
b=3Y=R/3
3. The root sign of chord length of 2 times cut on the line y = x 7: chord length = quarter root sign 7
Using the formula of distance from point to line, the distance from center of circle to line y = x is obtained
|A-b| / Radix 2
4. The line y = x intersects the chord, radius and center of the circle to the vertical line of the line y = X
Angle triangle. (using Pythagorean theorem) we get the following results
(| A-B | / Radix 2) ^ 2 = R ^ 2 - (seven quarter root sign) ^ 2 because: a = x = R
b=3Y=R/3
The result is: (R-R / 3) ^ 2 / 2 = R ^ 2-7 / 16
8 * 4 / 9R ^ 2 = 16r ^ 2-7
112R^2=7
R = 1 / 4 (positive or negative)
Because r = a = 3B = 1 / 4 (positive or negative)
Then B = 1 / 12 (positive or negative)

It is known that the circle C satisfies the following three conditions at the same time: 1 is tangent to the y-axis. 2 on the line y = x, the chord length is 2, and the circle center is 7.3. The equation of circle C on the line x-3y = 0 is obtained Please write the detailed process

Let the coordinates of the center of the circle (3a, a)
The circular equation is (x-3a) ^ 2 + (Y-A) ^ 2 = 9A ^ 2
Using the distance from the center of a circle to the line y = x, d = | 2a| / √ 2
The vertical diameter theorem is obtained
d^2+7=r^2
2a^2+7=9a^2
a^2=1
The equation of circle C (x-3) ^ 2 + (Y-1) ^ 2 = 9 or (x + 3) ^ 2 + (y + 1) ^ 2 = 9

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

3x-y=0
y=3x
C(a,3a),r=|3a|
Y=x
x-y=0
d^2=|a-3a|^2/2=2a^2
d^2+(2√7/2)^2=r^2
2a^2+7=(3a)^2
a=±1,r^2=9
(x±1)^2+(y±3)^2=9

Find the equation of the circle whose center is on the line 3x-y = O, tangent to the x-axis, and cut by the line X-Y = 0, whose chord length is twice the root 7