The straight line passing through the right focus x ^ 2-y ^ 2 = 1 intersects hyperbola at a, B, AB as the diameter of the circle. If the circle passes through the origin, what is the slope k?

In general, we don't go through the origin, and only certain conditions pass through the origin,
When OA is perpendicular to ob, the triangle AOB is a RT triangle and passes through the origin,
a=1,b=1,c=√2,
AB equation is: y = K (x - √ 2),
Let a (x1, Y1), B (X2, Y2),
The vector OA · ob = 0,
x1x2+y1y2=0,
y1=k(x1-√2),
y2=k(x2-√2),
（1+k^2)x1x2-√2k^2(x1+x2)+2k^2=0,(1)
x^2-k^2(x-√2)^2=1,
(1-k^2)x^2+2√2k^2x-2k^2-1=0,
According to Veda's theorem,
x1+x2=-2√2k^2/(1-k^2),
x1*x2=-(2k^2+1)/(1-k^2),
(1),
-(1+k^2)(2k^2+1)/(1-k^2)-√2k^2*[-2√2k^2/(1-k^2)]+2k^2=0,
-k^2-1=0,
The circle does not pass through the origin

Given the circle C: x ^ 2 + y ^ 2-2 * x + 4 * y-4 = O, is there a straight line m with slope 1, so that a circle with diameter AB cut by circle C passes through the origin? If it exists, find the equation of straight line m; if not, explain the reason

X ^ 2 + y ^ 2 + y ^ 2 * x + 4 * y-4 = O is 2x ^ 2 + (2B + 2) x + B ^ 2 + 4b-4-4 intersection a (x1, Y2) B (X2, Y2) vector vector vector OA * ob = 0 ﹥ x1x2 + y1y21y2 = 0 ﹥ x1x2 + y1y2y2 = 0 ﹥ 2x1x2 + B (x1 + x2) + B ^ 2 = 0 ﹤ B ^ 2 + 3 B ^ 4 = 0b = 1 = 1, - 4 again verify that M is y = x = x + 1 or y = x-4 I do what I do y = x + 1 or y = x-4 I do what I do, I do y = x-4 I do y = x-4 I do y = x + 1 or fan, have good, must send me

Let the vertex of the parabola be at the origin, and the focus is the center of the circle x ^ 2-4x + y ^ 2 = 0. The straight line passing through the focus and slope of 2 intersects with the parabola at a and B, and find the segment ab

The center of the circle x ^ 2-4x + y ^ 2 = 0 is (2,0), that is, the parabola focal point is (2,0), the equation is y ^ 2 = 8x, the straight line equation with slope 2 is y = 2X-4 simultaneous equations, the gain is greater than the loss of 4x ^ 2-24x + 16 = 0, and x ^ 2-6x + 4 = 0. Let the two equations be x1, X2, X1 + x2 = 6, X1 * x2 = 4, which are obtained from Veda's theorem

The circle C: x ^ 2 + y ^ 2-8y + 12 = 0. The straight line L: ax + y + 2A = 0 (1) When a is the value, the line L is tangent to the circle C (2) When the line L and circle C intersect with two points a and B, and ab = 2 * 2 ^ 1 / 2, the linear equation is solved

(1) If the circle C is transformed into the standard equation x ^ 2 + (y-4) ^ 2 = 4, then C (0,4), r = 2 because the straight line L is tangent to the circle C, the distance from C to the straight line is equal to the radius. Finally, a = - 3 / 4 (2) circle C: x ^ 2 + y ^ 2-8y + 12 = 0 ^ 2 + (y-4) ^ 2 = 4 center (0,4) radius 2 line L ax + y + 2A = 0, the chord length of plane geometry is 2

Known: circle C: x + Y - 8y + 12 = 0, straight line L: ax + y + 2A = 0 1. When a is the value, the line L is tangent to the circle C 2. When the line L and circle C intersect at two points a and B, and the absolute value of AB = 2, find the equation of line L

The straight line L: ax + y + 2A = 0 crossing point (- 2,0)
If the tangent slope exists
Let the tangent equation be y = K (x + 2)
Kx-y + 2K = 0
Simplify the circular equation x^2+ (y-4) ^2=4
The center coordinates (0,4) and the radius is 2
The distance from the center of a circle to the tangent line d = | kx-y + 2K | / √ 1 + K ^ 2 = | 4-2k | / √ 1 + K ^ 2 = 2
K = 3 / 4
The tangent equation is 3 / 4x-y + 3 / 2 = 0
3x-4y+6=0
If the tangent slope does not exist
The linear equation is x = - 2, which is the tangent line of the circle
So the tangent equation is 3x-4y + 6 = 0 and x = - 2

Given the circle C: x ^ 2 + y ^ 2-6x-8y = 0, if the longest chord and the shortest chord passing through point P (2,5) in circle C are a and B respectively, then the sum of slopes of straight lines a and B is?

Zero
The longest chord is the diameter, the slope is - 1, the shortest chord is the chord with vertical diameter passing through point P, the slope is 1, so the sum is 0

The center locus equation of a circle which is tangent to the Y-axis and x 2 + y-4x = 0

Let the center of the circle be (x, y),
If the circumscribed is on the right side of the y-axis, then:
(x+2)²=y²+(2-x)²
The trajectory equation of the center of the circle is obtained: y 2 = 8 x (x ≠ 0)
If the outer tangent is on the left side of the y-axis or inscribed, then the center of the circle is on the x-axis. The trajectory equation of the center of the circle is y = O (x ≠ 0,2)

Find the trajectory equation of the center of a moving circle tangent to the X axis and circumscribed to the circle x ^ 2 + y ^ 2 = 1

Since the two circles are circumscribed, the center line of the two circles passes through the tangent point. Therefore, the center coordinate of the moving circle is (x, y). Because the moving circle is tangent to the X axis, the radius of the moving circle is y, while the circle x ^ 2 + y ^ 2 = 1, and the radius is 1, so the length of the connecting center line is called y + 1
The square of the distance between two circles is calculated as x ^ 2 + y ^ 2
There is x ^ 2 + y ^ 2 = (y + 1) ^ 2
The equation of moving circle center is obtained by simplification, x ^ 2-2y-1 = 0

If the moving circle passing through the point (2,0) is circumscribed with the circle x ^ 2 + y ^ 2 + 4x + 3 = 0, then the trajectory equation of the center of the moving circle is____

Let B coordinate of moving circle center be (x, y) and radius of moving circle be r
The circle a: x ^ 2 + y ^ 2 + 4x + 3 = 0 is (x + 2) ^ 2 + y ^ 2 = 1, that is, the circle a has (- 2,0) as its center and R = 1 as its radius
Then the distance between the center of circle B and the center of circle a is constant: D = R + R '' '' '' ` 1
Moving a circle through a point (2,0)
Then the distance from the center of the moving circle to the point (2,0) is the radius of the moving circle R '' '' '' ''2
There are 1 and 2 equations
(x+2)^2+y^2=D^2=(R+1)^2
(x-2)^2+y^2=R^2
The elimination of R by simultaneous equations is as follows:
60x^2-4y^2=15
That's what you want

The moving circle passes through the point a (2,0) and is tangent to the fixed circle x ^ + 4x + y ^ - 32 = 0. The trajectory equation of the moving circle center m is obtained

If the center of a moving circle m (x, y), then the distance between the center of the circle and the point a (2,0) is equal to the radius, and the distance between the center of the moving circle and the center of the fixed circle (- 2,0) is equal to the difference of the radius of the two circles

The straight line passing through the right focus x ^ 2-y ^ 2 = 1 intersects hyperbola at a, B, AB as the diameter of the circle. If the circle passes through the origin, what is the slope k?