It is known that the circle x ^ 2 + y ^ 2 + 2x-6y + F = 0 and X + 2y-5 = 0 intersect at two points a and B, and O is the origin of coordinates. If OA is perpendicular to ob, then the value of F is? The answer is zero·

It is known that the circle x ^ 2 + y ^ 2 + 2x-6y + F = 0 and X + 2y-5 = 0 intersect at two points a and B, and O is the origin of coordinates. If OA is perpendicular to ob, then the value of F is? The answer is zero·

Did you notice that the center of the circle (- 1,3) is on a straight line?
So AB is the diameter of the circle,
From the vertical of OA and ob, we can know that AOB = 90 degrees,
The circumference angle of the diameter pair is exactly 90 degrees,
So the origin (0,0), not on the circle, where is it?
In this case, f = 0

The straight line L: x + 2Y + M = 0 and the circle C: x ^ 2 + y ^ 2 + 6x-6y = 0 have two intersection points a, B, O and OA ⊥ ob, then M=

Let point a (x1, Y2), B (X2, Y2) Y1 / x2 * Y2 / x2 = - 1x1x2 + y1y2 = 0, the origin (0,0) is on the circle, x 2 + y 2 + 6x-6y = 0, so AB is the diameter, that is, the straight line x + 2Y + M = 0 through the center of the circle x? + y? + 6x-6y = 0 (x + 3) 2 + (Y-3) 2 = 18 Center (- 3,3) is replaced by the line x + 2Y + M = 0 -

It is known that the circle x ^ 2 y ^ 2 x-6y M = 0 and the straight line x 2y-3 = 0 intersect at two points a and B, O is the origin, and OA is perpendicular to ob, so we can find the value of real number M

∵ A and B are on the straight line x + Y-1 = 0, that is, on y = 1-x, ᙽ let the coordinates of a and B be (a, 1-A), (B, 1-B) ᙽ the slope of OA = (1-A) / A, and the slope of OB = (1-B) / b. substituting y = 1-x into the equation of a given circle, x ^ 2 ＋ (1-x) ^ 2 ＋ X-6 (1-x) + M = 0, ∵ 2

It is urgently needed that "the circle x + y + x-6y + M = 0 and the straight line x + 2y-3 = 0 intersect at two points a and B, O is the origin, and the vector OA ⊥ vector ob is used to find the value of real number M The circle x + y + x-6y + M = 0 and the straight line x + 2y-3 = 0 intersect at two points a and B, O is the origin, and the vector OA ⊥ vector ob is used to find the value of real number M

Let a, B coordinates are (x1, Y1), (X2, Y2) respectively, let a, B coordinates are (x1, Y1), (X2, Y2), then from vector OA ⊥ vector ob, get: x1 × x2 + Y1 × y2 = 0, x = 3-2y into x + y + x-6y + M = 0, get x1 × x2 = (4m-27) / 5, y = (3-x) / 2 into x + y + x-6y + m + M = 0, get: Y1 × y2 = (12 + m) / 5 (4m-27) / 5 + (12 + m) / 5 + (12 + m) / 5 = 0 ͮ M = 3, solve the solution of M = 3, solve the solution, solve the solution, solve the solution of the solution, solve the solution of the solution of the solution of the solution of the solution finish

It is known that the circle x2 + Y2 + x-6y + M = 0 and the straight line x + 2y-3 = 0 intersect at different points of P and Q. if op ⊥ OQ (o is the origin of coordinates), then M=______ .

The equation of straight line and circle is obtained: (2y-3) 2 - (2y-3) + y2-6y + M = 0, then: 5y2-20y + (M + 12) = 0, then: Y1 + y2 = 4, Y1 · y2 = m + 125 ﹥ x1 · x2 = (- 2y1 + 3) · (- 2Y2 + 3) = 4y1y2y2-6 (Y1 + Y2) + 9 = 4 · m + 125-15. If op ⊥ OQ is known, Kop * koq = - 1, i.e., Y1 ·

It is known that the circle x2 + Y2 + x-6y + C = 0 and the straight line x + 2y-3 + 0 intersect at two points of PQ, and O is the origin of coordinates. If OP is perpendicular to OQ, find C

Let P (x0, Y0), q (x1, Y1) be perpendicular to vector OP, so x0x1 + y0y1 = 0; equation (3)
Equation (1) x2 + Y2 + x-6y + C = 0 equation (2) x + 2y-3 = 0
The equation (2) is transformed into x = 3-2y and y = (3-x) / 2, and are introduced into equation (1) respectively
The results are as follows: 5x2 + 10x-27 + 4C = 0 and 5y2-20y + 12 + C = 0;
(x 0, y 0) and (x 1, Y 1) are the solutions of equations (1) and (2)
The results show that the equation (3) has the following properties: x0x1 = (- 27 + 4C) / 5; y0y1 = (12 + C) / 5
C=3

The two points P and Q on the circle x2 + Y2 + x-6y + 3 = 0 satisfy the following requirements: (1) symmetry about the straight line kx-y + 4 = 0; ② op ⊥ OQ (1) Value K (2) Find the equation of straight line PQ

(1) The curve x2 + Y2 + x-6y + 3 = 0 can be changed into: (x + 12) 2 + (Y-3) 2 = (52) 2, the center of the circle (- 12,3) is obtained, and the radius is 52; because there are two points P and Q symmetrical about the straight line, the center of the circle is on the straight line, substituting (- 12,3) into kx-y + 4 = 0, and finding the slope of k = 2 (2) line PQ = - 1K = - 12

The circle x ^ 2 + y ^ 2 + 8x-6y + 21 = 0 and the straight line y = MX intersect at two points P and Q, O is the coordinate origin (1) to find OP, OQ (2) to find the equation of coordinates of midpoint m of chord PQ

(x+4)^2+(y-3)^2=4
A(-4,3),r=2
So OA = 5
If the tangent ob is made through O, OAB is a right triangle
OA=5,AB=r=2
So ob ^ 2 = 25-4 = 21
So OP * OQ = OA ^ 2 = 21
Put y = MX in
(m^2+1)x^2+(8-6m)y+21=0
x1+x2=-(8-6m)/(m^2+1)
x=(x1+x2)/2=(3m-8)/(m^2+1)
y=mx,m=y/x
So x = (3Y / X-8) / (y ^ 2 / x ^ 2 + 1) = (3xy-8x ^ 2) / (x ^ 2 + y ^ 2)
x^3+xy^2-3xy+8y^2=0

The two points P, Q on the curve X 2 + y 2 + x-6y + 3 = 0 satisfy 1. On the straight line kx-y + 4 = 0, symmetric 2. Op ⊥ OQ, find the linear PQ equation

After judging that the curve is a circle, then the straight line passes through the center of the circle (- 0.5,3) and is brought into the straight line

The straight line x + 2y-3 = 0 and the circle x ^ 2 + y ^ 2 + x-6y + C = 0 intersect at two points P and Q. if the circle with PQ diameter passes through the origin, the value of C is calculated

By rewriting the linear equation, we get: x = 3-2y. ∵ P and Q are all on the line x = 3-2y,