It is known that circle C: x ^ 2 + y ^ 2-2x + 4y-4 = 0. Is there any symmetry between two points a and B on circle C about the straight line y = kx-1, and the circle with diameter AB passing through the origin?

It is known that circle C: x ^ 2 + y ^ 2-2x + 4y-4 = 0. Is there any symmetry between two points a and B on circle C about the straight line y = kx-1, and the circle with diameter AB passing through the origin?

It is known that there are two symmetric points on the circle C: x2 + Y2 + mx-4 = 0 about the straight line X-Y + 3 = 0, then the value of the real number m ()
A. 8b. - 4C. 6D. Not sure

Because two points a and B on the circle are symmetrical about the straight line X-Y + 3 = 0, the straight line X-Y + 3 = 0 passes through the center of the circle (- m2, 0), so - M2 + 3 = 0, that is, M = 6

Given that circle C and circle x ^ 2 + y ^ 2-2x-1 are symmetric with respect to line 2x-y + 3 = 0, the equation of circle C is obtained

X & sup2; + Y & sup2; - 2x-1 = 0 becomes the standard form: (x-1) & sup2; + Y & sup2; = 2 Center (1,0)
Let the center of circle C be (a, b) symmetric with respect to the line 2x-y + 3 = 0
We can get 1 / 2 (a + 1) * 2-B / 2 + 3 = 0, B = 2A + 8
The slope of the line equation of two points symmetrical to the line is - (1 / k), so B / (A-1) = - 1 / 2, B = - A / 2 + 1 / 2
So a = - 3, B = 2, the center of circle C is (- 3,2) symmetric, and the radius remains unchanged
So the equation of circle C is (x + 3) & sup2; + (Y-2) & sup2; = 2

In the circle O with radius r, the chord AB is perpendicular to the chord CD and at the point P. it is proved that AP ∧ 2 + CP ∧ 2 + Pb ∧ 2 + PD ^ 2 is the fixed value

Let m be the midpoint of AB, and NPA * Pb be the midpoint of CD = am ^ 2 - PM ^ 2 = Ao ^ 2 - Po ^ 2 = R ^ 2 - Po ^ 2. Similarly, if PC * PD = R ^ 2 - Po ^ 2, then PA ^ 2 + Pb ^ 2 = AB ^ 2 - 2PA * Pb = 2 (R ^ 2 - om ^ 2) - 2 (R ^ 2 - Po ^ 2) = 2 (PO ^ 2 - om ^ 2) = 2pm ^ 2. Similarly, PC ^ 2 + PD ^ 2 = 2pn ^ 2 = > pa2 + Pb

In ⊙ o, P is the point on the chord AB, and AP = 3, Pb = 5, Op = 2, then the

Take the midpoint C of AB and connect OC
The drawing shows that OPC three points form a right triangle
cos∠OPB=1/2
The solution is ∠ OPB = 60 °
Connect OB and OBC to form a right triangle, and the hypotenuse is r
The solution is r = √ 19

If the line tangent to the circle C: x2 + y2-2x-2y + 1 = 0 intersects the positive half axis of X axis and Y axis at a and B and | OA | > 2, | ob | > 2, then the minimum area of triangle AOB is______ ．

Let a (a, 0), B (0, b), then the equation of line AB is XA + Yb = 1, that is, BX + ay AB = 0, the distance from center C (1, 1) to line AB is d = r = 1, that is, | B + a − ab | A2 + B2 = 1, and the two sides are square, then 2ab-2ab (B + a) + a2b2 = A2 + B2, ∧ ab ≠ 0, ∧ 2-2 (B + a) + AB = 0, ∧ (A-2) - b-2a + 4 = 2 Let A-2 = m ＞ 0, B-2 = n ＞ 0, and Mn = 2, so s △ AOB = 12ab = 12 (M + 2) (n + 2) = 12 (Mn + 2m + 2n + 4) ≥ 12 (Mn + 24MN + 4) = 3 + 22, if and only if M = n, i.e. a = B, take the equal sign. So the minimum area of triangle AOB is 3 + 22, so the answer is: 3 + 22

If the line y = ax intersects the circle (x-4) square + (Y-2) square = 4 at two points a and B, then the value of | OA | * | ob | is equal to a, 4, B and 16
A. 4 B, 16 C, 16 / (1 + a square) d, 16 radical (1 + a square)

|OA|*|OB|=r*r=2×2=4
Choose a

Y ^ 2 = 4x and y = 2X-4 intersect at two points a and B, a point P on the parabola, so that OA + ob = XOP, find the value of X
Online, etc

【1】 The simultaneous equations of parabola and line can be solved as follows: x = 4, y = 4, or x = 1, y = - 2. The point a (4,4), B (1, - 2).. the vector OA = (4,4), the vector ob = (1, - 2).. the vector OA + the vector ob = (5,2). [2] ∵ the point P is on the parabola y & sup2; = 4x, and the set point P (P & sup2;, 2P), P ∈ r.. The vector OP = (P & S

A straight line y = 2x + m and a circle x2 + y2 = 1 intersect two points a and B, with the positive direction of x-axis as the starting edge, OA as the ending edge, and ob as the ending edge

The simultaneous elimination of y = 2x + m and circle x2 + y2 = 1 leads to: 5x ^ 2 + 4mx + m ^ 2-1 = 0, 5Y ^ 2-2my + m ^ 2-4 = 0, then x1x2 = cos α cos β = (m ^ 2-1) / 5, y1y2 = sin α sin β = (m ^ 2-4) / 5, so cos (α + β) = cos α cos β - sin α sin β = (m ^ 2-1) / 5 - (m ^ 2-4) / 5 = 3 / 5, thus sin (α + β

Two circles, X2 + Y2 + 4x-4y = 0, X2 + Y2 + 2x-12 = 0, intersect at two points a and B,
If a and B intersect, the equation of a and B is

2x-4y + 12 = 0 question: answer: subtract the equations of two circles