The line y = - 4 3x + 4, x-axis and y-axis intersect at point a and B respectively. In the plane rectangular coordinate system, the distance between a and B points to the straight line a is 2, then the number of lines a satisfying the condition is () A. 1 B. 2 C. 3 D. 4

The line y = - 4 3x + 4, x-axis and y-axis intersect at point a and B respectively. In the plane rectangular coordinate system, the distance between a and B points to the straight line a is 2, then the number of lines a satisfying the condition is () A. 1 B. 2 C. 3 D. 4

∵ when x = 0, y = 4,
When y = 0, x = 3,
∴A（3，0），B（0，4），
As shown in the figure, and the line y = - 4
There are two intersecting lines of 3x + 4. The distance between point a and point B to the straight line a is 2. There are 2 such lines, 4 in total
Therefore, D

In the plane rectangular coordinate system as shown in the figure, the parabola y = - 4 3x2+8 3x + 4 intersects the x-axis at two points a and B (point B is on the right side of point a), and the intersection Y-axis is at point C. taking OC and ob as two sides, obdc is made, and CD intersects parabola at G (1) Find the length of OC and ob; (2) Let OE = m, PM = h, find the function relation between H and m, and find the maximum value of PM; (3) If PC is connected, is there such a point P in the parabola above CD, so that the triangle with P, C and F as vertices is similar to △ BEM? If it exists, calculate the value of M directly and judge the shape of △ PCM directly; if not, please explain the reason

(1) For y = - 43x2 + 83x + 4, when x = 0, y = 4; when y = 0, - 43x2 + 83x + 4 = 0, the solution X1 = - 1, X2 = 3; (2 points)

As shown in the figure, AB is the diameter of circle O, C is the midpoint of arc AE, CD is perpendicular to D, intersection AE and point F connect AC, try to explain AF = CF

Extend CD to intersect O and G, connect AG. So arc AC = arc CE, so angle AGC = angle CAE / x0dcg is perpendicular to AB, so arc AC = arc AG, so angle AGC = angle ACG / x0d, so angle CAE = angle ACG, we can get AF = CF

It is known that AB is the diameter of circle O, AE is the chord, C is the midpoint of arc AE, CD perpendicular to ab intersects point D, AE intersects point F, CB intersects AE at point g. it is proved that CF = FG

It is proved that: even EB ∵ AB is the diameter of circle O  AEB = 90  EGb + ∠ EBG = 90 ° then the vertex angle ? CGF + ∠ EBG = 90 ° - --- (1) ? CD ⊥ ab ? C + ∠ CBD = 90 ° - - - - (2) ∵ C is the circle angle opposite to the midpoint of arc AE, which is equal to the circle angle of equal arc ∠ EBG = ∠ CBD

Circle O is circumscribed circle of triangle ABC, AB is diameter, arc AC = arc CF, CD is perpendicular to AB and D, and intersection circle O is g, AF intersects CD in E, and AE = CE

∵ arc AC = arc FC
Ψ B = ∠ CaF (equal arc with equal circle angle)
∵ AB is the diameter
∴AC⊥BC
∴∠CAB+∠B=90°
∵∠CAB+∠ACD=90°
∴∠B=∠ACD
∵ B =  CaF (proved)
∴∠ACD=∠CAF
∴CE=AE

As shown in Fig. AB is the diameter of circle O, AE is the chord, C is the midpoint of arc AE, CD is perpendicular to point D, intersecting AE at point F, and BC intersecting AE at point g. it is proved that CF = GF

It is proved that: connect AC, extend CD intersection o to M
If CD is perpendicular to AB, then: arc am = arc AC = arc CE, ∠ ACM = ∠ CAE;
And ab is the diameter, ∠ ACB = 90 degrees. Therefore: ∠ FCG = ∠ FGC (the remainder of equal angle is equal)
So CF = GF

As shown in the figure, it is known that AB is the diameter of ⊙ o, and point C is The middle point of AE is the chord CD ⊥ AB through C, and AE is intersected with F. it is proved that AF = CF

Proof: connect AC,
∵ chord CD ⊥ AB, AB is the diameter of ⊙ o,
Qi
AC=
AD，
∵ point C is
The midpoint of AE,
Qi
AC=
CE，
Qi
AD=
CE，
∴∠ACD=∠CAE，
∴AF=CF．

In the plane rectangular coordinate system xoy, the intersection points of the curve y = x ^ 2-6x + 1 and the coordinate axis are all on the circle C In the plane rectangular coordinate system xoy, the intersection points of the curve y = x ^ 2-6x + 1 and the coordinate axis are all on the circle C. (1) find the equation of circle C? (2) if the tangent lines of circle C have the same intercept on the X and Y axes, find the tangent equation

(1) Solve the three intersection points and put them into the standard equation. Or find the center of the circle first
(2) Let the tangent be x / A + Y / a = 1 and y = - x, then we can get it by using the discriminant = 0

In the plane rectangular coordinate system xoy, the intersection points of the curve y = x? - 6x + 1 and the coordinate axis are all on the circle C (I) find the equation of circle C; (II) if the circle C and the straight line X-Y + a = 0 intersect two points a and B, and OA ⊥ ob, find the value of A

(1) Intersection point of curve y = x? - 6x + 1 and coordinate axis (3 ± 2 √ 2,0), (0,1)
Let C (3, a) be the center of the circle
（3+2√2-3）²+a²=3²+(a-1)²=r²
∴a=1,r=3
⊙ C: (x-3) 2 + (Y-1) 2 = 3?, that is, x? - 6x + y? - 2Y + 1=-
(2) X-Y + a = 0 leads to y = x + a
x²-6x+x²+2ax+a²-2x-2a+1=0
2x²+(2a-8)x+(a²-2a+1)=0
﹤ x1x2 = (a? - 2A + 1) / 2 X1 + x2 = 4-A △ = 4A? - 32A + 64-8a? + 16a-8 = - 4A? - 16A + 56 > 0, resulting in - 2-3 √ 2 ≤ a ≤ - 2 + 3 √ 2
y1y2=x1*x2+a(x1+x2)+a²=(a²-2a+1)/2+4a-a²+a²=(a²+6a+1)/2
∵OA⊥OB
∴y1/x1*y2/x2=-1
(a²+6a+1)/(a²-2a+1)=-1
∴a²+2a+1=0
∴a=-1

In the plane rectangular coordinate system, the focus of curve y = x2-6x + 1 and coordinate axis is on circle C. 1. Find the equation of circle C; 2. If the circle C intersects the straight line X-Y + a = 0 At a and B, and OA is perpendicular to ob, find the value of A

1. Let y = x ^ 2-6x + 1 = 0, two are (x1,0) (x2,0), the intersection point of curve and Y axis is (0,1) X1 = 3-2 radical sign 2 x2 = 3 + 2 radical sign 2, let the center of circle be (X3, Y3) obviously, the center of circle is on the middle line of (x1,0) (x2,0), X3 = (x1 + x2) / 2 = 3 = the distance from the center of circle to (0,1) = the distance from the center of circle to (3 + 2 root sign 2,0) = > (3) ^ 2 + (y3-1) ^ 2 = (2