Circle C: x ^ 2 + y ^ 2-dx-ey + 3 = 0, symmetric about the straight line x + Y-1 = 0, the radius of the center of the circle in the second quadrant is root 2 It is known that a straight line L which does not have the origin is tangent to a circle C, and the intercept on the x-axis and y-axis is equal, so the equation of the straight line is obtained

Circle C: x ^ 2 + y ^ 2-dx-ey + 3 = 0, symmetric about the straight line x + Y-1 = 0, the radius of the center of the circle in the second quadrant is root 2 It is known that a straight line L which does not have the origin is tangent to a circle C, and the intercept on the x-axis and y-axis is equal, so the equation of the straight line is obtained

L has four X + Y-3 = 0, x + y + 1 = 0, X-Y + 5 = 0, X-Y + 1 = 0
X ^ 2 + y ^ 2-dx-ey + 3 = 0, Center (D / 2, E / 2)
(D/2）^2+(E/2)^2=r^2+3
D/2+(E/2)-1=0
If e = 4 and d = - 2, the center of the circle is (- 1,2)
However, the line L of the origin is tangent to the circle C, and the intercept on the x-axis and y-axis is equal
Then l is parallel or perpendicular to the line x + Y-1 = 0
When parallel, the distance between X + Y-1 and X + Y-1 = 0 is positive and negative root sign 2,
Then I forgot some formulas, which I worked out geometrically. They are x + Y-3 = 0, x + y + 1 = 0
Vertical
Find out that the line passing through the center of the circle and perpendicular to x + Y-1 = 0 is X-Y + 3 = 0
Find out that the two lines parallel to X-Y + 3 = 0 and the distance is positive and negative root sign 2 are X-Y + 5 = 0, X-Y + 1 = 0

Given that the circle C: x + y + DX + ey + 3 = 0, the circle C is symmetric with respect to x + Y-1 = 0, the center of the circle is in the second quadrant, and the radius is √ 2, we can find the equation of the circle

Given the circle C: x ^ 2 + y ^ 2 + DX + ey + 3 = 0, the circle C is symmetric with respect to the straight line x + Y-1 = 0, the center of the circle is in the second quadrant, and the radius is the root sign 2. To solve the equation of circle C, it is known that the line L at the origin is tangent to the circle C, and the intercept on the x-axis and y-axis is the same. To solve the equation of line L, if the circle C is symmetric with respect to the straight line x + Y-1 = 0, then the center of the circle is on the straight line: - D / 2-e / 2-1 = 0, that is d + e + 2 = 0, and the radius is the root sign 2, Then: root sign (d ^ 2 + e ^ 2-12) / 2 = root sign 2, straighten out: D ^ 2 + e ^ 2 = 20, eliminate element to get d = 4, e = - 2, the equation of circle is: x ^ 2 + y ^ 2 + 4x-2y + 3 = 0, x0d2. Let the linear equation be x / A + Y / b = 1, that is, BX + ay AB = 0, under the root sign (a ^ 2 + B ^ 2) = root sign 2, and because the intercept of the line on the X axis and Y axis is the same, then a = B. straighten out: a = b = 1, so the equation of line L is x + Y-1 = 0
I hope it can help you

Given the circle C: x2 + Y2 + DX + ey + 3 = 0, the center of the circle is in the straight line x + Y-1 = 0, the center of the circle is in the second quadrant, and the radius is root 2, the general equation of the circle is solved

x²+y²dx++ey=0 .(x+d/2)²+(y+e/2)²=(d/2+e/2)=2,
Let x + Y-1 = 0 and x-axis and y-axis intersect m respectively,
N,M（1,0）,N（0.1）,O(1/2,1/2)
d=1.e=1
The general equation  + y is  + 178

The sum of the first n terms of an is Sn, and the limit of Sn is 1 / 2

an = a1q^(n-1)
Sn = a1( 1- q^n)/(1-q)
lim(n->∞) Sn = 1/2
a1/(1-q) = 1/2
a1 = (1/2)(1-q)
0

If LiMn →∞ an = 1, then A1 is equal to ()
A. 32B. 3C. 4D. 5

∵an-an-1=(-a12)•(-12)n-2(n∈N*，n≥2)∴a2-a1=(-a12)•(-12)2-2，a3-a2=(-a12)•(-12)3-2，… The superposition of an-an-1 = (- A12) · (- 12) n-2 yields an-a1 = (- A12) · [(- 12) 0 + (- 12) 1 + +(-12)n-2]∴an=a13[2+(-12)n-1]∵limn→∞an...

In known sequence {an}, A1 = 1, an + 1-an = 1 / 3 ^ (n + 1), then Liman (N ~ + infinity) =?
Such as the title

a(n+1)3^(n+1) = 3a(n)3^n + 1,
b(n)=a(n)3^n,
b(n+1) = 3b(n) + 1,
b(n+1) + x = 3[b(n) + x],1 = 3x-x=2x,x = 1/2.
b(n+1) + 1/2 = 3[b(n) + 1/2],
{B (n) + 1 / 2} is an equal ratio sequence with the first term B (1) + 1 / 2 = a (1) * 3 + 1 / 2 = 7 / 2 and the common ratio 3
b(n)+1/2=(7/2)*3^(n-1)=(7/6)3^n,
a(n) = b(n)/3^n = 7/6 - (1/2)*(1/3)^n,n=1,2,...
LIM (n - > + infinity) a (n) = 7 / 6

Find Liman with known sequence an satisfying A1 = 0 A2 = 1 an = (an-1 + An-2) / 2

2An=An-1 + An-2
2(An - An-1)=-(An-1 - An-2)
An - An-1=(-1/2)^(n-2)
An=(-1/2)^(n-2)+(-1/2)^(n-3)+…… +(-1/2)^0=(2/3)*(1-(-1/2)^(n-1))
limAn=2/3

Let {an}, A1 > 0, an = radical [3A (n-1) + 4], n-1 be the subscript, and prove: | an-4 | = 2); Liman = 4
Let {an}, A1 > 0, an = radical [3A (n-1) + 4], n-1 be the subscript, (n > = 2), and prove: | an-4 | = 2); Liman = 4

This is a good topic
For the sequence {an}, although the recurrence relation an = √ (3a (n-1) + 4) is clear, the first term A1 is not clear, so the sequence is uncertain and usually needs to be discussed
It is not difficult to find that when A1 = 4, A2 = A3 =... = an = 4, which means that the sequence {an} is a constant sequence and the general term an = 4
When 0

Lim [sin π / (√ n ＾ 2 + 1) + sin π / (√ n ^ 2 + 2) +... + sin π / √ n ^ 2 + n), N - > infinity

Note that sin is an increasing function, so that the first term in brackets is the largest, and the last term is the smallest, under nsin / root sign (n ^ 2 + n)

I can write sentences like this. 1. The yellow leaves are like small fans, which fan away the heat of summer______________ Like_____________ ，_______________________。______________ Like_____________ ，_______________________。 The autumn rain blows the golden trumpet___________________________________________________________

(1) ∵ f (θ) = Cos2 θ + (K-4) sin θ + 2k-9, when k = 3, f (θ) = Cos2 θ - sin θ - 3 = - (sin θ + 12) 2-74, when sin θ = - 12 & nbsp; i.e. θ = 7 π 6 or 11 π 6, f (θ) max = - 74, when sin θ = 1 & nbsp; i.e. θ = π 2, f (θ) min = - 4; (2) ∵ f (θ