Find the equation of the square of the sum circle: (x + 2) + (y-6) = 1 with respect to the symmetric circle of the line 3x-4y + 5 = 0 Please write down the process and steps, thank you very much!

Find the equation of the square of the sum circle: (x + 2) + (y-6) = 1 with respect to the symmetric circle of the line 3x-4y + 5 = 0 Please write down the process and steps, thank you very much!

Circle C (- 2,6) symmetry point P (m, n) intersection line 3x-4y + 5 = 0, slope K1 × 3 / 4 = - 1, K1 = - 4 / 3 PC: y = - 4 / 3x + B, passing through C point q is the intersection point of line y = - 4 / 3x + 10 / 3 and line 3x-4y + 5 = 0, the coordinates of Q point Q (1,2) P is CP midpoint (- 2 + m) / 2 = 1, (6 + n) / 2 = 2p (4, - 2)
Given the vector a = E1 + E2, B = 4E1 + 3e2, where E1 = (1,2), E2 = (0,1), calculate the value of a * B. | a + B | and find the cosine of the angle between a and B?
a=(1,3) b=(4,11)
a*b=1*4+3*11=37
|a+b|=221^(1/2)
cos=37/(1370^1/2)
You can do that
a=(1,3)
b=(4,11)
a*b=1*4+3*11=37
|a+b|=221^(1/2)
cos=37/(1370^1/2)
The correct solution of 2 / 3 x minus 1 minus 5 / 2 x plus 1 is less than or equal to 1!
Multiply both sides by 6 - 11x-5 = - 1
Factorization (x2 + y2-a2) 2-4x2y2
(x2 + y2-a2-a2) 2-4x2x2y2 = (X ; (; (x #35; (178;; + y #35;35;178; + y (x2 + Y2 + y2-a2-a2) 2-4x2y2 = (x ##35\\35\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ifthere is
If f (x) = x & # 178; - 2aX + A & # 178; + 2 is a decreasing function on (- ∞, 1), then the range of a is
A:
f(x)=x^2-2ax+a^2+2
f(x)=(x-a)^2+2
The opening of parabola f (x) is upward, and the axis of symmetry x = a
X=1
The solution is: a > = 1
f(x) =x^2-2ax+a^2+2
f'(x) = 2x-2a a>1
Given the vectors E1 and E2, the included angle is 60 degrees; now define the vector a = E1 + 3e2, B = 2E1; find the projection of a on B
=(E1 + 3e2) · 2E1 = 2 [E11 + 3e2] · 2E1 = 2 [E11 + 3e2] · 2E1 = 2 [E12 + 6e1 · E2 = 2 [e1se1] and [E12 + 6 [e1se1] and [6 [e1se1 |||124;||||||| [E2 ||||| ^ ^ 2 + 6e1 · E2 · E2 = 2 |||||||| [cos = a · B · B / | [[b] this is the projection in the projection of the projection in B direction in the direction of projection of the projection of the projection of the projection of the projection of the projection of this paper [[[a] a] a] a] objective objective objective objective objective objective objective objective objective] projection: the projection: the if the title is: E1 and E2 are unit vectors, then a is in B
The solution of two thirds plus one minus five sixths minus one equals one
Multiply both sides of the equation by 6
2（2x+1）-5x+1=6,x=-3
(2x+1)/3-(5x-1)/6=1
2(2x+1)-5x+1=6
4x+2-5x+1=6
-x=3
x=-3
-6
Zero
Factorization 9x2 - Y2 - 4Y - 4
Factorization 9x 2 - y 2 - 4Y - 4
9x-y-4y-4 =9x-(y+4y+4) =(3x)-(y+2) =[3x+(y+2)][3x-(y+2)] =(3x+y+2)(3x-y-2)
Let f (x) = 1 / 1-x & # 178; (x)
prove:
Let X1 and X2 be the two points of F (x) on the domain (- ∞, - 1), and X1 > X2, then
f(x1)-f(x2)=1/(1-x1²)-1/(1-x2²)=[(1-x2²)-(1-x1²)]/(1-x1²)(1-x2²)=(x1²-x2²)/(1-x1²)(1-x2²)
Because - 1 > X1 > x2
Therefore, 1-x1 & # 178;
Let x1f (x2)
And because of x1
Let the vectors E1 and E2 be two non collinear vectors. If a = E1 + λ E2 and B = - (2e1-3e2) are collinear, then the real number λ=
∵ a, B vectors are collinear
∴e1+λe2=-(2e1-3e2)
e1+λe2=3e2-2e1
∴-（e1+λe2）=3e2-2e1
-λe2-e1
∴3：-λ=2:1
∴λ=-3/2
∵ a, B vectors are collinear
∴m（e1+λe2）=-(2e1-3e2)
That is ME1 + m λ E2 = - 2E1 + 3e2
So m = - 2
So m λ = 3
So λ = - 3 / 2