After translating the vector a = (3, - 2) according to the vector (1,1), we get the vector B, then | B|=

After translating the vector a = (3, - 2) according to the vector (1,1), we get the vector B, then | B|=

The vector is invariant after translation,
The vector B is obtained by translating the vector a = (3, - 2) according to the vector (1,1)
Then B = a = (3, - 2)
∴ |b|=√(9+4)=√13

If the circle C: X & # 178; + Y & # 178; + 4Y = 5 is translated according to vector a = (- 3,4), then the radius and center coordinates of the circle C 'are

The solution is from the circle C: X & # 178; + Y & # 178; + 4Y = 5
That is X & # 178; + (y + 2) &# 178; = 9
C (0-2), radius 3
Then translate the circle C: X & # 178; + Y & # 178; + 4Y = 5 according to vector a = (- 3,4), then the radius to the circle C 'is 3
Let the center of circle C 'be C' (a, b), and the vector CC '= (a, B + 2)
Then the vector CC '= the vector a
That is, (a, B + 2) = (- 3,4)
That is, a = - 3, B + 2 = 4
That is, a = - 3, B = 2
That is, the center of circle C 'is C' (- 3,2),

After the circle x2 + y2 = 2 is translated according to the vector V = (2,1), it is tangent to the straight line x + y + λ = 0, then the value of real number λ is______ ．

After the circle x2 + y2 = 2 is translated according to the vector V = (2,1), it becomes (X-2) 2 + (Y-1) 2 = 2 and is tangent to the straight line x + y + λ = 0. There is | 2 + 1 + λ| 2 = 2, and the solution is λ = - 1 or λ = - 5, so the answer is: - 1; - 5

Translate point a (2,0) to point B according to vector A. if there is only one straight line L passing through point B tangent to circle x ^ 2 + y ^ 2-2x + 2y-6 = 0, find the equation of L when | a | is minimum

Translate point a (2,0) to point B according to vector A. if there is only one straight line L passing through point B tangent to circle x ^ 2 + y ^ 2-2x + 2y-6 = 0, find the equation of L when | a | is minimum
Because there is only one line L passing through point B tangent to the circle x ^ 2 + y ^ 2-2x + 2y-6 = 0, so point B is a point on the circle!
Then when | a | is minimum, the direction of vector a must be on the line between point a and the center of the circle
Circle x ^ 2 + y ^ 2-2x + 2y-6 = 0 can be changed into: (x-1) ^ 2 + (y + 1) ^ 2 = 4
The center coordinates are: O (1, - 1)
The slope of Ao is k = (- 1-0) / (1-2) = 1
The equation of line Ao is y = X-2
The intersection point of the line AO and the circle is (the equation can be solved by substituting y = X-2 into the circle)
x1=1+√2 ,x2=1-√2
It can be verified that X1 of the above two roots can make | a | the smallest, so X2 is rounded off
Therefore, the coordinates of point B are further obtained as: (1 + √ 2, √ 2-1)
It is easy to know that the tangent L is perpendicular to Ao, so the slope of L is KL = - 1 / 1 = - 1 and passes through point B
So the equation of L is y = - 1 (x-1 - √ 2) + √ 2-1
The results are as follows
L：y＝－x＋2√2

If the line 3x-y + C = 0 is tangent to the circle x2 + y2 = 10 after translation according to the vector a = (1, - 1), then the value of C is ()
A. 14 or - 6B. 12 or - 8C. 8 or - 12D. 6 or - 14

After the line 3x-y + C = 0 is translated according to the vector a = (1, - 1), we get 3 (x-1) - (y + 1) + C = 0, which is changed into 3x-y-4 + C = 0. From the meaning of the question, we can get | C − 4 | 10 = 10, and the solution is C = - 6 or 14

If the image of y = 2sin2x is translated according to vector a to get the image of y = sin (2x + π / 6) + 3, then vector a can

y-3=2sin[2(x+π/12)]
y-3=2sin{2[x-(-π/12)]}
So a = (- π / 12,3)

If the image y = 2cos (x / 3 + π / 6) is translated according to vector (- π / 4, - 2), the analytic expression of the translated image is

According to the vector translation and we usually up and down left and right translation writing is not the same
y=2cos[(x+π/4)/3+π/6]-2
=2cos(x/3+π/4)-2
If you don't understand this question, you can ask,

If the image of y = 2cos (x / 3 + Pai / 6) is translated according to the vector a = (- Pai / 4, - 2), the analytic expression of the translated image is
A.y=2cos（3/x+4/pai)-2
B.y=2cos(x/3-pai/4)+2
C.y=2cos(x/3-pai/12)-2
D.y=2cos(x/3,pai/12)+2

Y = 2cos [(x + π / 4) / 3 - π / 6] - 2 = 2cos (x / 3 - π / 12) - 2 choose C

If the image of y = Tan (x / 3 + Pie / 6) is translated according to the vector a = (negative pie, 2), the analytic expression of the translated image is
My expression is not very clear, sorry, please write down the process,

That is, X becomes X - (- π) = x + π
Y becomes Y-2
So Y-2 = Tam [(x + π) / 3 + π / 6]
y=tan(x/3+π/2)+2

Given the line y = - 12x + 1, how to translate the line along the X axis to cross the origin

Let the analytic expression of the line after translation be y = - 12x + B, and substitute the origin (0, 0) to get b = 0, that is, the analytic expression of the line after translation is y = - 12x, ∵ y = - 12x + 1 = - 12 (X-2), ∵ translate the line y = - 12x + 1 to the left along the X axis by 2 units, and get the image of y = - 12 (X-2 + 2), that is, y = - 12x