If x-2y = 7, xy = - 5, then the value of x2 + 4y2-4 is () A. 25 B. 35 C. 49 D. 65

If x-2y = 7, xy = - 5, then the value of x2 + 4y2-4 is () A. 25 B. 35 C. 49 D. 65

∵x-2y=7，
ν (x-2y) 2 = 49, that is, x2-4xy + 4y2 = 49,
∵xy=-5，
∴x2-4×（-5）+4y2=49，
∴x2+4y2=29，
∴x2+4y2-4=29-4=25．
Therefore, a

Given the square of (4x-2y-1) + | XY-2 | = 0, find the square of 2x, the square of y-xy and the value of XY

According to the meaning of the title
4x-2y-1=0
2(2x-y)=1
2x-y=1/2
xy-2=0
xy=2
2x^2y-xy2+xy
=xy(2x-y+1)
=2(1/2+1)
=3

If the square of X + the square of y-4x-2y + 5 = 0, find the value of the square of (root x-root y) + 4 × root x × root Y / x + root x × root y

The original formula = (X-2) ^ 2 + (Y-1) ^ 2 = 0
Because the square is greater than or equal to 0
So x = 2, y = 1
The square of (root x-root y) + 4 × root x × root Y / x + root x × root y
= (2 + 1-2 root 2) + 2 root 2 + root 2
= 3 + root 2

Let x 2 + 4x + y 2-2y + 5 = 0, find the value of X and y

x2+4x+y2-2y+5=0，
The deformation is: (x2 + 4x + 4) + (y2-2y + 1) = 0,
That is (x + 2) 2 + (Y-1) 2 = 0,
Because (x + 2) 2 and (Y-1) 2 are both nonnegative numbers,
So (x + 2) 2 = 0 and (Y-1) 2 = 0,
That is, x + 2 = 0, Y-1 = 0,
X = - 2, y = 1;
A: y = 1

Given the square of X + the square of Y + 4x-2y + 5 = 0, find the value of X + y. thank you,

Take 5 apart and make it into the form of square difference. X + y + 4x-2y + 5 = 0 x + 4x + y-2y + 5 = 0 x + 4x + 4 + y-2y + 1 = 0 (x + 2) + (Y-1) = 0, that is, x + 2 = 0, Y-1 = 0, x = - 2, y = 1, x + y = - 2 + 1 = 1

Given that the circle C1: x2 + Y2 + 2x + 3Y + 1 = 0 and C2: x2 + Y2 + 4x + 3Y = 0, then the positional relationship between circle C1 and circle C2 is___ .

Circle C1:x2+y2+2x+3y+1=0, converted to (x+1) 2+ (y+3
2）2=（3
2) The coordinates of the center of the circle are (- 1, - 3)
2) With a radius of 3
2；
The circle C2: x2 + Y2 + 4x + 3Y = 0 is reduced to (x + 2) 2 + (y + 3)
2）2=（5
2) 2, center coordinates (- 2, - 3)
2) With a radius of 5
2．
The center distance of the circle is: | - 1 - (- 2) | = 1,
Because 5
2-3
2 = 1, so two circles are inscribed
So the answer is: introversion

It is known that the circle C1: x ^ 2 + y ^ 2-4x-2y-5 = 0 and the circle C2: x ^ 2 + y ^ 2-6x-y-9 = 0, Find a point P on the plane, lead the tangent lines of two circles through the point P, and make their length equal to 6 √ 2

There are some troubles in the process. Here I will write my thoughts
Let P coordinate (x, y) use the Pythagorean theorem of right triangle composed of P, center and tangent point to calculate the distance between two centers of circle and point P respectively, and set it as D1 and D2, and then use Pythagorean theorem to list the equations D1 ^ 2 + R1 ^ 2 = (6 √ 2) ^ 2 and D2 ^ 2 + R2 ^ 2 = (6 √ 2) ^ 2
Two unknown numbers x y two equations can solve P (x, y)

The number of common tangents of the circle x? + y? + 4x-4y + 7 = 0 and the circle x? + y? - 4x + 10Y + 13 = 0

Four

How many common tangents of the circle x ^ 2 + y ^ 2 + 4x-4y + 4 = 0 and x ^ 2 + y ^ 2-4x-10y + 13 = 0?

First, the standard circle equation is reduced
It can be seen that the centers of the two circles are (2,5), (- 2,2), and the radii of the two circles are root 29 and root 2 respectively
Two circles intersect
Then there are two common tangents

How many lines are tangent to the two circles x + y + 4x-4y + 7 = 0 and X + y-4x-10y-13 = 0

Three