It is known that the circle ⊙ C1: x2 + Y2 + 2x + 2y-8 = 0 and ⊙ C2: x2 + y2-2x + 10y-24 = 0 intersect at two points a and B Find the equation of the circle with the smallest area passing through two points a and B

It is known that the circle ⊙ C1: x2 + Y2 + 2x + 2y-8 = 0 and ⊙ C2: x2 + y2-2x + 10y-24 = 0 intersect at two points a and B Find the equation of the circle with the smallest area passing through two points a and B

According to the problem, the circular system equation passing through two points is
x2+y2+2x+2y-8+λ（x2+y2-2x+10y-24)=0
The general equation of circle is x2 + Y2 + DX + ey + F = 0, the radius is 1 / 2 √ (D2 + e2-4f), and the formula can be used

If the square of (x + 2y-3) + the absolute value of 3x-2y-5 = 0

The square of (x + 2y-3) must be greater than or equal to 0, while the absolute value of 3x-2y-5 must be greater than or equal to 0, which requires that x + 2y-3 = 0, 3x-2y-5 = 0

If the absolute value of 2x-y + 1 + the square of 3x-2y-3 = 0, then the value of X-Y is?

The absolute value of 2x-y + 1 + the square of 3x-2y-3 = 0
So 2x-y + 1 = 0 (1)
3x-2y-3=0 (2)
(2)-(1)
x-y-4=0
x-y=4

If the absolute value of X-1 / 2 + the square of (2Y + 1) is 0, then what is the value of the square of x plus the cube of Y?

The absolute value of (x-1 / 2) and the square of (2Y + 1) are greater than or equal to 0,
The absolute value of (x-1 / 2) + the square of (2Y + 1) is 0,
So X-1 / 2 = 0,2y + 1 = 0
So x = 1 / 2, y = - 1 / 2
The square of X + the cube of y = (1 / 2) + (- 1 / 2) = 1 / 8

If the rational number x, y satisfies the equation (x + Y-2) 2 + | x + 2Y | = 0, then x2 + Y3=______

∵ the rational number x, y satisfies the equation (x + Y-2) 2 + | x + 2y| = 0,
Qi
x+y−2＝0
x+2y＝0 ，
The solution is,
x＝4
y＝−2 ；
∴x2+y3
=42+（-2）3
=16-8
=8；
So the answer is: 8

The absolute value of X-2 + the square of (2y-1) = the square of X + the square of Y=

|x-1/2|+(2y-1)²=0 x=1/2 y=1/2 x²+y²=1/2

∥ X-1 / 2 ∥ (2Y + 1) squared = 0, the square of x plus the cube of y =? ‖ represents the absolute value

∥ X-1 / 2 ∥ + (2Y + 1) square = 0, x square plus y cube = (1 / 2) square + (- 1 / 2) cube = 1 / 4-1 / 8 = 1 / 8

If the square of {X-1 / 2} + (2Y + 1) is equal to 0, then what is the value of the square of X + the cube of Y? (note, these two symbols are absolute value symbols)

∵ the square of {x-1/2} + (2y+1) =0
∴：x=1/2 y=-0.5
The value of the square of X + the cube of Y is:
23
（1/2） +(-0.5)
=0.25-0.125
=0.125

The absolute value of X + 2y-5 plus the square of x-2y + 3 is equal to zero. Find the values of X and y

x+2y-5=0
x-2y+3=0
x=1,y=2

If | x + 3 | + (Y-2) 2 = 0, then x-2y=______ .

∵|x+3|+（y-2）2=0，
∴x+3=0，y-2=0，
X = - 3, y = 2
∴x-2y=-3-2×2=-7．