A mathematical problem of circle The equation of the circle passing through point a (2, - 2) B (5,3) C (3, - 1), and the center and radius of the circle are obtained

A mathematical problem of circle The equation of the circle passing through point a (2, - 2) B (5,3) C (3, - 1), and the center and radius of the circle are obtained

Let (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2
Take three points into the equation
(2-a)^2+(2+b)^2=r^2①
(5-a)^2+(3-b)^2=r^2②
(3-a)^2+(1+b)^2=r^2③
① - 26 + 6A + 10B = 0.5
② (3) 24-4a-8b = 0.6
⑤ A = - 4, B = 5
Bring in ① to get r = √ 85

Solutions of bivariate quadratic equation
Square of X + X-6 = 0

x²+x-6=0
(x+3)(x-2)=0
x=-3,x=2

Solving quadratic equation of two variables
X square - 2XY + 3Y square = 9
4X square - 5xy + 6y square = 30

X ^ 2-2xy + 3Y ^ 2 = 9 --- (1) 4x ^ 2-5xy + 6y ^ 2 = 30 --- (2) multiply formula (1) by 10 to get 10x ^ 2-20xy + 30y ^ 2 = 90. Multiply formula (2) by 3 to get 12x ^ 2-15xy + 18y ^ 2 = 90. Subtract the two formulas to get 2x ^ 2 + 5xy-12y ^ 2 = 0. Then (x + 4Y) (2x-3y) = 0, then x = - 4Y or x = 3Y / 2. If x = - 4Y, then (1)

Find the minimum positive period of function f (x) = √ 3sin (2x - π / 6) + 2Sin (x - π / 12) and the set to obtain the maximum value x?
It's urgent!

F (x) = √ 3sin (2x - π / 6) + 2Sin (x - π / 12) = √ 3sin (2x - π / 6) + 1-cos (2x - π / 6) = 2 (√ 3 / 2Sin (2x - π / 6) - 1 / 2cos (2x - π / 6)) + 1 = 2 (sin (2x - π / 6) cos π / 6-cos (2x - π / 6) sin π / 6) + 1 = 2Sin (2x - π / 6) + 1 = 2Sin (2x - π / 3)

Divide 12 into the sum of two positive integers and get the maximum value of the product of two positive integers

Let one integer be x and the other be 12-x
x(12-x)
=12x-x²
=-(x²-12x+36)+36
=-(x-6)²+36
The maximum value is 36 when x = 6
The other number is 12-6 = 6

On the problem of quadratic equation
It is known that a, B and C are the lengths of three sides of the triangle ABC and satisfy the following conditions
(2a^2)/(1+a^2)=b (2b^2)/(1+b^2)=c (2c^2)/(1+c^2)=1
Finding the area of triangle ABC

From (2C ^ 2) / (1 + C ^ 2) = 1
c=1
Substitute C = 1 into (2B ^ 2) / (1 + B ^ 2) = C to get
b=1
Let B = 1 (2a ^ 2) / (1 + A ^ 2) = b
a=1
So there is
a=b=c
So a triangle is an equilateral triangle
S = (1 / 2) * (1 * radical 3 / 2) = radical 3 / 4

Bivariate quadratic equation X-Y / 2 = 2,2x + 3Y = 12
Binary linear equations X-Y / 2 = 2,2x + 3Y = 12

x-y/2=2,（1）
2x+3y=12 （2）
(2) - (1) * 2
4y=8
y=2
Substituting: x = 3
Therefore, the solution of the equations is: x = 3; y = 2

How can these two binary quadratic equations be brought into and solved: X-Y = 63y + 2x = 72

x-y=6
y=x-6
Substituting
3(x-6)+2x=72
3x-18+2x=72
5x=90
x=18
y=x-6=12

The bivariate quadratic equation (2x + 3) (3y-2) = 0 has solutions

When x = - 3 / 2, y has innumerable solutions
When y = 2 / 3, X has innumerable solutions

How to solve binary linear equations. Find m (8x = 2m (- 5x = M-8)
（8x=2m
（-5x=m-8
6x-10y=4m（1）
6x+21y=3m-24（2）
Then (1) - (2)
y=

The general idea is as follows: (1) 8x = 2m, M = 4x
(2) - 5x = M-8 gives m = - 5x + 8
(3) That is 4x = - 5x + 8 to get x = 8 / 9
（4）M=4X =4*8/9=32/9