The relationship between polar coordinate equation and circle equation

The relationship between polar coordinate equation and circle equation

ρ = a, that is, x ^ 2 + y ^ 2 = a ^ 2, ρ - polar radius of the circle. A - constant. 2. ρ = ACOS θ, that is, x ^ 2 + y ^ 2 = ax, (x-a / 2) ^ 2 + y ^ 2 = (a / 2) ^ 2. (θ - angle between ρ and X axis positive direction) 3. ρ = asin θ, that is, x ^ 2 + y ^ 2 = ay, x ^ 2 + (Y-A / 2) ^ 2 = (A / 2) ^ 2.4

6X ^ 2-13xy + 6y2 4x ^ 2-37x ^ 2Y ^ 2 + 9y ^ 4 4m ^ 2 + 8MN + 3N ^ 2 (a ^ 2-A) x - (AB + 2b) x-6b ^ 2 3x ^ 2-2 (2a + 1) x + A ^ 2-1 x ^ 4-3x ^ 3-28x ^ 2 (x ^ 2 = 2x) ^ 2-7 (x ^ 2 + 2x) - 8 it's better to be more detailed,

1. Primitive = (3x-2y) (2x-3y) 2. Primitive = 4x ^ 4-12x ^ 2Y ^ 2 + 9y ^ 4-25x ^ 2Y ^ 2 = (2x ^ 2-3y ^ 2) ^ 2 - (5xy) ^ 2 = (2x ^ 2 + 5xy-3y ^ 2) (2x ^ 2-5xy-3y ^ 2) = (x + 3Y) (2x-y) (x-3y) (2x + y) 3. Primitive = (M + 3n) (2m + n) 4. Primitive = [(A-1) X-3B)] (AX + 2b) = (...)

Write out any three solutions of the quadratic equation 3x-5y = - 7

x = 1,y=2
x=2,y=13/5
x=3,y=16/5

2X plus 25 equals 3x times 3 (x is the one to solve the equation, not a multiplication sign)!

2X plus 25 is 3x times 3
2x+25=9x
7x=25
x=25/7

The distance from a point in the image of positive scale function y = KX to the coordinate axis is 3,
The area of the line between the point and the perpendicular foot, the x-axis and the figure enclosed by the image is 6. The analytic expression of the positive scale function is obtained
It's a process
jijijijijijijijijij

Because the distance from this point to the axis is 3
If it is to the x-axis, then the coordinates are (3 / K, 3) or (- 3 / K, 3)
Then | 3 / K | * | 3 | / 2 = 6
|9/k|=12
K = 3 / 4 or K = - 3 / 4
If it goes to the y-axis, the coordinates are (3,3k) or (- 3, - 3K)
|3|*|3k|/2=6
K = 4 / 3 or K = - 4 / 3
So the analytic formula is y = 3x / 4 or y = - 3x / 4 or y = 4x / 3 or y = - 4x / 3

Solving a system of linear equations of two variables
Solving {2x + 2Y = 8} by adding and subtracting
2x-2y=4

4X = 12 from 1 + 2
x=3
If we take x = 3 into (1), we get y = 1
The solution of the original equations is {x = 3}
y=1

Calculation of Ohm's law of closed circuit
As shown in the figure (lamp L1 is connected in series with the power supply, and connected in parallel with lamp L2 of the switch), the resistance R1 of lamp L1 is 6 Ω. From the switch opening to closing, the total current of the circuit becomes 2 times of the original, while the current through lamp L1 becomes 1 / 2 of the original. Calculate the resistance of lamp L2 and the internal resistance of the battery
The process should be complete and clear

Suppose that the total unclosed current is I1, and the current in L2 is 3 / 2 of the original current after closing, so R1 = 3R2 R2 = 2 Ω, and the resistance of R2 in parallel with R1 is 1.5 Ω
There are (1.5 + R) · 2i1 = (6 + R) · I1
The solution is r = 3 Ω
There may be other methods applied to parallel resistors where the current is inversely proportional to the resistance and an equation

Factorization of the following formula: (a + b) ^ 2 + 3 (a + b) - 18 =?

Multiplication by cross phase
(a+b)^2+3(a+b)-18
a+b 6
a+b -3
therefore
(a+b)^2+3(a+b)-18
=(a+b+6)(a+b-3)

November 1, 2009 is a Sunday. If we calculate by classes from Monday to Friday and rest on Saturdays and Sundays, how many days of classes and rest in November?
Write the formula

21 days of classes, 9 days off

A math problem (be sure to write the calculation process and formula)
The number of students in class A is 6 more than 4 / 7 of that in class B. If one person is transferred from class B to class A, the number of students in class A is 5 more than 6 / 7 of that in class B
It's more than six out of five

There are x people in class A and y people in class B
4/5 Y+6=X
6/7 (Y-1)+5=X+1
X=46 Y=50
There are 46 in class A and 50 in class B