Given that point a (- 2,0) B (0,2), C is any point on the circle x ^ 2 + y ^ 2 + 2x = 0, then the maximum area of triangle ABC is?

Given that point a (- 2,0) B (0,2), C is any point on the circle x ^ 2 + y ^ 2 + 2x = 0, then the maximum area of triangle ABC is?

As shown in the following figure, let the tangent y = x + B and the circle x ^ 2 + y ^ 2 + 2x = 0 be tangent to point C
That is, the equation x ^ 2 + (x + b) ^ 2 + 2x = 0 has only one real root
x²+(b+1)x+b²/2=0
(b+1)²-2b²=0
-b²+2b+1=0
(b-1)²=2
B = 1 - √ 2 (positive value 1 + √ 2 is rounded off)
|AB|=2√2
The distance from C to AB is the distance between the line AB and the tangent y = x + B
=(2-b)/√2 = (2-1+√2)/√2= (√2+2)/2
Maximum ABC area = & nbsp; (1 / 2) * 2 √ 2 * & nbsp; (√ 2 + 2) / 2 = 1 + √ 2

The equation y = f (x) denotes a curve. In the new coordinate system, the equation is y '= f (x' + 1) + 2, then the origin of the new coordinate system is the equation in the original coordinate system
If we know that "m = 1" is a complex number and m ^ 2-1 + I is a pure imaginary number, what is the condition?

① y=f(x)
y'=f(x'+1)+2
∴ y'-2=f'(x'+1)
The transformation formula is y = y '- 2, x = x' + 1
The original coordinates of the new origin are (1, - 2)
② Is there a condition that M is a real number
If not, "m = 1" is a sufficient and unnecessary condition for the complex number m ^ 2-1 + I to be "pure imaginary"

If the equation (m-1) x & sup2; + 2Y & sup2; + M & sup2; - 2m-3 = 0 denotes an ellipse, then the range of real number m is?

m²-2m-30
m-1≠2
Solution
-1

If the equation x + 2y-2m √ (x + 2Y) + m + 2 = 0 represents a straight line, then the value range of real number m is?

x+2y-2m√(x+2y)+m+2=0
[√(x+2y)-m]^2=(m+1)*(m-2)≥0
m≤-1,m ≥2