Given that the point P (2,1) is on the circle C: x ^ 2 + y ^ 2 + ax-2y + B = 0, and the symmetric point P 'of point P about the straight line X-Y = 0 is also on the circle C, then a + B =?

Given that the point P (2,1) is on the circle C: x ^ 2 + y ^ 2 + ax-2y + B = 0, and the symmetric point P 'of point P about the straight line X-Y = 0 is also on the circle C, then a + B =?

If two points on a circle are symmetrical with respect to a line, then the line passes through the center of the circle
(x+a/2)^2+(y-1)^2=a^2/4+1-b
So the center of the circle (- A / 2,1) is on X-Y = 0
-a/2-1=0
a=-2
Substituting P into a circle
4+1-4-2+b=0
B=1
a+b=-1
We know that P is a moving point on the line 3x + 4Y + 8 = 0, PA and Pb are two tangents of the circle XX + yy-2x-2y + 1 = 0,
A. B is the tangent point, C is the center of the circle, then the minimum area of the quadrilateral PACB is? (please give an explanation,)
The area of quadrilateral can be divided into two RT △ CPA and RT △ CPB. The height is radius 1, and the two bottom edges are equal. When the bottom edges PA and Pb are the smallest, the area is the smallest. At this time, PC is the smallest, when PC ⊥ is the straight line, PC is the smallest. The center coordinates of circle (1,1), distance d = 15 / 5 = 3, so the minimum of bottom edge is 2 √ 2, so the minimum area of quadrilateral is 2 √ 2
It is known that P is the moving point on the straight line 3x + 4Y + 8 = 0, PA and Pb are the two tangent lines of the circle C: X & sup2; + Y & sup2; - 2x-2y + 1 = 0, and a and B are the tangent points
C is the center of the circle, find the minimum area of the quadrilateral PACB
The minimum area is required. PABC can be divided into two triangles, APC and BPC. Because PA and Pb are tangent lines, the heights of the two triangles have been determined. AC and BC are equal to the far radius 1. What is required is the minimum area, that is, AP and BP reach the minimum. At the same time, because the two triangles
Zero
Finding the inequality x ^ 2-ax-12a ^ 2 about X
(x-4a)(x+3a)
x²-ax-12a²
Given that the solution set of quadratic inequality ax & # 178; + BX + C ＜ 0 is {x | x ＜ 1 / 2 or X ＞ 1 / 3}, find the solution set of Cx & # 178; - BX + a ＞ 0
We know that the solution set of the quadratic inequality ax & # 178; + BX + C ＜ 0 is {x | x ＜ 1 / 2 or X ＞ 1 / 3}. We show that the two roots of a ＜ 0, ax & # 178; + BX + C = 0 are x = 1 / 2, x = 1 / 3. According to Weida's theorem: X1 + x2 = 1 / 2 + 1 / 3 = 5 / 6 = - B / ax1 * x2 = 1 / 2 * 1 / 3 = 1 / 6 = C / AC = A / 6, B = - 5A / 6cx & # 178; - BX + a = ax
If the solution set of inequality ax & # 178; + BX + 2 > 0 is (- 1 / 2,1 / 3), then the value of a + B is
In the solution set of quadratic inequality, the two short points are the two roots of the corresponding quadratic equation,
The simplest way is WIDA's theorem: X1 + x2 = - 1 / 6 = - B / A, X1 * x2 = - 1 / 6 = 2 / A
So: a = - 12, B = - 2
So: a + B = - 14
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It's easy to learn the Vader theorem
That is, the relationship between root and coefficient
According to the solution set, A0
=>(x^2+b/a*x+2/a)a=-12
b=-2 a+b=-14
Hope to help you
Given a ≠ 0, we solve the quadratic inequality ax & # 178; + (a + 2) x + 2 > 0
ax²+(a+2)x+2=(ax+2)(x+1)>0
The discussion is divided into three cases
Solve the quadratic inequality of one variable about X: ax ^ 2 + (A-1) X-1 > 0
① When a = 0
Inequality into
-x-1>0
x+10
Let ax ^ 2 + (A-1) - 1 = 0
The solution is X1 = - 1, X2 = 1 / A
So x ∈ (negative infinity, - 1) ∪ (1 / A, positive infinity)
③ When a = - 1
Change to inequality
-x^2-2x-1>0
x^2+2x+1
To solve the inequality ax + 56x & # 178; < A & # 178;
∵56x²+ax＜a²
And △ 0
So the equation has two solutions,
That is, X1 = - A / 7, X2 = A / 8
When a = 0, it is an empty set
When a > 0, that is X1 < X2, then the solution set of the equation is (- A / 7, a / 8)
When a ＜ 0, that is, x1 ＞ X2, the solution set of the equation is (A / 8, - A / 7)
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The solution of inequality about X: 56x2-ax-a2 > 0
56x2-ax-a2 ＞ 0 can be changed into (7x-a) (8x + a) ＞ 0. ① when a ＞ 0, - A8 ＜ A7, ｜ x ＞ A7 or X ＜ - A8; ② when a ＜ 0, - A8 ＞ A7, ｜ x ＞ - A8 or X ＜ A7; ③ when a = 0, X ≠ 0