Through the point P (- 2, - 3), make two tangents of the circle C: (x-4) ^ 2 + (Y-2) ^ 2 = 9, the tangent points are a ` B respectively. Find the square of the circle passing through the center C and the tangent point is A.B

Through the point P (- 2, - 3), make two tangents of the circle C: (x-4) ^ 2 + (Y-2) ^ 2 = 9, the tangent points are a ` B respectively. Find the square of the circle passing through the center C and the tangent point is A.B

PA^2=PC^2-9=36+25-9=52.
The equation for a circle with P as its center and PA as its radius is
(x+2)^2+(y+3)^2-52=0,
Let the equation of the circle passing a and B be
(x-4) ^ 2 + (Y-2) ^ 2-9 + m [(x + 2) ^ 2 + (y + 3) ^ 2-52] = 0, it passes through point C, so - 9 + 9m = 0, M = 1
The equation of the circle is x ^ 2 + y ^ 2-2x + y-14 = 0

If the tangent of the circle C: x ^ 2 + y ^ 2 = 9 is made through the point P (1,3), then the length of the tangent is longer__________
(shouldn't the tangent be a straight line? How can there be a length? Is the topic wrong? Instead, find the tangent equation!)

We can first calculate the distance from P to the center of the circle (set as a), and at the same time, let the "tangent length" be D, and the radius of the circle be r. we can get d from Pythagorean square A-R square = D square

Through the point P (3,4) to make the tangent of the square of the circle x plus the square of y equal to 1, the tangent point is a and B to find the length of ab

Let the center of the circle be o,
Then OP = 5
PA = Pb = 2 radical 6
AB = (2 radical 6 * 1 / 5) * 2 = 4 / 5 (radical 6)

Let x, y ∈ R, then the plane area of the point P (x, y) satisfying the condition x + 2Y ≥ 0, x-3y-5 ≤ 0. X ^ 2 + y ^ 2-4x + 2y-4 ≤ 0 is?

The plane region represented by inequality (x + 2y-1) (X-Y + 3) > 0 is ()
A. B. C. D.

Inequality (x + 2y-1) (X-Y + 3) > 0 can be changed into x + 2Y − 1 > 0x − y + 3 > 0, or x + 2Y − 1 < 0x − y + 3 < 0; in the same coordinate plane, make the plane region represented by these two inequality groups, as shown in the figure; from the figure, the plane region represented by the original inequality is C

Suppose that the variables X, y ∈ (0, + ∞), the parameters a, B ∈ (0, + ∞), and 2x + y + a = 6, x + 2Y + B = 6, then the area of the plane region formed by (x, y) satisfying the requirements is?
A. 12 b, 6 C, 4 D, infinity

D
How to use X-Y to get X-Y = A-B, and because a and B belong to 0 to positive infinity, we can know that the area of plane region is infinite

If the variables X, y satisfy x-2y + 1 ≤ 0, 2x-y ≥ 0, X ≤ 1, then the point P (2x-y, x + y) represents the area of the region

Let m = 2x-y, n = x + y; because x-2y + 1 ≤ 0, 2x-y ≥ 0, X ≤ 1
So M-N + 1 ≤ 0, m ≥ 0, M + n = 3x ≤ 3
In this way, it is transformed into a plane linear programming problem. The intersection of the line M-N + 1 = 0 and the line M + n = 3 is (1,2)
So the point P (2x-y, x + y) indicates that the area of the region is 1

If a ≥ 0, B ≥ 0, and X ≥ 0, y ≥ 0 x + y ≤ 1, ax + by ≤ 1, then the area of plane region formed by point P (a, b) with a, B as coordinates is ()
A. 12B. π4C. 1D. π2

∵ a ≥ 0, B ≥ 0t = ax + by, the maximum value is obtained at the upper right of the region, that is, it must be obtained at the point (0, 1) or (1, 0), so by ≤ 1 or ax ≤ 1, 0 ≤ B ≤ 1 or 0 ≤ a ≤ 1, and the plane region formed by P (a, b) with a and B as coordinates is a square, so the area is 1

If a > = 0, b > = 0, and x > = 0, Y > = 0, x + y

This is the problem in my homework today. The following is what I want to find. I think it's right,
y

Given that P (x, y) is the last moving point on the circle (x-1) 2 + (y + 1) 2 = 4, then the value range of x2 + Y2 is obtained-----------

Let x = 2cosa + 1, y = 2sina-1
x^2+y^2=(2cosa+1)^2+(2sina-1)^2
=-4sina+4cosa+6
=-4√2sin(a-π/4)+6
Therefore, the value range of x ^ 2 + y ^ 2 is [6-4 √ 2,6 + 4 √ 2]