Given that the equation of a circle is x2 + y2-2x + 6y + 8 = 0, then the equation of a straight line with a diameter of the circle is () A. 2x-y+1=0B. 2x-y-1=0C. 2x+y+1=0D. 2x+y-1=0

Given that the equation of a circle is x2 + y2-2x + 6y + 8 = 0, then the equation of a straight line with a diameter of the circle is () A. 2x-y+1=0B. 2x-y-1=0C. 2x+y+1=0D. 2x+y-1=0

Because the equation of the circle is x2 + y2-2x + 6y + 8 = 0, so the coordinates of the center of the circle (1, - 3), substituting the option, we can see that C is correct

The equation of a circle with radius 101 and tangent to point P (10,10) is obtained
The answer is (X-2) ^ 2 = (y-4) ^ 2 = 100, or (x-18) ^ 2 + (Y-16) ^ 2 = 100
The radius is 10

It is easy to get the equation of a line perpendicular to a known line and passing through a known point: y = (3 / 4) x + 5 / 2
Let the coordinates of the center of the circle (on the straight line) be (x, (3 / 4) x + 5 / 2), then the distance to the known straight line is:
|4X + 3 ((3 / 4) x + 5 / 2) - 70 | = 50, the solution is X1 = 2, X2 = 18, thus the answer can be obtained
The radius of the original question should be 10

The equation for a circle with radius 10 and tangent to (10,10) with 4x-3y-70 = 0

The distance from the center of the circle to the tangent is the radius
Let (a, b) be the center of the circle
Then | 4a-3b-70 | / √ (4 & sup2; + 3 & sup2;) = 10
|4a-3b-70|=50
The radius of tangent point is vertical 4x-3y-70 = 0
So the slope is - 3 / 4
So (B-10) / (A-10) = - 3 / 4
therefore
a=58/5,b=44/5
a=138/5,b=-16/5
(x-58/5)²+(y²-44/5)²=100
(x-138/5)²+(y²+16/5)²=100

The equation for finding the circle with radius 1 and tangent to the point (1,1) of the line 4x + 3y-7 = 0

The center of the circle is (1 / 5,2 / 5) or (9 / 5,8 / 5),

It is known that the center of a circle with radius 5 is on the x-axis. The abscissa of the center is an integer and tangent to the straight line 4x + 3y-29 = 0. The equation of a circle is obtained

Because the center of the circle is on the x-axis, the center of the circle is set to (x, 0) because it is tangent to the line 4x + 3y-29 = 0, then the distance from the center of the circle to the line r = | 4x + 3 * 0-29 | / √ (4 ^ 2 + 3 ^ 2) = 5 | 4x-29 | 254x-29 = 25 or 4x-29 = - 254x = 54 or 4x = 4x = 13.5 (because it is an integer, so it is rounded off) x = 1, so the equation of the circle is (x-1) ^ 2 + y ^

Given that the center of circle C with radius 5 is on the x-axis, the abscissa of the center is an integer and tangent to the line 4x + 3y-29 = 0, the equation of circle C is obtained

Let C (m, 0) be the center of the circle and m be an integer
If the circle C is tangent to the line 4x + 3y-29 = 0, the distance from the center of the circle C to the line 4x + 3y-29 = 0 is equal to the radius
That is | 4a-29 | / √ (16 + 9) = 5
|4m-29|=25
The solution is m = 27 / 2 or M = 1
Because m is an integer, M = 1
Then the equation of circle C is (x-1) &# 178; + Y & # 178; = 25

It is known that the abscissa of the center of a circle with radius 5 on the x-axis is an integer and tangent to the straight line 4x + 3y-29. The circular equation is solved

If the center of the circle is on the axis, let the coordinates of the center of the circle be o (a, 0)
The circle O is tangent to the straight line 4x + 3y-29 = 0, and the radius is 5, that is, the distance from the center O (a, 0) to the straight line is 5
That is: D = │ 4 * a + 3 * 0-29 / √ (4 & sup2; + 3 & sup2;) = │ 4 * a-29 / 5 = 5
4 * a-29 = 25 or 4 * a-29 = - 25
A = 13.5 or a = 1
∵ A is an integer
∴a=1
The center coordinates o (1,0)
The circular equation is: (x-1) & sup2; + Y & sup2; = 25

Given that the plane region represented by the inequality system y ≤ − x + 2Y ≥ KX + 1x ≥ 0 is a triangle with area equal to 1, then the value of real number k is ()
A. -1B. −12C. 12D. 1

The plane region represented by the system of inequalities y ≤ − x + 2Y ≥ KX + 1x ≥ 0 is shown as follows: the plane is a triangle, so it passes through point (2, 0), ∵ y = KX + 1, and the intersection point with X axis is (- 1K, 0), ∵ 1K = 2, ∵ k = - 12, at this time, s = 12 × 1 × 2 = 1, so B

The area of the plane region represented by the constraint x ≥ 0y ≥ 0x + y ≤ 2 is______ ．

The plane region represented by the inequality system is shown in the figure, and the solution is a (2,0), B (0,2), O (0,0), so s △ ABO = 12 × 2 × 2 = 2; the area of the plane region represented by the inequality system is: 2, so the answer is: 2

The area of the plane region bounded by X ≥ 0, y ≥ 0 and X + y ≤ 4 is______ ．

Let a straight line x + y-4 = 0 intersect the x-axis at point a (4, 0) and the y-axis at point B (0, 4). Therefore, let a plane region bounded by X ≥ 0, y ≥ 0 and X + y ≤ 4 be obtained. As shown in the figure, ∫ OA | = 4, | ob | = 4, ∫ s △ ABO = 12 ×| OA | ×| ob | = 8, that is, the area of the plane region bounded by X ≥ 0, y ≥ 0 and X + y ≤ 4 is 8