The equation of tangent between passing point (- 1,2) and the square of circle (X-2) + (y + 1) = 25 is?

The equation of tangent between passing point (- 1,2) and the square of circle (X-2) + (y + 1) = 25 is?

Let the coordinates of the center of the circle be (- 1,2)
Then the radius of the circle r = 5 - [(2, - 1) to (- 1,2) distance] = 7 open arithmetic square root
The equation is (x + 1) * (x + 1) + (Y-2) (Y-2) = 7
y=3/2x+7/2 y=2/3x+8/3
Center (2, - 1), radius = 5
(- 1,2) distance to center = √ 18
Find the linear equation that passes through point (5, - 5) and is tangent to circle (x-1) ^ 2 + (y + 2) ^ 2 = 25?
Theoretically, there are two straight lines. Why did they lose their roots in the end?
Let y + 5 = K (X-5) and kx-y-5k-5 = 0
∵(x-1)^2+(y+2)^2=25
The center of the circle is (1, - 2) and the radius is 5
∴|k+2-5k-5|/√k^2+1=5
9k^2-24k+16=0
（3k-4）^2=0
∴k=4/3
∴y+5=4/3(x-5)
Y = 0-4x-35
Hee hee, OK~
Let y + 5 = K (X-5) and kx-y-5k-5 = 0
The center of the circle is (1, - 2) and the radius is 5
The distance from the center of the circle to the straight line is 5
|k+2-5k-5|/√k^2+1=5
arrangement
9k^2-24k+16=0
We find that K has two values
Given the circle C: (x-1) ^ 2 + y ^ 2 = 25, point a (6,6), find the equation of point a and tangent to the circle
The solution shows that there are two straight lines passing through the point (6,6) and the circle C: (x-1) ^ 2 + y ^ 2 = 25, one of which has no slope and is a straight line x = 6. Let the equation of the other straight line be y-6 = K (X-6). The distance from the center (1,0) of the circle C: (x-1) ^ 2 + y ^ 2 = 25 to the straight line y-6 = K (X-6) is 5, that is / 6-5k / √ (1 + K & # 178;) =
Because the tangent line goes through (6, 6)
So let y-6 = K (X-6)
That is kx-y-6k + 6 = 0
And the line is tangent to the circle, so the distance from the center of the circle to the line is equal to the radius! The center coordinates are (1,0)
The formula of distance from point to line:
|k－6k＋6|/√(1＋k∧2)＝5
The solution is k = 1 / 12
So the tangent equation is:
Y / y = 12 x / 11
Because the tangent line goes through (6, 6)
So let y-6 = K (X-6)
That is kx-y-6k + 6 = 0
And the line is tangent to the circle, so the distance from the center of the circle to the line is equal to the radius! The center coordinates are (1,0)
The formula of distance from point to line:
|k－6k＋6|/√(1＋k∧2)＝5
The solution is k = 1 / 12
So the tangent equation is:
Y = 1 / 12x + 11 / 2
According to the circle equation, the coordinates of point C at the center of the circle are: C (1,0), radius = 5,
Let the tangent point coordinate be B (m, n)
⊥ from ab ⊥ CB:
①、[﹙6－n﹚／﹙6－m﹚][﹙0－n﹚﹙1－m﹚]=－1
∵ B is on the circle C, ∵ we get that:
②、﹙m－1﹚²＋n²=25
From the simultaneous equations of (1) and (2)
m1=6，m2=6／61
∴n1=0，n2=300／61
... unfold
According to the circle equation, the coordinates of point C at the center of the circle are: C (1,0), radius = 5,
Let the tangent point coordinate be B (m, n)
⊥ from ab ⊥ CB:
①、[﹙6－n﹚／﹙6－m﹚][﹙0－n﹚﹙1－m﹚]=－1
∵ B is on the circle C, ∵ we get that:
②、﹙m－1﹚²＋n²=25
From the simultaneous equations of (1) and (2)
m1=6，m2=6／61
∴n1=0，n2=300／61
The coordinates of point B are B (6,0) or B (6 / 61300 / 6)
The AB line equation can be obtained from a and B coordinates
X = 6 or y = (1 / 50) x + 294 / 50
The solution set of inequality 3x + 2 < 3 is______ .
Let f (x) = 3x, because 3 ＞ 1, so f (x) is a monotone increasing function on R, and f (x + 2) = 3x + 2, f (1) = 3, so the inequality 3x + 2 ＜ 3 is f (x + 2) ＜ f (1). According to the monotonicity of F (x), there is x + 2 ＜ 1, that is, X ＜ - 1. The solution set of the original inequality is: (- ∞, - 1). So the answer is: (- ∞, - 1)
Known inequality system x + 2 > a X-1
x+2>a,x>a-2,
X-1
If a, B, a = 1, B = - 2, then 1
If there are exactly three integers in the solution set of inequality (2x-1) < ax about X, then the value range of real number a?
When the inequality is transformed, it becomes that (1 / X-2) ^ 2 approaches 4 when (1 / X-2) ^ 2 is infinite, so x takes the negative number (1 / X-2) ^ 2 and the minimum is greater than 4. When x takes the positive integer, (1 / X-2) ^ 2 must be less than 4, so x must be a positive integer and X is a positive integer (1 / X-2) ^ 2 increasing, so the positive integer value of X should be 1,2,3, so parameter a should make x take 3 but not 4, that is (1 / 3-2) ^ 2
If there is exactly one integer in the solution set of inequality (2x-1) ^ 2 < ax ^ 2 about X, then the value range of real number a is?
Obviously, a > 0,
Square difference formula of term shift
The results show that [(2 - √ a) X-1] [(2 + √ a) X-1] < 0
We know that the two inequalities are quadratic (a = - 2) / (a = - 2)
The solution set is 1 / (2 + √ a) ＜ x ＜ 1 / (2 - √ a),
Obviously, 0 ＜ 1 / (2 + √ a) ＜ 1,
It is easy to see from the number axis that the two integers are 1 and 2 respectively,
So that there are exactly two integers in the solution set,
It only needs 2 ＜ smaller root ≤ 3,
That is, 1 / 3 ≤ 2 - √ a < 1 / 2
The results are as follows: 3 / 2 ＜ a ≤ 5 / 3,
That is, a ∈ (9 / 4,25 / 9]
A is greater than or equal to 4
When a < 0, the solution set of the inequality (ax-b) with absolute value ≥ 1 is
The inequality is equivalent to ax-b > = 1 or ax-b
Absolute value of solving inequality AX-1
Sub situation:
1. When a > = 1, because 1-A
|ax-1|
The solution set of known inequality ax ^ 2 + BX + C0 is
Because ax ^ 2 + BX + C0 can be reduced to - 5AX ^ 2 + ax + 6A > 0,
Divide the two sides by the positive number - 5A
x^2-1/5*x-6/5>0 ,
(x + 1) (X-6 / 5) > 0,
Therefore, the solution set of the inequality is {x | X6 / 5}
The solution set is X3 equivalent to
(X-2) (x-3) > 0, that is, A0
-5ax^2+ax+6a>0
5x^2-x-6>0
(5x-6)(x+1)>0
x> 6 / 5 or X