Find the linear equation which passes through point a (3.4) and is tangent to circle C: (X-2) &# 178; + (Y-1) = 1

Find the linear equation which passes through point a (3.4) and is tangent to circle C: (X-2) &# 178; + (Y-1) = 1

The center of the circle is (2,1) and the radius is 1
Obviously, x = 3 is a vertical tangent
Let another oblique tangent be y = K (x-3) + 4
Then the distance from the center of the circle to the straight line is r
That is, 1 = | - K + 4-1 | / √ (1 + K ^ 2)
That is 1 + K ^ 2 = k ^ 2-6k + 9
k=4/3
So all the other lines are: y = 4 / 3 * (x-3) + 4 = 4x / 3
X = 3 or y = x + 1
Let the circle C with the center of the circle on the line x = 3 and the line y = X-1 be tangent to the point a (2,1), and solve the equation of circle C
The straight line perpendicular to y = X-1 is x + y = 3. This straight line must pass through the center of the circle and stand with the straight line x = 3. Then we can get the coordinates of the center of the circle C (3,0) and the radius length is equal to ca. the equation of the circle is (x-3) & sup2; + Y & sup2; = 2
It's so simple. Read more books by yourself
。。。。。。。。 It's really simple
(x－3)²＋y²=2
Find the equation of the circle C: x ^ 2 + y ^ 2-2x-2y-2 = 0 circumscribed at point P (3,1) and tangent to the line L: x + 2Y = 0 at point m (6, - 3)
"It's better to teach people to fish than to teach them to fish"
I'll tell you the main idea, and you can solve it yourself
Main idea: the radius can be obtained by calculating the center of the circle
The center of the circle is on the line L 'passing through the point m and perpendicular to L, and on the middle vertical line of PM, so the center of the circle is the intersection of the two. Just find the intersection
Given that the inequality x ^ 2 + 4mx-4 > 0 holds when x belongs to [- 1,4], find the value range of M
When x = 0 - 4 > 0 does not hold, so this is wrong
On the inequality MX & # 178; - 4mx-2 < 0 for X ∈ R, then the range of M is_____
Discussion by situation
M = 0, - 2 ＜ 0, the constant holds
M > 0 means that the opening of quadratic function on the left side is upward, < 0, and it is impossible to hold
M < 0, the opening of quadratic function is downward, as long as the maximum value of the function is less than 0, or the discriminant is less than 0, it can be true for X
To sum up, there are three cases
It is known that the solution set of the inequality ax to the power of 3-3x + 6 greater than 4 is XB, and the solution of the inequality ax-b is x-C > 0
The solution set of ∵ ax & # 179; - 3x + 6 > 4 is XB,
X = 1 and x = B are the roots of the equation AX & # 179; - 3x + 6 = 4
Substituting x = 1 into the equation AX & # 179; - 3x + 6 = 4
A = 1
Solving X & # 179; - 3x + 6 = 4, we can get x = 1 or x = - 2 [x & # 179; - 3x + 6 = 4, the specific process of solving the equation: X & # 179; - 3x + 2 = 0, X & # 179; - 1-3x + 2 = 0,
(x-1)(x²+x+1)-3(x-1)=0,(x-1)(x²+x-2)=0 (x-1)²(x+2)=0
[x = 1 or x = - 2]
∴b=-2
The inequality ax - (x / b) - C > 0 is equivalent to
x+(x/2)-c>0
∴x>(2/3)c
When a is greater than 0, the quadratic power of the solution to the inequality ax about X - (a + 1) x + 1 is greater than 0
y=ax^2-(a+1)x+1>0
When ax ^ 2 - (a + 1) x + 1 = 0
X = [(A-1) ^ 2 under (a + 1) plus minus radical] / 2
When a is greater than 0 and less than 1
X = 1 / A or x = 1
y> 0
x> 1 / A or x0
X is equal to (a + 1) / 2A
When a > 1
X = 1 / A or x = 1
y> 0
x> 1 or X1 / A or x1
x> 1 or X
ax2--(a+1)x+1>0
（ax-1)(x-1)>0
1. AX-1 > 0 and X-1 > 0
01, x > 1
The solution of a = 1 is x ≠ 1
2. ax-1
If the solution set of inequality ax quadratic - x + C > 0 is (- 1,2 / 3), find the value of AC
That is, the solutions of ax & # 178; - x + C = 0 are - 1 and 2 / 3
So - 1 + 2 / 3 = 1 / A
-1×2/3=c/a
So a = - 3
C=2
Because the solution set of ax quadratic - x + C > 0 is (- 1,2 / 3)
therefore
x=-1,x=2/3
Is the root of the equation: ax quadratic - x + C = 0
therefore
a+1+c=0
4/9a-2/3+c=0
Solution
a=-3
C=2
A0;
-1+2/3=1/a;
a=-3;
-2/3=c/a;
c=2;
ac=-6;
a＜0
On two X1 = - 1, X2 = 2 / 3 of ax ^ 2-x + C = 0
According to Weida's theorem:
x1+x2=1/a=-1/3
x1x2=c/a=-2/3
C=2
a=-3
ac=-6
Solve the following inequality: (1) the power of (half) x2-2x > (half) 10x-32
(2) The x power of 9-12 × 3 + 27 > 0
1. Because 00
x> 2 or X
To solve the inequality: 1 / 3 log to the power of 2 (- 3x-4) > 1 / 3 log to the power of 2 to the power of 10
x<-14