The equation of a line passing through point P (2,3) and tangent to circle x2 + y2 = 4 is () A. 2x+3y=4B. x=2C. 5x-12y+26=0D. 5x-12y+26=0x=2

The equation of a line passing through point P (2,3) and tangent to circle x2 + y2 = 4 is () A. 2x+3y=4B. x=2C. 5x-12y+26=0D. 5x-12y+26=0x=2

The point P (2,3) is outside the circle x2 + y2 = 4. When the slope of the tangent does not exist, the linear equation is x = 2. When the slope of the straight line exists, the tangent equation is Y-3 = K (X-2). According to the tangent between the line and the circle, the distance from the center of the circle (0,0) to the straight line d = | 3 − 2k| 1 + K2 = 2K = 512
Find the circle equation with the center of the circle on the line y = - 4x and tangent to the line x + Y-1 = 0 at P (3. - 2)
The problem description is the same as the title
Since the center of the circle O is tangent to the line x + Y-1 = 0, OP is perpendicular to the line, and P (3, - 2), then OP sits on the line X-Y = 5, so only the intersection of the two lines is needed as the center of the circle coordinate solution, O (1, - 4) r = OP = 2 √ 2, then the circle equation is (x-1) ^ 2 + (y + 4) ^ 2 = 8
Solving the linear equation of point (1, - 7) and tangent to circle x ^ 2 + y ^ 2 = 25
Let the tangent equation be y + 7 = K (x-1), that is, kx-y-k-7 = 0, the center of circle x ^ 2 + y ^ 2 = 25 is O (0,0), the radius is 5O, and the distance from the tangent is equal to the radius 5 | - k-7 | / √ (K & # 178; + 1) = 5 | (K + 7) &# 178; = 25 (K & # 178; + 1), that is, 12K & # 178; - 7k-12 = 0
Solve inequality: log2x + log (1 / 2) (X-2) > 1.2 and 1 / 2 are base numbers!
x-2>0,x>2
log2x+log（1/2）(x-2)>1
log(2)x+log(2)1/(x-2)>1
log(2)x/(x-2)>1
x/(x-2)>2
x>2x-4
X
log2x+log（1/2）(x-2)>1
lgx/lg2+lg(x-2)/lg1/2>1
lgx/lg2-lg(x-2)/lg2>1
lgx-lg(x-2)>lg2
lgx/x-2>lg2
x/x-2>2
x-2x+4/x-2>0
-x+4/x-2>0
x-4/x-2
The process and answer of solving log with a as the base bracket, the x power of a minus 1 and 0 under the root sign of inequality
If your 1 is outside the log, it will be changed into x > 1. If it is inside, it will be divided into a > L or 01) = > x > log (a 2) when l > a > O, log (a 2)
Let a > 0, a ≠ 1, and the function f (x) = a has the maximum value of LG (x ∧ 2-2x + 3), then the inequality log is based on a（
Let a ≠ 0, a ≠ 1
Then the inequality of Λ (log + 2) = Λ (log + 2) Λ (log + 2) Λ (log + 2) Λ (log + 2) Λ (log + 2) Λ (log + 2) Λ (log + 2) Λ (log + 2) Λ (log + 2) Λ (log + 2)
A ^ t: the power of a to t
x²－2x＋3=(x－1)²＋2
Then:
LG (X & # 178; - 2x + 3) has a minimum LG2
Because the title tells you: A ^ [LG (X & # 178; - 2x + 3)] has the maximum, then:
Zero
The function y = log is based on (X-2) + 8 constant crossing point a and point a is on the x power image of function y = a. the X power-4x-5 ＜ 1 solution set of inequality a is? Speed solution
Idea: use the logarithmic function to pass through the fixed point (1,0) to find out the coordinates of point a,
The power of x-4x-5 ＜ 1 of a is equivalent to a ^ X
The third power of x minus 6x minus 5 is greater than 0
x³-6x+5>0
That is X & sup3; - X & sup2; + X & sup2; - 6x + 5 > 0
That is X & sup2; (x-1) + (x-1) (X-5) > 0
That is, (x-1) (X & sup2; + X-5) > 0
According to the number axis root method, the solution is
x> (- 1 + √ 21) / 2 or (- 1 - √ 21) / 2
x³-6x-5>0
(x³-x)-(5x+5)>0
x(x²-1)-5(x+1)>0
x(x-1)(x+1)-5(x+1)>0
[x(x-1)-5](x+1)>0
(x²-x-5)(x+1)>0
Find the three roots of (X & sup2; - X-5) (x + 1) = 0 as
x1=(1+√21)/2，x2=(1-√21)/2，x3=-1
The solution of the inequality is obtained by the method of number axis root
(1-√21)/2
Solving inequality: X fifth power - 6x fourth power + 8x third power ≥ (5x square + 6x) (x square - 6x + 8)
The third power of X is extracted from the left side, and the factorization is (x-4) (X-2), which can be eliminated with (x-6x + 8) on the right side. The right side (5x square + 6x) is extracted as X (5x + 6), and X and the x cube on the left side are eliminated
It's time to go down
Solve the inequality x (X & # 178; - 5x + 6) (the third power of X + 1) > 0
X (X & # 178; - 5x + 6) (the third power of X + 1)
=(the third power of X + 1) x (X-2) (x-3)
x> 3, or
2> X > 0, or
-1>x