Find the equation of the straight line passing through the point (5, - 5) and tangent to the circle x ^ 2 + y ^ 2 = 25

Find the equation of the straight line passing through the point (5, - 5) and tangent to the circle x ^ 2 + y ^ 2 = 25

Let y + 5 = K (X-5) kx-y-5k-5 = 0 Center (0,0), the distance from radius 5 center to tangent is equal to radius, so | 0-0-5k-5 | / √ (k ^ 2 + 1) = 55 | K + 1 | = 5 √ (k ^ 2 + 1) (K + 1) ^ 2 + K ^ 2 + 12K + 1 = 1K = 0, so a tangent
Obviously, the center coordinates are (0,0) and the radius is 5,
By drawing, we can know that there are two straight lines passing through points (5, - 5) and tangent to circle x ^ 2 + y ^ 2 = 25: they are straight lines perpendicular to X axis and Y axis, passing through points (5,0), (0, - 5),
So these two lines are x = 5 and y = - 5
The equation of line passing through point (- 3,4) and tangent to circle (x-1) 2 + (Y-1) 2 = 25 is______ .
From the equation of the circle, find out that the coordinates of the center of the circle are (1,1), radius r = 5, so the distance from the point (- 3,4) to the center of the circle d = (1 + 3) 2 + (4 − 1) 2 = 5 = R, then the point (- 3,4) is on the circle, so the slope of the straight line passing through the radius of the point is 4 − 1 − 3 − 1 = - 34, so the slope of the tangent equation is 43, and then passing through (- 3,4), then
The equation of straight line passing through point (0,6) and tangent to circle (x-1) &# 178; + (Y-1) &# 178; = 1
Because the straight line passes through the point (0,6), let the linear equation be y = KX + 6, because the center of the circle is (1,1) and the radius is 1, so the distance from the center of the circle to the straight line is 1. The formula of the distance from the point to the straight line can be listed to get k = 12 / 5, which is brought into the linear equation, 12x-5y + 30 = 0
And don't forget the other one is x = 0
Example 3. Let a > 0, b > 0, solve the inequality about X: | AX-2 | ≥ BX
The original inequality | AX-2 | ≥ BX can be reduced to AX-2 ≥ BX or AX-2 ≤ - BX, (1) for the inequality AX-2 ≤ - BX, that is, (a + b) x ≤ 2 & nbsp; & nbsp; Because a > 0, b > 0, that is: X ≤ 2A + B. (2) for inequality AX-2 ≥ BX, that is, (a-b) x ≥ 2. (1) when a > b > 0, we get x ≥ 2A − B from ①, at this time, the original inequality solution is: X ≥ 2A − B or X ≤ 2A + B; when a = b > 0, we get x ∈ϕ from ①, at this time, the original inequality solution is: X ≤ 2A + B; when 0 < a < B, we get x ≤ 2A − B from ①, at this time, the original inequality solution is: X ≤ 2A + B In conclusion, when a > b > 0, the original inequality solution set is (− ∞, 2A + b] ∪ [2A − B, + ∞), when 0 < a ≤ B, the original inequality solution set is (− ∞, 2A + b]
The absolute value inequality f (x) = ax ^ 2 + BX + C, (a, B, C ∈ R), when x ∈ [- 1,1], there is always | f (x) | ≤ 1, the proof of | B | ≤ 1
Like the title,
From the meaning of the title,
|f(1)|=|a+b+c|=
Draw a picture
Given the set a = {a | equation x2-ax + 1 = 0 with real roots}, B = {a | inequality ax2-x + 1 ＞ 0 holds for all x ∈ r}, find a ∩ B
If the equation in the set a has real roots, we can get △≥ 0, that is, A2-4 ≥ 0, and (a + 2) (A-2) ≥ 0, then the solution of a + 2 ≥ 0A − 2 ≥ 0 or a + 2 ≤ 0A − 2 ≤ 0 can get a ≥ 2 or a ≤ - 2; the inequality ax2-x + 1 > 0 in the set B holds for all x ∈ R, according to the image and properties of quadratic function, we can get a > 0, △ = 1-4a < 0, and the solution is a > 14. So a ∩ B = {a | a ≥ 2}
The known proposition p: the equation x ^ 2 + ax + a = 0 of X has no real root; the solution of the inequality x + | x-2a | > 1 of X is r, if Q or P is true, Q and P is false,
Find the value range of real number a
The true solution of P proposition is: Δ 1-x or x-2a
I suggest you ask your teacher, or you can see me, but I don't think it's reliable!
Given the set a = {a | equation x2-ax + 1 = 0 with real roots}, B = {a | inequality ax2-x + 1 ＞ 0 holds for all x ∈ r}, find a ∩ B
If the equation in the set a has real roots, we can get △≥ 0, that is, A2-4 ≥ 0, and (a + 2) (A-2) ≥ 0, then the solution of a + 2 ≥ 0A − 2 ≥ 0 or a + 2 ≤ 0A − 2 ≤ 0 can get a ≥ 2 or a ≤ - 2; the inequality ax2-x + 1 > 0 in the set B holds for all x ∈ R, according to the image and properties of quadratic function, we can get a > 0, △ = 1-4a < 0, and the solution is a > 14. So a ∩ B = {a | a ≥ 2}
The equations ax, B are known to be real numbers
The solution set from inequality system is - 23
The solution set of square + BX + C > 0 of quadratic inequality ax is all real numbers if ()
Why do we need △ = B & sup2; - 4ac
ax²+bx+c>0
Conditions:
A>0
△=b²-4ac