Passing through the circle x ^ 2 + y ^ 2 = R ^ 2 (r > 0), a point P (x0, Y0) outside the circle is taken as two tangent lines of the circle, and the tangent points are m and N respectively. It is proved that the equation of the straight line Mn is x0x + y0y = R ^ 2

Passing through the circle x ^ 2 + y ^ 2 = R ^ 2 (r > 0), a point P (x0, Y0) outside the circle is taken as two tangent lines of the circle, and the tangent points are m and N respectively. It is proved that the equation of the straight line Mn is x0x + y0y = R ^ 2

Suppose the center of circle C, then: CP is perpendicular to Mn straight line, CP slope k: k = Y0 / x0, distance from center of circle C to straight line D, has: D / r = R / CPD = R ^ 2 / √ (x0 ^ 2 + Y0 ^ 2) because Mn straight line slope K1 = - 1 / k = - x0 / Y0: Mn: x0x + y0y-y0b = 0 | 0 + 0-y0b | / √ (x0 ^ 2 + Y0 ^ 2) = D = R ^ 2 / √ (x0 ^ 2 + Y0 ^ 2), so: y0b = R ^ 2, so

If P (x0, Y0) is on the circle x square + y square (r > 0), then the tangent equation with P as the tangent point is x0x + y0y = R square
Thank you very much. There are no friends in the class, only opponents

x*x+y*y=r^2 P(x0,y0)
Replace x with x0 and y with Y0 in the original equation

Given that the equation of circle C is x ^ 2 + y ^ 2 = R ^ 2, and the point m (x0, Y0) is a point outside the circle, then the position relationship between the line x0x + y0y = R ^ 2 and the circle is zero

Because x ^ 2 + y ^ 2 = R ^ 2 is a circle with the origin as the center and R as the radius
M (x0, Y0) is a point outside the circle, with x0 ^ 2 + Y0 ^ 2 > R
The distance from the center of the circle (0,0) to the straight line x0x + y0y-r ^ 2 = 0 is d = - R ^ 2 / √ (x0 ^ 2 + Y0 ^ 2) < R ^ 2
So the position relation of circle and diameter intersects

1. The distance between the moon and the earth is about 3.8 times the octave of 10 m. on a clear night, what is the distance between the moon and the water surface if one can see the image of the moon in a basin 10 cm deep; 2. Standing by the calm lake, you can see the reflection of the scenery on the Bank of the lake in the water. Is the principle of the reflection here the same as that of the shadow in the sun? Why

1. The moon in the basin of water is due to the reflection of moonlight on the water surface, which has nothing to do with the amount of water. So the distance to the water surface is 3.8 times 10 octave M
2. Not the same. The reflection of the lake is formed by the reflection principle of the water surface. The shadow is because the light propagates in a straight line. When it encounters obstacles and cannot pass through, it leaves the shadow of the object underground

The quadratic inequality ax ^ 2-bx + C is known

From the meaning of the problem, we get the solution of the quadratic inequality ax ^ 2-bx + C = 0 with a > 0 and x = - 2 and x = 3
So let's put x = - 2 and x = 3 into the equation and get a = B, C = - 6A
So the inequality BX ^ 2-ax + C

If the point representing the number x on the number axis is to the left of the origin, what is the result of simplifying | x | + √ X & # 178?
A.2x.B.-2x.C.0.D.x.

The point of X is to the left of the origin, X

Given that the tangent of function f (x) = x3-3x2 + ax + B at x = - 1 is parallel to X axis, (1) find the value of a and the monotone interval of function f (x); (2) if the image of function y = f (x) and parabola y = 32x2-15x + 3 have exactly three different intersections, find the value range of B

(1) From the known f ′ (x) = 3x2-6x + A, the tangent at x = - 1 is parallel to the X axis, and the solution is a = - 9. Then f ′ (x) = 3x2-6x-9 = 3 (x + 1) (x-3). From F ′ (x) > 0, the solution is x ＞ 3 or X ＜ - 1. From F ′ (x) ＜ 0, the solution - 1 ＜ x ＜ 3

Are 2 (a + b) & 178; and 3 (a + b) & 178; similar?

2 (a + b) & 178; and 3 (a + b) & 178; are similar terms

Find the range of y = log2 x + log2 (1-x)

y=log2 X+log2 （1-x）
=log2(x(1-x))
also
x*(1-x)x=1/2
=>y

How to draw the function image of F (x) = x + | x | / x?
Can we draw the image of function f (x) = x + 1 + X-1 and function f (x) = x + 1 - X-1

De absolute sign
When x > 0, f (x) = x + | x | / x = x + 1, the image is a part of the line y = x + 1, that is, the part of x > 0
When x < 0, f (x) = x + | x | / x = X-1, the image is a part of the line y = X-1, that is, the part of x < 0
There is no x = 0 in the definition
The absolute sign is still removed
When x > 1, f (x) = x + 1 + X-1 = 2x
When - 1 ≤ x ≤ 1, f (x) = x + 1 + X-1 = 2
When x ＜ - 1, f (x) = x + 1 + X-1 = - 2x
The absolute sign is still removed
When x > 1, f (x) = x + 1 - X-1 = 2
When - 1 ≤ x ≤ 1, f (x) = x + 1 - X-1 = 2x
When x ＜ - 1, f (x) = ｜ x + 1 - ｜ X-1 = - 2, you can draw the figure yourself, which is the content of junior high school