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Given the circle C; x 2 + y 2 = 12 and the straight line 4x + 3Y = 25, what is the probability that the distance between any point a on circle C and line L is less than 2?
The distance between the center of the circle and the known straight line is 5. The radius of the circle is 2 √ 3 ≈ 3. 46 ＜ 5. Therefore, there is a small arc on the circle, and the distance between the point on the arc and the straight line is less than 2
Given the circle C: x2 + y2 = 12, the straight line L: 4x + 3Y = 25, the probability that the distance between any point a on circle C and the line L is less than 2 is () A. 1 Six B. 1 Three C. 1 Two D. 1 Four
The experiment involves taking a point randomly from the circle, corresponding to the arc length of the whole circle, satisfying the condition that the distance from the straight line L is less than 2, making a straight line crossing the line L and a point, ∵ the distance from the center of the circle to the straight line is 255 = 5, ᙽ in this
It is known that the probability that the distance between any point a on the circle x 2 + y 2 = 12, 4x + 3Y = 25 is less than 2
Distance from center of circle to line 4x + 3y-25 = 0
=|0+0-25|/5
=5
find a point with a distance of 3 from the center of the circle on the radius perpendicular to the straight line L to make a vertical line of radius,
According to the chord center distance, radius and chord length, the right triangle formed by chord center distance, radius and chord length is obtained, and the arc length corresponding to the condition is 60 degrees
According to the probability formula of geometric probability, P = 60 ° / 360 ° = 1 / 6
Given the circle C; X? + y? = 12 straight line L; 4x + 3Y = 25, the distance from the center of circle C to the straight line L is? Write down the specific point, and write down the formula. Thank you
Center (0,0)
Linearization is 4x + 3y-25 = 0
Distance d = | 4 * 0 + 3 * 0-25 | / √ (4? + 3?) = 25 / 5 = 5
If the point (m, n) straight line ax + by + C = 0
The formula of distance from a point to a straight line d = | am + BN + C | / √ (a | + B?)
Please ask: "find a point on the square of the parabola y equals - x, so that it has the minimum distance to the line 4x + 3y-8 = 0"
Suppose the coordinates of the points are (x, - x2), and the distance formula can obtain the absolute value / root of D = 4x-3x2-8, A2 + B2 = 3x2-4x + 8 / 5, 3x2-4x + 8 = 3 (x2-4 / 3x) + 8 = 3 (x2-4 / 3x) + 8 = 3 (x2-4 / 3x + 2 / 3 square) + 8 = 3 (X-2 / 3) square + 20 / 3
If the square of x plus the square of Y equals 4, what is the maximum value of 4x + 3Y? Why x = 2cosa, y = 2sina? And 4x + 3Y = 8cosa + 6sina = 10sin (a + θ)? I don't understand all this? Who can give me a detailed account?
x²+y²=4
Let x = 2cosa, y = 2sina
4x+3y
=8cosa+6sina
=10sin(a+θ)
So the maximum value of 4x + 3Y is 10
The square of x plus the square of y plus 4 times of XY minus 13 equals 0. Find 4x + 3Y =?
According to the meaning of the title, x ^ 2 + y ^ 2 + 4xy-13 = 0
The answer is not unique, because we can't find anything related to 4x + 3Y from the above equation,
So we can use a special value instead, for example: when x = 0, y = positive and negative root sign 13
When y = 0, x = 13
When x = 1, y = 2 or - 6
So: 4x + 3Y =? There are countless filling methods!
Given that the square of 4x plus the square of 9y plus 12Y minus 4x equals 5, find 2x minus 3Y
4x²+9y²+12y-4x=5
This is not right. It should be - 5
Divide 5 into 1 + 4
(4x²-4x+1)+(9y²+12y+4)=0
(2x-1)²+(3y+2)²=0
If the square is greater than or equal to 0, the sum is equal to 0
If one is greater than 0, the other is less than 0
So both are equal to zero
So 2x-1 = 0,3y + 2 = 0
2x=1,3y=-2
So 2x-3y = 1 - (- 2) = 3
Find the equation of the circle C: (x + 2) square + (y-6) square = 1 on the symmetric circle of the straight line 3x-4y + 5 = 0
Let C '(x, y) be symmetric with respect to the straight line 3x-4y + 5 = 0, i.e. 3 × (X-2) / 2-4 × (y + 6) / 2 + 5 = 0 (y-6) / (x + 2) × 3 / 4 = - 1. The simultaneous solution of the two formulas leads to x = 4, y = - 2, so c' (4, - 2) therefore, the equation (x-4) square + (y +...) of the circle is obtained
Find the equation of circle whose center is on the straight line 3x + 4y-1 = 0 and crosses the intersection of two circles x2 + y2-x + Y-2 = 0 and X2 + y2 = 5
According to the meaning of the title, let the equation of the circle be (x2 + y2-x + Y-2) + m (x2 + y2-5) = 0, then we can get: (1 + m) x2 + (1 + m) y2-x + y-2-5m = 0, that is, X2 + y2-11 + MX + 11 + MY-2 + 5m1 + M = 0, the coordinates of the circle center are (12 (1 + m), - 12 (1 + m)), and the center of the circle is on the straight line 3x + 4y-1 = 0, ﹤ 3 · 12 (1 + m) -
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