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Given the circle C: x ^ 2 + y ^ 2-2x-4y + 1 = 0 and the straight line L: ax-y + 3 = 0, if the circle C is symmetric about the line L, find the value of A
Because the circle C is symmetric about the line L
So the center of the circle is in a straight line
Because the circle C: x ^ 2 + y ^ 2-2x-4y + 1 = 0
So the center of the circle (1,2)
So we bring in the solution of the linear equation
a=-1
It is known that there are only two points on the circle x ^ 2 + y ^ 2 = 4, and the distance from the straight line 12x-5y + C = 0 is 1
The center of the circle is (0,0), and the radius is 2
When the distance from the center of the circle to the straight line is greater than or equal to 3, there is at most one point on the circumference, and the distance from the straight line is 1
When the distance from the center of the circle to the straight line is less than or equal to 1, there are at least 3 points on the circumference and 4 points at most, and the distance from the straight line is 1
Therefore, the distance from the center of the circle to the straight line d = | C | / √ (12 | + 5 ﹤) = | C | / 13
And 1 < d < 3
1＜|c|/13＜3,13＜|c|＜39
13 ＜ C ＜ 39, or - 39 ＜ C ＜ - 13
It is known that there are only four points on the circle x * x + y * y = 4, and the distance from the straight line 12x-5y + C = 0 is 1. Find the value range of real number C
The problem of the distance between the early lines on a circle is divided into: if the lines are outside the circle, there are at most two. If the lines are tangent to the circle, there are two or three. Therefore, four must intersect the circle. The radius of the circle is 2. The distance is 1. So the distance from the center of the circle (0,0) to the straight line is less than 1
If there are only 4 points on the circle x? +y? =r? (r>0) and the distance from the straight line L:x-y-2=0 is 1, then the real number R is taken What is the value range?
If there are four points on circle O whose distance from L to L is 1, then l intersects with circle O, and there are two points on both sides of L. when the distance from L to L is equal to 1, the line is tangent to the circle, and only two points on the left meet the condition r = √ 2 + 1
In the plane rectangular coordinate system xoy, it is known that there are only three points on the circle x2 + y2 = 4 whose distance from the straight line 12x-5y + C = 0 is 1, then the value of real number C is______ .
From the equation of circle x2 + y2 = 4, the coordinates of the center of the circle are (0, 0), and the radius of the circle is r = 2,
∵ the distance from the center of the circle to the straight line 12x-5y + C = 0 d = 1,
∴d=|c|
122+(−5)2=|c|
13 = 1, that is, | C | = 13,
C = ± 13
So the answer is: ± 13
The equation of the circle is known as x ^ 2 + y ^ 2-6x-8y = 0. If the longest chord and the shortest chord of the circle passing through the point (3,5) are AB and CD respectively, then the sum of slopes of line AB and CD is? Wrong, wrong, it's the longest string over point (2,5)
The equation of circle: (x-3) ^ 2 + (y-4) ^ 2 = 25,
Center (3,4),
The slope of the line where the longest chord AB crosses (2,5) = (5-4) / (2-3) = - 1
Because the line of the longest chord is perpendicular to the line of the shortest chord
The slope of the line passing through (2,5) shortest chord CD is 1
Therefore, the sum of slopes of line AB and CD is - 1 + 1 = 0
Given that the equation of circle is x2 + y2-6x-8y = 0, if the longest chord and the shortest chord passing through the point (2,5) in the circle are AB and CD respectively, then the sum of slopes of line AB and CD is______ .
If the coordinates of the center of the circle x2 + y2-6x-8y = 0 are m (3,4), and the point (2,5) is n, then
The longest chord ab of the circle passing through the point n (2,5) passes through the center of the circle, so the slope is 5 − 4
2−3=-1；
The shortest chord is perpendicular to Mn, so the slope is 1
The sum of slopes of line AB and CD is 0
So the answer is: 0
The slope of the line k and the circle help It is known that the slope of a straight line y = KX + Z (k exists) and intersects a circle (x + a) square + (y + b) square = R square. Is there the simplest way to find the chord length? (only need to find X1 and x2) can we find the chord without finding the y value of the intersection point?
Chord length = | X1 - x2 | √ (1 + k), followed by the square of 1 plus K under the root sign
Given the circle C: x2 + y2-2x + 4y-4 = 0, whether there is a straight line L with slope of 1, so that the chord length AB cut by circle C is the circle with diameter passing through the origin. If there is an equation L for finding the straight line, if there is no equation, explain the reason
The standard equation of circle C is (x-1) 2 + (y + 2) 2 = 9. Suppose that there is a circle m with ab as its diameter, the coordinates of its center m are (a, b)
∵ cm ⊥ L, that is, KCM · KL = B + 2
a−1×1=-1
∴b=-a-1
The equation of the line L is y-b = x-a, that is, x-y-2a-1 = 0
∴|CM|2=（|1+2−2a−1|
2）2=2（1-a）2
∴|MB|2=|CB|2-|CM|2=-2a2+4a+7
∵|MB|=|OM|
ν - 2A2 + 4A + 7 = A2 + B2, a = - 1 or 3
2，
When a = 3
2, B = - 5
The equation of the line L is X-Y-4 = 0
When a = - 1, B = 0, the equation of the line L is X-Y + 1 = 0
Therefore, such a straight line L exists, and the equation is X-Y-4 = 0 or X-Y + 1 = 0
In the plane rectangular coordinate system xoy, we know that C: x ^ 2 / 3 + y ^ 2 = 1, the slope is K (k > 0), and the line L intersects the ellipse C at two points a and B, the midpoint of line AB is e, the ray OE intersects the ellipse C at point G, and the intersection line x = - 3 at point d (- 3, m) (1) find the minimum value of m ^ 2 + K ^ 2; （2） If OG^2=OD*OE, verify: straight line L passes through the fixed point;
(1) Let l be: y = KX + B (B ≠ 0) then: x ^ 2 + 3Y ^ 2 = 3, that is: x ^ 2 + 3 (KX + b) ^ 2 = 3, so there are: XA + XB = - 6KB / (1 + 3K ^ 2), ya + Yb = 2B / (1 + 3K ^ 2), Ya + Yb = 2B / (1 + 3K ^ 2) ray OE, intersection ellipse C at point G, intersection line x = - 3 at point d (- 3, m), we can get: (Ya + Yb) / (XA + XB) = m / (- 3) get: km = 1m ^ 2 + K ^ 2 ≥ 2km = 2km = 1m ^ 2 + K ^ 2 ≥ 2km = 2km = 2km = 1m ^ 2 + K ^ 2 ≥ 2km = 2km = 1km = 1m ^ 2 + K 2
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