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It is known that the radii R1 and R2 of the two circles are two of the equations x2-7x + 10 = 0, and the distance between the centers of the two circles is 7, then the position relationship between the two circles is______ .
∵x2-7x+10=0，
∴（x-2）（x-5）=0，
∴x1=2，x2=5，
That is, the radii R1 and R2 of the two circles are 2 and 5 respectively,
∵ 2 + 5 = 7, and the center distance between two circles is 7,
The relationship between the two circumscribed positions
So the answer is: circumcision
Given that the line L is perpendicular to the line 3x + 4y-9 = 0, and the distance between point a (2,3) and line L is 1, the equation of line L is solved Specific process
Because it is vertical, let l equation be 4x-3y + C = 0
Using the formula of distance from point to line:
Absolute value (4 * 2-3 * 3 + C) / root (square of 4 + square of 3) = 1
C = - 4 or 6
So the equation of L is: 4x-3y + 6 = 0 or 4x-3y-4 = 0
The minimum distance between the point on the circle x2 + y2 = 1 and the straight line 3x + 4y-25 = 0 is______ .
∵ the distance from the center of the circle (0, 0) to the straight line 3x + 4y-25 = 0 d = 25
5＝5
The minimum distance from the point on the circle x2 + y2 = 1 to the line 3x + 4y-25 = 0 is AC = 5-r = 5-1 = 4
So the answer is: 4
The equation of the line with distance 2 to the line 3x-4y-1 = 0 is () A. 3x-4y-11=0 B. 3x-4y-11 = 0 or 3x-4y + 9 = 0 C. 3x-4y+9=0 D. 3x-4y + 11 = 0 or 3x-4y-9 = 0
Let the distance to the straight line 3x-4y-1 = 0 be 2, the equation of the line is 3x-4y + C = 0, which is obtained from the distance formula between two parallel lines
|c+1|
5 = 2, C = - 11, or C = 9. ν the equation of a line with a distance of 2 from the line 3x-4y-1 = 0 is 3x-4y-11 = 0,
Or 3x-4y + 9 = 0,
Therefore, B
Given that the line L is perpendicular to the line 3x + 4y-9 = 0, and the distance between point a (2,3) and line L is 1, the equation of line L is solved How to solve problems
Let the slope of the straight line l be K, because the line L is perpendicular to the line 3x + 4y-9 = 0, then k * (- 3 / 4) = - 1 can get k = 4 / 3, and the equation of line L is y = 4 / 3 * x + m, and m is any number. According to the calculation formula of the distance from point to line, 1 = (8-9 + 3M) / 5, M = 2. Therefore, the equation of L is: 4x-3y + 6 = 0
The minimum distance between the point on the circle x2 + y2 = 1 and the straight line 3x + 4y-25 = 0 is______ .
∵ the distance from the center of the circle (0, 0) to the straight line 3x + 4y-25 = 0 d = 25
5＝5
The minimum distance from the point on the circle x2 + y2 = 1 to the line 3x + 4y-25 = 0 is AC = 5-r = 5-1 = 4
So the answer is: 4
Find the equation of the circle whose center is on the straight line x + y = 0 and passes through the intersection point of two circles x? 2 + y? 2 + 10y-24 = 0, x? 2 + y? 2 + 2x + 2y-8 = 0
I don't know if you have heard of "circle family". Given two circle equations, all circles passing through the intersection point of two circles can be expressed in a very simple form. For example, the equations of all circles passing through the intersection points of the two circles can be expressed as (x? + y? + 10y-24) + k * (x? + y? + 2x + 2y-8) = 0
Find the equation of the circle whose center is on the straight line x + y = 0 and passes through the intersection point of two circles x? + y? + 10y-24 = 0, x? + y? + 2x + 2y-8 = 0
Through the intersection of two circles, we can get the following results:
x²+y²+10y-24=x²+y²+2x+2y-8
That is 2x-8y + 16 = 0
Comprehensive, get
x+y=0
{
2x-8y+16=0
The equation of the circle whose center is on the straight line 2x + 4Y = 1 is obtained through the intersection point of two circles x ^ 2 + y ^ 2-4x + 2Y = 0 and x ^ 2 + y ^ 2-2y-4 = 0
The center of a circle is [2 / (1 + λ), (λ - 1) / (1 + λ)] at 2x + 4Y = 14 / (1 + λ) + 4 (λ - 1) / (1 + λ) = 14 λ = 1 + λ = 1 / 3 Center (3 / 2, - 1 / 2)
Given the parallel lines 3x + 2y-6 = 0 and 6x + 4y-3 = 0, find the locus of the point whose distance is equal to the two parallel lines Please write down the problem-solving process, thank you
f1:3X+2Y-6=0
f2:6X+4Y-3=0
The result must be a straight line with a slope of - 3 / 2
Let the equation be F3: y = - 3 / 2x + a
Because F3 is the same distance from F1, F2,
So a = (- 6 + (- 3)) / 2 = - 9 / 2
f3:y=-3/2x-9/2
Both 3x + 2Y + 9 = 0
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