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The position relation between circle x ^ 2 + y ^ 2-4x + 2Y + 4 = 0 and circle x ^ 2 + y ^ 2 + 2x-6y-26 = 0
(x-2)^2+(y+1)^2=1
Circle center (2, - 1), R1 = 1
(x+1)^2+(y-3)^2=36
Center (- 1,3), R2 = 6
Center distance d = √ [(2 + 1) 2 + (- 1-3) 2] = 5
So d = r2-r1
So it's introversion
Determine the position relationship between the two circles x ^ 2 + y ^ 2 + 2x-6y-26 = 0 and x ^ 2 + y ^ 2-4x + 2y-4 = 0 (x + 2) ^ 2 + (Y-2) ^ 2 = 13 and (x-4) ^ 2 + (y + 2) ^ 2 = 13 X ^ 2 + y ^ 2 = 9 and (X-2) ^ 2 + y ^ 2 = 1 It's a question of two lines,
The first is circumcision, and the second is introversion
Determine the position relationship between the following two circles x ^ 2 + y ^ 2-4x-6y + 9 = 0 x ^ 2 + y ^ 2 + 2x-2y-2 = 0
formula
(x-2)²+(y-3)²=4
C1(2,3),r1=2
(x+1)²+(y-1)²=4
C2(-1,1),r2=2
Then the center distance | C1C2 | d = √ (3 | 2 |) = √ 13
So | R1-R2|
The position relation between circle a: x2 + Y2 + 4x + 2Y + 1 = 0 and circle B: x2 + y2-2x-6y + 1 = 0 is () A. Intersection B. Separation C. Tangency D. Contains
In this paper, the circle x2 + Y2 + 4x + 2Y + 1 = 0 and X2 + y2-2x-6y + 1 = 0 are transformed into standard equations, and the results are as follows: (x + 2) 2 + (y + 1) 2 = 4, (x-1) 2 + (Y-3) 2 = 9, so the coordinates of the center of the circle are (- 2, - 1) and (1,3), the radii are r = 2 and R = 3 respectively, ∵ the distance between the centers d = (1 + 2) 2 + (3 + 1) 2 = 5, R + r = 5, then
Find the equation of the circle passing through the intersection of two circles x ^ 2 + y ^ 2 + 6x-4 = 0 and x ^ 2 + y ^ 2 + 6y-28 = 0, and the center of the circle is on the straight line X-Y-4 = 0 Plagiarized words do not even review the title?
By solving the equations of the two circles, the intersection points are (- 1,3) and (- 6, - 2). The specific process is as follows
So we can get two solutions by substituting y = 2 or - 4
Let the center of the circle be (a, b)
The equation is as follows (- 1-A) (- 1-A) + (3-B) (3-B) = (- 6-A) (- 6-A) + (- 2-B) (- 2-B)
a-b-4=0
A = 1 / 2, B = - 7 / 2, radius 89 / 2
The equation (x + 2 / 2) is (x + 2 / 2) + 2 / 2
Through the intersection of two circles x ^ 2 + y ^ 2 + 6x-4 = 0 and x ^ 2 + y ^ 2 + 6y-28 = 0, and the equation of the center of the circle on the straight line X-Y-4 = 0 is
I don't know if you have learned it
Circular system equation:
Circle C1: x 2 + y 2 + D1X + e1y + F1 = 0
Circle C2: x 2 + y 2 + d2x + e2y + F2 = 0
If two circles intersect, the equation of circle system passing through the intersection point is:
x²+y²+D1x+E1y+F1+λ(x²+y²+D2x+E2y+F2)=0
Where λ is the parameter,
When λ = - 1, it is the linear equation of the common chord of two circles
Let the two circles pass through x 2 + y 2 + 6 x-4 = 0 and x 2 + y 2 + 6y-28 = 0
The equation of the circle at the intersection point is x? + y? + 6x-4 + λ (x? + y? + 6y-28) = 0
That is (1 + λ) x 2 + (1 + λ) y 2 + 6x + 6 λ y-4-28 λ = 0
The coordinates of the center of the circle are (- 3 / (1 + λ), - 3 λ / (1 + λ))
∵ the center of the circle is on the line X-Y-4 = 0
If there is 3 / (1 + λ) - 3 λ (1 + λ) + 4 = 0, the solution is λ = - 7
the equation of the circle is x y 6x-4-7 (x y 6y-28) = 0
That is, x 2 + y 2 - x + 7y-32 = 0
The equation of the circle whose center is on the straight line X-Y-4 = 0 and x ^ 2 + y ^ 2 + 6y-28 = 0 and x ^ 2 + y ^ 2 + 6y-28 = 0 are obtained
(x-1/2)(x-1/2)+(y+7/2)(y+7/2)=89/2
By solving the equations of the two circles, the intersection points are (- 1,3) and (- 6, - 2). The specific process is as follows
So we can get two solutions by substituting y = 2 or - 4
Let the center of the circle be (a, b)
The equation is as follows (- 1-A) (- 1-A) + (3-B) (3-B) = (- 6-A) (- 6-A) + (- 2-B) (- 2-B)
a-b-4=0
A = 1 / 2, B = - 7 / 2, radius 89 / 2
The equation (x + 2 / 2) is (x + 2 / 2) + 2 / 2
The equation of two circles x + y + 6x-5 = 0 and X + y + 6y-7 = 0 and the center of the circle is on the straight line X-Y = 4
First of all, your question may be wrong
In this paper, we have solved the equation of two circles x ^ 2 + y ^ 2 + 6x-5 = 0 and x ^ 2 + y ^ 2 + 6y-7 = 0, and the circle center is on the straight line X-Y = 4
Then start to solve the problem
This is a circle system problem
There are formulas in the problem of circular system
(X^2+y^2+6x-5)+ λ(x^2+y^2+6y-7)=0 ………… J remember the formula
It is concluded that: (1 + λ) x ^ 2 + (1 + λ) y ^ 2 + 6x + 6 Λ y-5-7 λ = 0
The center of the circle is: (- 3 / (1 + λ), - 3 λ / (1 + λ))
Substituting the center of a circle into a straight line, the result is: [- 3 / (1 + λ)] - [- 3 λ / (1 + λ)] = 4
The solution is: λ = - 7
Substituting λ = - 7 into (1 + λ) x ^ 2 + (1 + λ) y ^ 2 + 6x + 6 Λ y-5-7 λ = 0
The equation for the required circle is:
3X^2+3Y^2-3X+21Y-22=0
(PS: I'm afraid I'll make mistakes if I do it myself. I'll do it again.)
Find the equation of the circle whose center is on the straight line X-Y-4 = 0 and passes through the intersection of two circles x ^ 2 + y ^ 2 + 6x-4 = 0 and x ^ 2 + y ^ 2 + 6y-28 = 0
According to the knowledge of circle system, the equations of all circles (except one case) passing through the intersection point of two circles can be set as follows:
X^2+Y^2+6X-4+λ(X^2+Y^2+6Y-28)=0
The coordinates of the center of the circle: (- 3 / 1 + λ, - 3 λ / 1 + λ) are brought into the solution of the linear equation and λ = - 7
Then the equation of the circle is: x ^ 2 + y ^ 2-x + 7y-192 = 0
Find the equation of the circle passing through the intersection of two circles x 2 + y 2 + 6x-4 = 0 and X? 2 + y? 2 + 6y-28 = 0, and the center of the circle is on the straight line X-Y-4 = 0 Don't use the answer of Baidu before ah, I read, is not understand, so to ask again
This problem is the most convenient combination of number and shape. First, reasoning: ∵ the circle to cross the intersection of two circles (there are two points) ᙽ the distance between the center of the circle and the two points must be the same ᙽ the center of the circle should draw a sketch on the vertical bisector of the two points. Obviously, the line connecting the centers of the two circles is the vertical bisector of the above two points
RELATED INFORMATIONS
- 1. Given the circle C1: x square + y square + 2x + 6y + 9 = 0 and circle C2: x square + y square - 6x + 2Y + 1 = 0, find the common tangent equation of circle C1 and circle C2
- 2. The linear equation of chord length 8 passing through point a (- 1,10) and cut by circle x2 + y2-4x-2y-20 = 0 is______ .
- 3. If two circles x2 + y2-10x-10y = 0 and X2 + y2-6x + 2y-40 = 0 intersect at two points, the equation of the line where their common chord is located is______ .
- 4. Find a point in the ellipse C: x ^ 2 + 4Y ^ 2 = 16. Find the equation of the line where the chord is located with a point a (1, - 1) in the ellipse x * 2 + 4Y * 2 = 16 as the midpoint
- 5. If the chord ab of the circle x ^ 2, y ^ 2 = 20 is made through the point P (2, - 3), and P bisects AB, then the equation of the line where the chord AB is located is
- 6. The straight line kx-y + 1 = 0 and the circle x? 2 + y? 2 = 1 intersect a and B, and find the locus of the midpoint of chord ab
- 7. Find the equation of the circle whose center is on the line 3x-y = 0, tangent to the x-axis, and cut by the line X-Y = 0
- 8. If x ^ 2 + y ^ 2-2y = 0, the midpoint of chord AB is (- 1 / 2,3 / 2), then ab=
- 9. It is known that the circle C passes through point 10 and the center of the circle is on the positive half axis of the X axis. The straight line x + Y-1 = 0 is cut by the circle C, and the chord length is 2, and the circle standard is found equation
- 10. Given that the circle C: x 2 + y 2 = 5, the straight line L: ax-y-2a = 0 and circle C intersect at two points AB, the trajectory equation of the midpoint of chord AB is obtained
- 11. The diameter of a circle is 5 meters. What is its circumference? The ratio of circumference is 3
- 12. The circumference of a circle is the same as that of a square. The side length of the square is 6.28 cm. What is the area of the circle?
- 13. Draw the largest circle in a square with a circumference of 32 decimeters. What is the area of the circle?
- 14. The circumference of a circle is fixed, and its ratio is inversely proportional to its diameter. Right or wrong? The judgment of midterm exam, I hit X
- 15. PI 1 to 100
- 16. If the perimeter of an equilateral triangle is equal to that of a circle, then their area ratio is equal to
- 17. An equilateral triangle has a circumference of 2.4 decimeters and a height of 0.6 decimeters. Its area is () decimeters
- 18. It is known that the radius of circumscribed circle of regular hexagon is 2, and the area of regular hexagon is?
- 19. Formula for calculating the area of regular hexagon?
- 20. As shown in the figure, ∠ AOC: ∠ cod: ∠ BOD = 2:3:4, and a.o.b is on a straight line, OE.OF Divide ∠ AOC and ∠ BOD equally, og bisect ∠ EOF, and calculate ∠ GOF