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The position relation of two circles C1: x2 + Y2 + 2x-6y-26 = 0 C2: x2 + y2-4x + 2Y = 0 I'm sorry the title is like this. I just missed the position relationship between two circles C1: x2 + Y2 + 2x-6y-26 = 0 C2: x2 + y2-4x + 2y-4 = 0
C1:(X+1)^2+(Y-3)^2=36
C2:(X-2)^2+(Y+1)^2=5
The center distance is 5
The sum of radii is 6 + radical 5
So it intersects
What is the positional relationship between the two circles x2 + y2-6y = 0 and X2 + y2-8y + 12 = 0?
The center (0,3) (0,4) distance d = 1 radius R1 = 3 R2 = 2 D = R1-R2, so it is inscribed
0
The intersection equation of two circles is 2x + y + 3 = 0, the intersection point x = 9 / 5 or - 1 (- 1,5), (9 / 5, - 3 / 5). Let the center of the circle (a, A-4), and the distance from the center to the two points is equal. This method is troublesome. Let me use the circle system to do it in another way. Let x2 + y2-4x-6y + K (x2 + y2-4y-6) = 0, the center of the circle is (2 / K + 1,3 +)
The position relation between the straight line 2x-y-5 = 0 and the circle x ^ 2 + y ^ 2-4x + 2Y + 2 = 0
The standard equation for a circle is:
(x－2)²＋(y＋1)²=3
Then the distance between the center of the circle (2, - 1) and the straight line 2x-y-5 = 0 is d = | 4 + 1-5 | / √ 5 = 0, that is, the straight line passes through the center of the circle
So the line intersects the circle
The position relationship between two circles C1: x ^ 2 + y ^ 2-4x + 2Y + 4 = 0 and C2: x ^ 2 + y ^ 2 + 2x-6y-26 = 0?
(x-2)²+(y+1)²=1
(x+1)²+(y-3)²=36
The center of the circle is (2, - 1), (- 1,3)
So the center distance D is 5
r1=1
r2=6
So d = r2-r1
So it's introversion
X ^ 2 + y ^ 2-4x-6y + 9 = 0 and x ^ 2 + y ^ 2 + 12x + 6y-19 = 0 judge the position relationship between the two circles
analysis
Circle (1) x? + y? - 4x-6y + 9 = 0
(x-2)²-4+(y-3)²-9+9=0
(x-2)²+(y-3)²=4
Circle (2) x 2 + y 2 + 12x + 6y-19 = 0
(x+6)²-36+(y+3)²-9-19=0
(x+6)²+(y+3)²=64
The center distance of two circles O1O2 = (- 6 - 3) - (23)
=(-8 -6)
|O1O2|=10
Radius of circle (1) 2
Radius of circle (2) 8
Center distance 10
So exotomy
Determine the position relationship of the following pairs of straight lines. If intersect, find the coordinates of the intersection point (1) 2x-3y = 7 4x + 2Y = 1 (2) 2x-6y + 4 = 0, y = x / 3 + 2 / 3 (Radix 2-1) x + y = 3 x + (Radix 2 + 1) y = 2
By solving the equations 2x-3y = 7 4x + 2Y = 1, we can get x = 17 / 16, y = - 13 / 8, so this pair of straight lines intersect, intersect at (17 / 16, - 13 / 8) and solve the equations 2x-6y + 4 = 0, y = x / 3 + 2 / 3, so the equations are the same line, that is, coincidence
The position relation between circle x2 + Y2 + 2x + 6y + 9 = 0 and circle x2 + y2-6x + 2Y + 1 = 0 is () A. Intersection B. Exotomy C. Separation D. Endonuclease
In this paper, the circle x2 + Y2 + 2x + 6y + 9 = 0 and the circle x2 + y2-6x + 2Y + 1 = 0 are transformed into standard equations, and the results are as follows: (x + 1) 2 + (y + 3) 2 = 1, (x-3) 2 + (y + 1) 2 = 9, so the coordinates of the center of the circle are (- 1, - 3) and (3, - 1), the radii are r = 1 and R = 3, respectively
Given that the equations of two circles C1 and C2 are C1: x2 + Y2 + 4x-6y + 5 = 0 and C2: x2 + y2-6x + 4y-5 = 0, how many common tangents of C1 and C2? Rt,3Q
This key depends on the position relationship between the two circles
C1 (x + 2) 2 + (Y-3) 2 = 8, center C1 (- 2,3), radius R1 = 2 √ 2
C2 (x-3) 2 + (y + 2) 2 = 18, center C1 (3, - 2), radius R2 = 3 √ 2
Circle center distance C1C2 = 5 √ 2 = R1 + R2
Two circles circumscribed,
So there are three common tangents
Let C1: x ^ 2 + y ^ 12-4x-6y + 9 = 0 and C2: x ^ 2 + y ^ 2 + 12x + 6y + k = 0 to find the value of K RT
(x-2)^2+(y-3)^2=4
(x+6)^2+(y+3)^2=45-k
If it is circumscribed, the center distance is equal to the sum of the radii
The center of the circle is (2,3), (- 6, - 3)
So the center distance = √ [(2 + 6) ^ 2 + (3 + 3) ^ 2] = 10
Radius sum = √ 4 + √ (45-k) = 10
√(45-k)=8
45-k=64
k=-19
RELATED INFORMATIONS
- 1. C 1: x 2 + y 2-2x-6y-6 = 0, and the number of common tangents of circle C2: x2 + y2-4x + 2Y + 4 = 0
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- 3. The common chord length of circle x2 + y2-4 = 0 and circle x2 + y2-4x + 4y-12 = 0 is () A. Two B. Three C. 2 Two D. 3 Two
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