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Given 4x-3y-6z = 0, x + 2y-7z = 0, XYZ ≠ 0, find 2x2 + 3y2 + 6z2 X2 + 5y2 + 7z2
4x-3y-6z = 0 ①, x + 2y-7z = 0 ②, ② × 4, 4x + 8y-28z = 0, ③, ③ - ①, 11y-22z = 0, that is, y = 2Z. By substituting y = 2Z into ②, x + 4z-7z = 0, i.e. x = 3Z, y = 2Z, y = 2Z, x = 3Z, the original formula = 2 (3Z) 2 + 3 (2Z) 2 + 6z2 (3Z) 2 + 5 (2Z) 2 + 7z2 = 36z236z2 = 1
4X + 5Y = 105,2x + 3Y = 178, x, y is equal to
Divide them into Formula 1 and formula 2 respectively. Multiply formula 2 by 2 to get 4x + 6y = 356. Record it as formula 3. Equation 3 - 2 gives y = 251. Bring y = 251 into Formula 1 or formula 2 to get X
Find the equation of the circle whose center is on the straight line x + y = 0 and passes through the intersection of two circles x ^ 2 + y ^ 2-2x + 10y-24 = 0, x ^ 2 + y ^ 2 + 2x + 2y-8 = 0
X²+Y²-2X+10Y-24=0.X²+Y²+2X+2Y-8=0
The former formula minus the latter formula
-4X+8Y-16=0
X=2Y-4
Substituting the former formula (2y-4) 2 - 2 (2y-4) + y 2 + 10y-24 = 0
5Y? - 10Y = 0, y = 0 or y = 2 corresponds to x = - 4 or x = 0 (0,2) (- 4,0)
Center of circle (a, - a)
(X-A)²+(Y+A)²=R²
The formulas are as follows: a 2 + (2 + a) 2 = R 2 and (- 4-A) 2 + a 2 = R 2
That is a 2 + a 2 + 4 A + 4 = a 2 + 8 A + 16 + a 2
4A=-12 A=-3
R²=9+1=10
(X+3)²+(Y-3)²=10
The equation of the circle whose center is on the straight line x + y = 0 is obtained through the intersection of two circles x ^ 2 + y ^ 2-2x + 10y-24 = 0 and x ^ 2 + y ^ 2 + 2x + 2y-8 = 0
C1: x square + y square - 2x + 10y-24 = 0
C2: x square + y square + 2x + 2y-8 = 0
Two points are obtained: x = - 4, y = 0 and x = 0, y = 2
Namely (- 4,0) and (0,2)
Let the center of the circle be (x, - x)
The distance from the center of the circle to two points is equal and the radius is long
（x+4)^2+x^2=x^2+(-x-2)^2
X = - 3
The square of the radius is (x + 4) ^ 2 + x ^ 2 = 10
So the equation of the circle is (x + 3) ^ 2 + (Y-3) ^ 2 = 10
The linear equation of the common chord of circle C1: x ^ 2 + y ^ 2-12x-2y-13 = 0 and circle C2: x ^ 2 + y ^ 2 + 12x + 16y-25 = 0 is
You can solve two equations by subtracting them directly
x^2+y^2-12x-2y-13=0
x^2+y^2+12x+16y-25=0
That is, 24x + 18y-12 = 0
That is 4x + 3y-2 = 0
In this paper, the equation of the circle with the common chord of the circle C1: x2 + y2-12x-2y-13 = 0 and C2: x2 + Y2 + 12x + 16y-25 = 0 is obtained
∵ circle C1: x2 + y2-12x-2y-13 = 0 and C2: x2 + Y2 + 12x + 16y-25 = 0,
In this paper, the equation of the circle with the common chord of the circle C1: x2 + y2-12x-2y-13 = 0 and C2: x2 + Y2 + 12x + 16y-25 = 0 is obtained
∵ circle C1: x2 + y2-12x-2y-13 = 0 and circle C2: x2 + Y2 + 12x + 16y-25 = 0,
The common string equation obtained by subtracting two circles is l: 4x + 3y-2 = 0
And ∵ the coordinates of the center of the circle C1: x2 + y2-12x-2y-13 = 0 are (6, 1), and the radius is 5
2；
The coordinates of the center of circle C2: x2 + Y2 + 12x + 16y-25 = 0 are (- 6, - 8), and the radius is 5
5，
The equation of C1C2 is 3x-4y-14 = 0
Ψ LIANLI
4x+3y−2＝0
3x − 4Y − 14 = 0, the center coordinates of the circle whose common chord is the diameter are (2, - 2),
∵ (6,1) the distance to the common chord is 5
The radius of the circle whose common chord is the diameter is 5
The equation of circle with common chord as diameter is (X-2) 2 + (y + 2) 2 = 25
Circle C1: x ^ 2 + y ^ 2 + 2x + 2y-8 = 0 and circle C2: x ^ 2 + y ^ 2-2x + 10y-24 = 0 intersect at two points a and B (1) Find AB equation of straight line (2) Find the equation of the circle with the smallest area passing through two points a and B
C1: x ^ 2 + y ^ 2 + 2x + 2y-8 = 0 (1) C2: x ^ 2 + y ^ 2-2x + 10y-24 = 0 (2) two equations are combined, (1) - (2) AB: x-2y + 4 = 0 (3) substitute (3) into (1) to get 5Y ^ 2-10y = 0, the intersection point (- 4,0) and (0,2) the minimum circle is the circle of diameter, (x + 2) ^ 2 + (Y-1) ^ 2 = 5
Circle C1: x + y + 2x + 2y-8 = 0 and circle C2: x + y-2x + 10y-24 = 0 intersect at points a, B Find: (1) the linear equation of common chord AB; (2) the length of common chord AB; (3) find the circle equation with the smallest area passing through a and B
(1) Two equations minus (2) to solve the intersection point (3) is the diameter of AB circle
It is known that the centers of two circles C1: x? + y? - 2Y = 0 and C2: x? + (y + 1) 2 = 4 are C1 and C2 respectively, and P is a moving point And the slope product of the straight line PC1 and PC2 is - 1 / 2 (1) the equation for finding the trajectory m of the moving point p; (2) whether there is a line L passing through the point a (2,0) and the trajectory m intersecting at different two points c and D, so that ic1ci = ic1di? If there is, find the equation of line I; if not, please explain the reason
(1) C1 (0,1), C2 (0,1), C1 (0,1), C2 (0, - 1), set P (x, y), according to the meaning of the topic (Y-1) (y + 1) / x ^ = - 1 / 2, ∈ x ^ / 2 + y ^ = 1, X ≠ 0, ① this is the equation of the trajectory m of the moving point P.. (2) let L: x = my + 2, ② put in ① * 2, get m ^ y ^ 4 + 4my + 4 + 2Y ^ = 2, (m ^ + 2) y ^ + 4my + 2 = 0, set C (x1, Y1), D (X2, Y2), Y1 ≠ Y2, then Y1, Y2, then Y1, Y2, then Y1 + Y2, Y2, then Y1 + Y2, y2= - 4m /
RELATED INFORMATIONS
- 1. Given that two circles C1: x? + y? = 4, C2: x? + y? - 2x-4y + 4 = 0, and the straight line L: x + 2Y = 0, find the passing circle C1 and C Find the equation of the circle which passes through the intersection of C1 and C2 and tangent to the line L Please don't copy the answer from others on Baidu. I said that I can't understand why he set the equation of circle like that. Our teacher has never taught us to set it like that
- 2. It is known that the circle ⊙ C1: x2 + Y2 + 2x + 2y-8 = 0 and ⊙ C2: x2 + y2-2x + 10y-24 = 0 intersect at two points a and B Find the equation of the circle with the smallest area passing through two points a and B
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