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Given that the straight line passes through the point m (2,1), and the chord length cut by the circle C: (x-1) 2 + (y + 1) 2 = 4, the chord length is 2 √ 2, and the equation of the straight line is obtained
The solution (x-1) 2 + (y + 1) 2 = 4C (1, - 1), r = 2 because the chord length = 2 √ 2, the distance from the center of the circle to the straight line d = √ 2 is obtained. Let the straight line L: Y-1 = K (X-2) kx-y + 1-2k = 0. The formula of the distance from point to line is (K + 1) ^ 2 = 1, and the solution is k = - 2 or K = 0, so l: y = 1 or 2x + Y-5 = 0. When K does not exist, l: x =
Given the ellipse x? 2 / 36 + X? 2 / 9 = 1, find the line equation of the chord with point P (4,2) as the midpoint? Given the ellipse x? 2 / 36 + y? 2 / 9 = 1, find the linear equation of chord with point P (4,2) as the midpoint?
Substituting the point P into the elliptic equation, we get
16/36+4/9
=16/36+16/36
=32/36
Elective course 1-1] given the ellipse x 2 / 36 + y 2 / 9 = 1, find the linear equation of the chord with point P (4,2) as the midpoint
Let the endpoint of a chord be a (x1, Y1) and the elliptic equation of B (X2, Y2) be x? 2 / 36 + y? 2 / 9 = 1, that is, x? + 4Y? = 36 X1 + x2 = 8, Y1 + y2 = 4A, B are all on the ellipse x1 ﹣ 1; 4Y1 36 ① x2 + 4y2 36 ② ① - ② (x1 -...)
Given the ellipse x ^ 2 / 36 + y ^ 2 / 9 = 1, the midpoint of chord AB is m [3.1] to find the linear equation of chord ab
Let a (x1, Y1), B (X2, Y2), if a (x1, Y1), B (X2, Y2), substitute the elliptic equation to get X1 ^ 2 / 36 + Y11 ^ 2 / 9 = 1 (1) x2 ^ 2 / 36 + Y2 ^ 2 / 9 = 1 (2) two formula subtracti, get (x2 + x1) (x2-x1) (x2-x1) / 36 + (Y2 + Y1) (y2-y1) (y2-y1) / 9 = 0, because X1 + x2 = 6, Y1 + y2 = 2, substitute the above formula, get (x2-x1) / 6 + 6 + 2 (y2-y1) / 9 = 0, the solution is (y2-y1) / / / / (y2-y1) / (y2-y1) / (y2-y1-y1) / (y2-y1x2 -
It is known that the midpoint of the two-point chord AB is (1,1) 1. Find the equation 2 of the straight line and the length of ab As the title
This problem is characterized by a number of (1) and (1) the number of (1) is (1) 1 (1) and (1) the number of (1) is (1) 1 (1) and (1) the number of (1) is (1) 1 (1) and (1) the number of (1) is (1) 1 (1) and (1) the (1) is (1) 1 (1) and (2) the (1) is (1) 1 (1) and (2) the (1) is (1) 1 (1) and (2) the (1) is: (1) the (1) is: (1) the (1) is: (1) the (1) is: (1) the (1) is (1) is (1) 1 (1) is (1) 1 (1) 1 (1) 1 (1) 1) Y2 + Y2
Given that the ellipse x ^ 2 / 36 + y ^ 2 / 9 = 1 and the midpoint m (3,1) of chord AB, AB equation is solved
If AB is parallel to the y-axis, then x = 3
So let AB: y = K (x-3) + 1
xx/36+(kx-3k+1)^2/9=1
(4kk+1)xx+4(-6kk+2k)x+4(3k-1)^2-36=0
3=(x1+x2)/2
6=x1+x2=4(6kk-2k)/(4kk+1)
-8k=6
k=-4/3
The chord length of the circle x2 + y2-4x + 4Y + 4 = 0 is equal to x-y-5 = 0
To solve this problem, we must first know the distance formula between point and line. The formula of distance from point P (x0, Y0) to the line ax + by + C = 0 is: D = [absolute value of ax0 + by0 + C] / under the root sign (a ^ 2 + B ^ 2), it will be simple, the formula of circle will be (X-2) ^ 2 + (y + 2) ^ 2 = 4, know that the center of the circle is (2, - 2), the radius is 2, use the point to
Find the length of the common chord of the circle x? + y? = 4 and the circle x? + y? - 4x + 4y-12 = 0
Two circle equations are subtracted,
We get 4x-4y + 8 = 0, that is, X-Y + 2 = 0, which is the equation of the line where the common chord is located, which is recorded as L
The equation of the previous circle can be reduced to x ^ 2 + y ^ 2 = 4,
The distance between L and the center of the circle is 2 under the root sign
Therefore, if the common chord length is 2 *, then (4-2) = 2
The chord length of circle x2 + y2-4x + 4Y + 6 = 0 is equal to () A. Six B. 5 Two Two C. 1 D. 5
Given that the circle x2 + y2-4x + 4Y + 6 = 0, the center of the circle is (2, - 2), and the radius is
2．
The center of the circle is (2, - 2) to the straight line x-y-5 = 0
Two
2．
Using geometric properties, the chord length is 2
(
2)2−(
Two
2)2=
6．
Therefore, a
If any point on the circle x 2 + y 2 + 2x-4y + 1 = 0 is still on the circle with respect to the line 2aX by + 2 = 0 (a, B ∈ R +), then the minimum value of 1 / A + 2 / B is 4 √ 2.2 √ 2.3 + 2 √ 2.3 + 4 √ 2
C
RELATED INFORMATIONS
- 1. If the circle x2 + Y2 + 2x-4y + 1 = 0 is symmetric about the straight line 2aX by + 2 = 0 (a, B ∈ R), then the value range of AB is () A. (−∞，1 4] B. [1 4，+∞) C. (−1 4，0) D. (0，1 4)
- 2. If the common chord length of circle x2 + y2 = 4 and circle x2 + Y2 + 2ay-6 = 0 (a > 0) is 2 3, then a is equal to () A. 1 B. Two C. Three D. 2
- 3. It is known that the length of the two chords of a circle AB CD is two of the equation x ^ 2-42x + 432 = 0, and ab is parallel to CD, and the distance between the two chords is 3. Find the radius
- 4. As shown in the figure, AB and CD are the two parallel chords in the circle O, M is the midpoint of AB, and the extension line of DM intersects the circle O at e. it is proved that O, m, e and C are in the same circle
- 5. As shown in the figure, the radius of ⊙ o is 1cm, and the lengths of strings AB and CD are respectively 2 cm, 1 cm, then the acute angle α between the strings AC and BD=______ Degree
- 6. It is known that AC is the diameter of circle 0, PA is vertical AC, connecting OP, xuancb is parallel OP, the straight line Pb intersects the straight line AC at point D, BD = 2PA
- 7. P is a point outside the circle O, Pb is the tangent line of circle O, a and B are the tangent points, BC is the diameter of circle O, so AC is parallel to Po As the title
- 8. Given that the circle C (x + radical 3) ^ 2 + y ^ 2 = 16, point a (radical 3,0) q is a moving point on the circle, the vertical bisector of AQ intersects CQ at point m, and the trajectory equation of M is e Find the equation of E. urgent! The area of the l-intersection locus e of the straight line passing through the point P (1,0) at two different points a, B and △ AOB is 4 / 5
- 9. As shown in the figure, AB is the diameter of ⊙ o, C is ⊙ o, and the tangent line passing through point C is perpendicular to each other, and the perpendicular foot is D, so as to verify the AC bisection angle DAB As shown in the figure, AB is the diameter of ⊙ o, C is the point on ⊙ o, CD cuts ⊙ o at point C, [label: ab CD, diameter, CD]
- 10. In the triangle ABC, ab = ac.f on AC, AE = AF is intercepted on the extension line of Ba to verify EF vertical BC
- 11. Through a point P (2,0) in the ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1, the chord AB is induced and the trajectory equation of the midpoint m of the chord is obtained Thinking
- 12. 0
- 13. (1) Find the equation of straight line passing through the intersection point of the line l1:7x-8y-1 = 0 and l2:2x + 17Y + 9 = 0 and perpendicular to the line 2x-y + 7 = 0 (2) The straight line L passes through the point P (5, 5) and intersects the circle C: x2 + y2 = 25, and the chord length is 4 5. Find the equation of L
- 14. It is known that the center of circle C is tangent to the line l2:4x + 3Y + 14 = 0 on the line L1: x-y-1 = 0, and the straight line l3:3x + 4Y + 10 = 0 is obtained. The chord length is 6, and the equation of circle C is obtained
- 15. Finding the minimum area of the intersection of the two circles of the sum of the two Y 2-2x-2x-2y = 0-2x-2
- 16. If P9 (2, - 1) is the midpoint of the chord ab of the circle (x-1) 2 + y2 = 25, what is the equation of AB
- 17. 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
- 18. The common length of the two circles obtained by + 1y + 10Y + 2x = 0-2
- 19. 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
- 20. 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