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It is known that the line L passes through the point P (- 2,5), and the slope is - 3 / 4. The equation of the circle where the line L is tangent to the point (2,2) and the center of the circle is on the line x + Y-11 = 0 is obtained Who knows? Reply
The equation (X-5) ^ 2 + (y-6) ^ 2 = 25 for a circle with center (5,6) r = 5
The equation of the ellipse is x ^ 2 / 4 + y ^ 2 / 3 = 1. If the line L passing through the point (0,1) intersects the ellipse at two points AB, and the circle with diameter AB just passes through F1, the slope of the line is calculated
Let a (x1, Yi), B (X2, Y2). According to the meaning of the problem, the slope of L exists. If K is set, then the equation of L is y = KX + 1 and the equation of ellipse is simultaneous, and (3 + 4K ^ 2) x ^ 2 + 8kx-8 = 0 (*). It is easy to know that X1 and X2 are the roots of the equation (*), so X1 + x2 = - 8K / (3 + 4K ^ 2), x1x2 = - 8 / (3 + 4K ^ 2) the circle with ab as the diameter just passes F1
Try to find the equation of the circle C1: (x-1) &# 178; + Y & # 178; = 1 phase circumscribed and tangent to the line x + √ 3 = 0 at the point Q (3, - 3)
Is it a straight line x + √ 3Y = 0?
RELATED INFORMATIONS
- 1. Given that two circles C1: (x + 4) 2 + y2 = 2, C2: (x-4) 2 + y2 = 2, the moving circle m is tangent to both circles C1 and C2, then the trajectory equation of the moving circle center m is () A. X = 0b. X22-y214 = 1 (x ≥ 2) C. x22-y214 = 1D. X22-y214 = 1 or x = 0
- 2. Given circle C1: (x + 3) ^ 2 + y ^ 2 = 1 and circle C2: (x-3) ^ 2 + y ^ 2 = 9, the moving circle m is circumscribed with circle C1 and C2 at the same time The orbit equation of moving circle center M
- 3. If the parabola C1 and the parabola C2: y ^ 2 = - 4x are symmetric with respect to the straight line x + y = 2, then the focal coordinate of the parabola C1 is
- 4. Let the equation of the ellipse be x square + y square / 4 = 1, the straight line passing through M (0,1) intersects the ellipse at two points AB, O is the coordinate origin, Op vector = 1 / 2 (OA vector + ob vector). When l rotates around point m, the trajectory equation of the moving point P is obtained
- 5. The linear equation with ellipse (x ^ 2 / 9) + (y ^ 2 / 4) = 1 intersecting at two points a and B, and the midpoint of line AB is m (1,1) is, using the point difference method
- 6. If the midpoint of line a and B is (1,1), then the equation of L is?
- 7. It is known that m (4,2) is the midpoint of line AB cut by ellipse x2 + 4y2 = 36, then the equation of line L is______ ．
- 8. A. B is two points on hyperbola x ^ 2-y ^ 2 / 2 = 1, and point (1,2) is a line segment ab. the midpoint of the line AB equation is obtained
- 9. The line passing through point P (8,1) intersects hyperbola x2-4y2 = 4 at two points a and B, and P is the midpoint of line ab. the equation of line AB is obtained
- 10. Given the hyperbola x ^ 2-4y ^ 2 = 4 and the point m (8,1), the straight line passing through the electric m intersects the hyperbola at two points a and B, and M is the midpoint of the line AB, the equation of the straight line is obtained
- 11. If a straight line passes through the point P (- 3, - 32), and the distance between the center of circle x2 + y2 = 25 and the straight line is 3, then the equation of the straight line is___ ．
- 12. The equation of the line parallel to the line 2x-y + 1 = 0 and tangent to the circle x2 + y2 = 5 is () A. 2x-y + 5 = 0b. X2-y-5 = 0C. 2x + y + 5 = 0 or 2x + Y-5 = 0d. 2x-y + 5 = 0 or 2x-y-5 = 0
- 13. Find the equation of a straight line passing through point a (3,4) and tangent to circle x square + y square = 9 Such as the title
- 14. It is known that the ellipse takes the symmetry axis as the coordinate axis, and the major axis is three times of the minor axis, and passes through the point (3,0)
- 15. Find the equation of the circle which passes through the point m (3, - 1) and the circle C: x2 + Y2 + 2x-6y + 5 = 0 and is tangent to the point n (1,2). Why can't we use the equation of the circle system? Isn't there an intersection point between the tangent and the circle?
- 16. A beam of light is emitted from a (- 2,3), reflected by x-axis, and tangent to the circle C (x-3) 2 + (Y-2) 2 = 1. The equation of the straight line of the reflected light is obtained Method 1: the symmetrical point of point a about the x-axis is on the line where the reflected light is located PS: it's a clue to the title, but I'm not good at it
- 17. Find the tangent equation of the circle x 2 + y 2 = 4 from point a (2,4)
- 18. Tangent equation of P (- 5, - 2) and circle x ^ 2 + y ^ 2 = 25 RT
- 19. Find the tangent equation passing through point a (- 4,3) and tangent to circle (x + 1) square + (y + 1) square = 25 Find the tangent equation passing through point a (- 4,3) and tangent to circle (x + 1) square + (y + 1) square = 25
- 20. If the line x + y = C is tangent to the circle X & # 178; + Y & # 178; = 2, then C = () 3Q