Home
Class 12
MATHS
The straight line 2x – 3y = 1 divides th...

The straight line 2x – 3y = 1 divides the circular region `x^(2) + y^(2) le 6` into two parts. If
`S = {(2, (3)/(4)), ((5)/(2), (3)/(4)), ((1)/(4), -(1)/(4)), ((1)/(8), (1)/(4))}`, then the number of point(s) in S lying inside the smaller part is

Text Solution

Verified by Experts

The correct Answer is:
`0002`
Promotional Banner

Topper's Solved these Questions

  • CIRCLE

    MOTION|Exercise Exerise - 4 | Level - I (Previous Year|JEE Main|10 Videos
  • BINOMIAL THEOREM

    MOTION|Exercise Exercise -4 (Level - II) ( Previous Year )|7 Videos
  • COMPLEX NUMBER

    MOTION|Exercise EXERCISE - 4 (LEVEL -II) PREVIOUS YEAR - JEE ADVANCED|33 Videos

Similar Questions

Explore conceptually related problems

The straight line 2x-3y=1 divides the circular region x^(2)+y^(2)<=6 into two parts.If S={(2,(3)/(4)),((5)/(2),(3)/(4)),((1)/(4),-(1)/(4)),((1)/(8),(1)/(4))} then the number of point(s) in s lying inside the smaller part is

S_(1)={2},S_(2)={(3)/(2),(4)/(2)},S_(3)={(4)/(4),(5)/(4),(6)/(4)},S_(4)={(5)/(8),(6)/(8),(7)/(8),(8)/(8)}, then the sum of numbers in S_(20) is

The length of the straight line x-3y=1 intercept by the hyperbola x^(2)-4y^(2)=1 is

If the straight line x=b divide the area enclosed by y=(1-x)^(2),y=0 " and " x=0 " into two parts " R_(1)(0le x le b) " and " R_(2)(b le x le 1) " such that " R_(1)-R_(2)=(1)/(4). Then, b equals

Let the straight line x = b divide the area enclosed by y = (1-x)^(2), y = 0 and x = 0 into two parts R_(1) (0 le x le b) and R_(2) (b le x le 1) such that R_(1) - R_(2) = 1/4 . Then b equals

S={3x^2 le 4y le 6x+2y} find the area

S=((1)/(2))^(2)+(((1)/(2))^(2)1)/(3)+(((1)/(2))^(3)1)/(4)+(((1)/(2))^(4)1)/(5)+......, then (a)S=ln8-2( b) S=ln((4)/(e))(c)S=ln4+1(d) none of these

The lines (x)/(1)=(y)/(2)=(z)/(3)and(x-1)/(-2)=(y-2)/(-4)=(z-3)/(-6) are

The d.c.s.of the line 2x+1=3-2y=4z+5 are

MOTION-CIRCLE-Exercise - 4 | Level - II (Previous Year | JEE Advanced
  1. Consider the two curves C1 ; y^2 = 4x, C2 : x^2 + y^2 - 6x + 1 = 0 the...

    Text Solution

    |

  2. Consider, L(1) : 2x + 3y + p – 3 = 0 , L(2) : 2x + 3y + p + 3 = 0, whe...

    Text Solution

    |

  3. A circle C of radius 1 is inscribed in an equilateral triangle PQR. Th...

    Text Solution

    |

  4. A circle C of radius 1 is inscribed in an equilateral triangle PQR. Th...

    Text Solution

    |

  5. A circle C of radius 1 is inscribed in an equilateral triangle PQR. Th...

    Text Solution

    |

  6. Tangents drawn from the point P(1,8) to the circle x^(2) + y^(2) - 6x ...

    Text Solution

    |

  7. The centres of two circles C(1) and C(2) each of unit radius are at a...

    Text Solution

    |

  8. Two parallel chords of a circle of radius 2 are at a distance sqrt3 + ...

    Text Solution

    |

  9. The circle passing through the point (-1,0) and touching the y-axis at...

    Text Solution

    |

  10. The straight line 2x – 3y = 1 divides the circular region x^(2) + y^(2...

    Text Solution

    |

  11. The locus of the mid-point of the chord of contact of tangents drawn f...

    Text Solution

    |

  12. A possible equation of L is (A) x 3y 1 (B) x 3y 1 (C) x 3y 1 (D) x...

    Text Solution

    |

  13. A tangent PT is drawn to the circle x^2+y^2=4 at the point P(sqrt3,1)....

    Text Solution

    |

  14. A possible equation of L is (A) x 3y 1 (B) x 3y 1 (C) x 3y 1 (D) x...

    Text Solution

    |

  15. A circle s passes through the point ( 0,1) and is orthogonal to the ci...

    Text Solution

    |

  16. Let , RS be the diameter of the circle x^(2) + y^(2) = 1 , where S...

    Text Solution

    |

  17. For how many values of p, the circle x^2+y^2+2x+4y-p=0 and the coordin...

    Text Solution

    |

  18. Let T be the line passing the points P(–2, 7) and Q(2,–5). Let F(1) be...

    Text Solution

    |

  19. PARAGRAPH "X" Let S be the circle in the x y -plane defined by the eq...

    Text Solution

    |

  20. Let P be a point on the circle S with both coordinates being positive....

    Text Solution

    |