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need maths whiz help

  • 19-06-2006 2:12pm
    #1
    MarieDuffy
    Closed Accounts Posts: 8 ✭


    S is the circle x squared + y squared + 4x + 4y - 17 = 0
    and K is the line4x+3y=12.

    i. Show that the line K does not intersect S.
    ii. Find the co-ordinates of the point on S that is closest to K.


    Any of you out there able to solve this for me...im not having any luck!


Comments

  • #2
    Attractive Nun
    Closed Accounts Posts: 1,203 ✭✭✭Attractive Nun


    For part (i) just get the radius of the circle, and show that it is not equal to the perpendicular distance from K to the centre. Or do simultaneous equations and show they cannot work out. (i.e. b^2-4ac < 0)

    Part (ii) involves differentiating and letting something equal to zero or something I think. The answer is (2,1)


  • #3
    Zoodlebop
    Closed Accounts Posts: 172 ✭✭Zoodlebop


    To find the point at which the line is closest to the circle, find the slope of the line. Invert this and multiply by -1. This is the slope of the line (called line-z from here) from the center of the circle to the line. Then find the equation of line-z. Do simultaneous equations to find the point where the two perpendicular lines intersect. (2, 1)


  • #4
    Attractive Nun
    Closed Accounts Posts: 1,203 ✭✭✭Attractive Nun


    Zoodlebop wrote:
    To find the point at which the line is closest to the circle, find the slope of the line. Invert this and multiply by -1. This is the slope of the line (called line-z from here) from the center of the circle to the line. Then find the equation of line-z. Do simultaneous equations to find the point where the two perpendicular lines intersect. (2, 1)

    Ah!!! D'Oh for me in the LC!


  • #5
    Nichololas
    Registered Users, Registered Users 2 Posts: 1,186 ✭✭✭Nichololas


    Zoodlebop wrote:
    To find the point at which the line is closest to the circle, find the slope of the line. Invert this and multiply by -1. This is the slope of the line (called line-z from here) from the center of the circle to the line. Then find the equation of line-z. Do simultaneous equations to find the point where the two perpendicular lines intersect. (2, 1)

    Actually you've got that wrong. It's the point on the CIRCLE that's closest to the line, not the point on the line closest to the circle.

    You do everything Zoodlebop says up until the simultaneous equations, you sub the values of z into the circle. Or something.. it's a blur now.


  • #6
    Zoodlebop
    Closed Accounts Posts: 172 ✭✭Zoodlebop


    Enlil_Nick wrote:
    Actually you've got that wrong. It's the point on the CIRCLE that's closest to the line, not the point on the line closest to the circle.

    You do everything Zoodlebop says up until the simultaneous equations, you sub the values of z into the circle. Or something.. it's a blur now.

    Oh bother, your right. Perhaps I did it right in the exam.....perhaps......not:(


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  • #7
    MarieDuffy
    Closed Accounts Posts: 8 ✭MarieDuffy


    SWEETTTTT....appreciate the help


  • #8
    MarieDuffy
    Closed Accounts Posts: 8 ✭MarieDuffy


    I got the slope of K to be -3/4 but how do i got about get'n the equation of -Z?


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