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-   -   Coordinate geometry(equation of a circle) (https://www.askmehelpdesk.com/showthread.php?t=428162)

  • Dec 25, 2009, 03:52 PM
    Ebudo
    coordinate geometry(equation of a circle)
    please can you explain to me on how to solve this questions...
    1, find the equation of the diameter of the circle x^2+y^2-8x+6y+21=0 which when produced passes through the point(2,1)...
    2, find the equation of the circle which passes through the point (1,1)has a radius of 10 and whose centre lies on the line y=3x-7...
  • Dec 25, 2009, 04:47 PM
    galactus
    1 Attachment(s)
    Quote:

    2, find the equation of the circle which passes through the point (1,1)has a radius of 10 and whose centre lies on the line y=3x-7...
    To find the x-coordinate of the center, we can use the distance formula.

    Since we are told that the center lies on y=3x-7, and the circle passes through , we have



    The distance from (1,1) to the line at the center of the circle is then:



    Solve for x, then sub that into the line equation to find y. This will be the coordinates of the center of the circle.
  • Dec 25, 2009, 06:12 PM
    galactus
    Quote:

    1, find the equation of the diameter of the circle x^2+y^2-8x+6y+21=0 which when produced passes through the point(2,1).
    The find the equation of the circle from the given standard form, complete the square.



    This has center (4,-3) and radius 2.

    By 'equation of the diameter', I assume you mean the equation of the line that passes through the center of the circle, (4,-3) and
    (2,1). I will leave you to find that. That is easy to do now.
  • Dec 26, 2009, 04:17 AM
    Unknown008

    Small typo galactus :)

  • Dec 26, 2009, 08:24 AM
    galactus

    Thanks for the heads up, Unknown. I tried the reutation thing, but it wouldn't let me... as usual.
  • Dec 26, 2009, 01:37 PM
    Ebudo

    Thanks a lot.. galactus... its so easy d way u explained it... nd unknown, thnks 4 d correction... lol...

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