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Math Help - points of tangency (plz help)

  1. #1
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    points of tangency (plz help)

    1.)for the circle x^2+y^2+6x-4y+3=0
    find the equation of the tangent line at (-2,5)

    2.)a circle is tangent to the y-axis at y=3 and has one x-intercept at x=1
    determine the other x-int and deduce the equation of the circle

    any help or hints on how to solve this problem would be GREATLY appreciated

    Thanks in advance
    Last edited by saar4ever; August 23rd 2009 at 06:28 PM.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by saar4ever View Post
    1.)for the circle x^2+y^2+6x-4y+3=0
    find the equation of the tangent line at (-2,5)
    Edit: is the answer y=-x/3-17/3?

    2.)a circle is tangent to the y-axis at y=3 and has one x-intercept at x=1
    determine the other x-int and deduce the equation of the circle

    any help or hints on how to solve this problem would be GREATLY appreciated

    Thanks in advance
    1) Using implicit differentiation,

    2x+2y\frac{dy}{dx}+6-4\frac{dy}{dx}=0 -------- (1)

    2(-2)+2(5)\frac{dy}{dx}+6-4\frac{dy}{dx}=0

    -4+10\frac{dy}{dx}+6-4\frac{dy}{dx}=0

    6\frac{dy}{dx}=-2

    \frac{dy}{dx}=-\frac{1}{3}
    Last edited by alexmahone; August 23rd 2009 at 12:03 PM.
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  3. #3
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    thanks so much alexmahone but im still stumped at the second problem
    btw i did it a bit diff way
    i found the slope of (-2,5) to (-3,2)<-- Center
    and found out it was m=3
    so reciprical of that is -1/3
    and then i plugged it into the point slope form and got
    y-5=-1/3(x+2)
    so y=-x/3+13/3
    right?
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  4. #4
    Senior Member nikhil's Avatar
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    Lightbulb here is the solution

    let the coordinate of other intercept be (a,0) (its on x axis)
    the coordinate of middle point of (1,0) and (a,0) will be ((a+1)/2,0)(let us name it as A).its magnitude will be equal to the length of radius.
    a line drawn perpendicular to this middle point and y intercept will intersect at centre of the circle (let centre of circle be O).
    now it can be easily observed that coordinate of centre of circle will be ((a+1)/2,3)
    let intercept (1,0) be B
    so triangle AOB is a right angle triangle where AO=3 (length of y intercept),OB=length of radius=(a+1)/2,AB=[(a+1)/2]-1=(a-1)/2
    now applying pythagores theorem we have
    [(a+1)/2]^2=[(a-1)/2]^2+9
    solving this we get a=9
    so the coordinate of other intercept will be (9,0) and radius will be 5 units snd coordinate of centre will be (5,3)
    and equation of circle will be
    (x-5)^2+(y-3)^2=25 or
    x^2+y^2-10x-6y+9=0
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  5. #5
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    Thank you soooo much for helping me on the second problem, nikhil.
    You are the best
    I really like the way u solved it step by step
    thx so much
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