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Math Help - Help with a seemingly simple math problem!

  1. #1
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    Help with a seemingly simple math problem!

    A circle is tangent to the y-axis at y=3 and has one x-intercept at x=1.

    (a) Determine the other x-intercept
    (b) Deduce the equation of the circle.





    My answer: EH? please help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by konnie View Post
    A circle is tangent to the y-axis at y=3 and has one x-intercept at x=1.

    (a) Determine the other x-intercept
    (b) Deduce the equation of the circle.





    My answer: EH? please help!
    First do a diagram, see attachment:

    RonL
    Attached Thumbnails Attached Thumbnails Help with a seemingly simple math problem!-gash.jpg  
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  3. #3
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    Quote Originally Posted by konnie View Post
    A circle is tangent to the y-axis at y=3 and has one x-intercept at x=1.

    (a) Determine the other x-intercept
    (b) Deduce the equation of the circle.

    My answer: EH? please help!
    The standard equation of a circle is
    (x-h)^2 +(y-k)^2 = r^2
    where
    (h,k)is the center.

    "The circle is tangent to the y-axis at y=3."
    So a radius, r, is perpendicular to the y-axis at y=3.
    Therefore, r = h, and k=3.
    Draw the figure on paper.
    Draw the horizontal radius h from (0,3) to (h,3) to see the relations.

    Draw the radius h from (1,0) to (h,3).
    By distance between two points,
    h = sqrt[(1-h)^2 +(0-3)^2]
    h^2 = (1 -2h +h^2) +9
    2h = 10
    h = 5

    Draw the vertical radius to intersect the x-axis.
    The horizontal chord, lying on the x-axis, of the circle is bisected by this vertical radius.
    Since h=5, the intersection of this vertical radius and this horizontal chord is at (5,0).
    Therefore, the (1,0) is 4 units to the left of (5,0).
    Hence, the other end of the chord is at (9,0). <----the other x-intercept, answer.

    Since (h,k) = (5,3), and r=h=5, the equation of the circle is
    (x-5)^2 +(y-3)^2 = 25 ---------------answer.

    -----------------
    Check if the (9,0) satisfies the equation.
    (9-5)^2 +(0-3)^2 =? 25
    16 +9 =? 25
    25 =? 25
    Yes, so, OK.
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  4. #4
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    Hello, konnie!

    This is a variation of ticbol's solution.
    . . Refer to Captain Black's diagram.


    A circle is tangent to the y-axis at y=3 and has an x-intercept at x=1.

    (a) Determine the other x-intercept
    (b) Deduce the equation of the circle.

    The center of the circle is: . C(h,3)
    The y-intercept is: . A(0,3)
    The x-intercept is: . B(1,0)

    We find that . CA\,=\,h . and . CB \:=\:\sqrt{(h-1)^2 + 3^2}

    Since CA and CB are radii: . h \:=\:\sqrt{(h-1)^2 + 9}

    . . and we have:. . h^2 \:=\;h^2 - 2h + 1 + 9\quad\Rightarrow\quad 2h \:=\:10\quad\Rightarrow\quad h \,=\,5

    Hence, the center is C(5,3) and  r\,=\,5

    (b) Therefore: . \boxed{{\color{blue}(x-5)^2 + (y-3)^2 \:=\:25}}


    Let y = 0\!:\;\;(x-5)^2 + (0-3)^2 \:=\:25\quad\Rightarrow\quad (x-5)^2 + 9 \:=\:25

    . . (x-5)^2\:=\:16\quad\Rightarrow\quad x-5 \:=\:\pm4\quad\Rightarrow\quad x \:=\:1,\,9

    (a) The other x-intercept is: . \boxed{{\color{blue}(9,\,0)}}

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  5. #5
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    Thank you ^___^
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