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

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

Thanks for helping!

  • math -

    Well, if it is tangent to the y axis at (0,3), then the center must be somewhere on the horizontal line y = 3 because the radius is perpendicular to the tangent.
    So if center is at (a,3)
    (x-a)^2 + (y-3)^2 = r^2
    we know when x = 0, y = 3 so
    a^2 = r^2
    so
    (x-a)^2 + (y-3)^2 = a^2
    now when y = 0, x = 1 and something else
    (1-a)^2 +9 = a^2
    1 -2a +a^2 +9 = a^2
    a = 5 = r
    so center is at (5,3)
    (x-5)^2 + (y-3)^2 = 25
    is the equation
    x intercepts:
    (x-5)^2 +9 = 25
    x^2 - 10 x + 25 + 9 = 25
    x^2 -10 x = -9
    x^2 - 10 x + 25 = 16
    (x-5)^2 = 16
    x-5 = +/- 4
    x = 1 or x = 9
    which you could have seen from a sketch once you knew the radius was 5 and the center at (5,3)

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