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Math Help - Problem in Coordinate geometry

  1. #1
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    Problem in Coordinate geometry

    Iíve two question please help me to solve.

    1. The line y = 0 is a tangent to the circle (x - 8)^2 + (y - a)^2 = 16. Find the value of a.

    2. The circle, center (p, q) radius 25, meets the x-axis at (-7, 0) and (7, 0), where q > 0.
    a) Find the value of p and q.
    b) Find the coordinates of the points where the circle meets the y-axis.
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  2. #2
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    Quote Originally Posted by geton View Post
    I’ve two question please help me to solve.

    1. The line y = 0 is a tangent to the circle (x - 8)^2 + (y - a)^2 = 16. Find the value of a.

    2. The circle, center (p, q) radius 25, meets the x-axis at (-7, 0) and (7, 0), where q > 0.
    a) Find the value of p and q.
    b) Find the coordinates of the points where the circle meets the y-axis.
    1. The distance of the centre of the circle to the x-axis (y = 0) has to equal the radius. (Why?) Therefore ......

    2. (a) By symmetry, p = 0 (draw a rough sketch to see why). Then x^2 + (y - q)^2 = 25^2. Now sub either of (-7, 0) or (7, 0) and solve for q.

    (b) (0, q + 25) and (0, q - 25) (why?)
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    1. The distance of the centre of the circle to the x-axis (y = 0) has to equal the radius. (Why?) Therefore ......

    2. (a) By symmetry, p = 0 (draw a rough sketch to see why). Then x^2 + (y - q)^2 = 25^2. Now sub either of (-7, 0) or (7, 0) and solve for q.

    (b) (0, q + 25) and (0, q - 25) (why?)
    By this way,
    1. I found a^2 = - 48. But answer is a = 4.
    2. a. q^2 = - 24. But answer is q = 24.



    I'm totally confused.
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  4. #4
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    Quote Originally Posted by geton View Post
    By this way,
    1. I found a^2 = - 48. But answer is a = 4. Mr F says: Remind me - what did I say about the distance of the centre from the x-axis? Hmmmm ..... And the radius is 4 .... Hmmmmm ......

    2. a. q^2 = - 24. But answer is q = 24. Mr F says: x^2 + (y - q)^2 = 25^2. I'll sub (7, 0) and solve for q: (7)^2 + (-q)^2 = 25^2 => 49 + q^2 = 25^2 => q^2 = 25^2 - 49 = (25 - 7)(25 + 7) = (18)(32) therefore \, q = 3\sqrt{2} \times 4\sqrt{2} = 24 . Or you could have recognised and exploited the pythagorean triple (7, 24, 25). Why use a calculator when you can exercise your brain.



    I'm totally confused.
    Without seeing any of your working it's difficult to know which of the many potential mistakes you might have made. Tip for the future: Show your working, not just your answers.
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