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Math Help - CENTRE OF CIRCLE: missed lesson plz help

  1. #1
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    CENTRE OF CIRCLE: missed lesson plz help

    PLZ GIVE EXPLANATIONS AND ANSWERS ASAP

    questions..
    The circle C with centre (a,b) and radius 5 touches x-axis at (4,0)

    need to find the values of a and b

    find cartesian equation for C




    A tangent to the circle, drawn from point p(8,17) touches the circle T

    find the length of PT to 3sf

    2)

    the point A has cooridinates 2,5 and point b has coordinates (-2,8)
    find in cartesian form and equation of the circle with diamitar AB





    PLZ GIVE EXPLANATIONS AND ANSWERS ASAP
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  2. #2
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    Quote Originally Posted by sasha9193 View Post
    PLZ GIVE EXPLANATIONS AND ANSWERS ASAP

    questions..
    The circle C with centre (a,b) and radius 5 touches x-axis at (4,0)

    need to find the values of a and b

    find cartesian equation for C
    All circles can be written as x^2 + y^2 = r^2.
    This is a circle centered about the origin (0,0) with radius r.
    A circle of at (3,-1) would be (x-3)^2 + (y+1)^2 = r^2
    The same circle with diameter 2 would be (x-3)^2 + (y+1)^2 = 4

    In your problem, you should draw out the circle.
    There are two possible circles.
    One that is below the x-axis, but touches at (4,0), and one that is above the x-axis, but touches at (4,0).
    You have a radius of 5, so for one circle, go up 5 in the y-direction, and for the other, go down 5 in the y-direction.

    The centers (a,b) would be (4,5) and (4,-5).
    You know how to write these now:
    A circle at (4,5) would be (x-4)^2 + (y-5)^2 = r^2
    A circle at (4,-5) would be (x-4)^2 + (y+5)^2 = r^2
    And you know the radius is 5, so r^2 equals (5^2) equals 25.
    A circle at (4,5) with radius 5 would be (x-4)^2 + (y-5)^2 = 25
    A circle at (4,-5) with radius 5 would be (x-4)^2 + (y+5)^2 = 25
    ^Cartesian equations.
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  3. #3
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    So y=0 is a tangent line to the circle. With means y=0 is perpendicular to its radius so the radius has the equation x=k for some number k. Since it intersects at (4,0) it means the perpendicular radius is x=4. So on the line x=4 we can find the center (a,b). Thus, a=4 because x is fixed at 4 and we need to solve for b. Now (a,b)=(4,b) must be 5 units away from (4,0) as the problem says so b=5 or -5. Thus, (4,5) or (4,-5) are the possible centers.
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  4. #4
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    the circle has a piccy on my question, its centre sits above the x axis and right from the y axis
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  5. #5
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    ok plz help with rest of question, i can do up to the second one, the equation need to know rest
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  6. #6
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    Quote Originally Posted by sasha9193 View Post
    the circle has a piccy on my question, its centre sits above the x axis and right from the y axis
    Then use this:
    A circle at (4,5) with radius 5 would be (x-4)^2 + (y-5)^2 = 25
    (it is above the x-axis and right of the y-axis)
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  7. #7
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    Quote Originally Posted by sasha9193 View Post
    ok plz help with rest of question, i can do up to the second one, the equation need to know rest
    the point A has cooridinates 2,5 and point b has coordinates (-2,8)
    find in cartesian form and equation of the circle with diamitar AB

    Make a diagram. Plot point A and point B.

    Note that the distance between point A and point B is:
    SquareRoot(( [2-(-2)]^2 + [8-5]^2 ))
    SquareRoot(( 4^2 + 3^2 ))
    SquareRoot(( 16 + 9 ))
    SquareRoot(( 25 ))
    5

    So the diameter is 5. Therefore the radius is 5/2.

    Find the midpoint between A and B.
    (-2,8) and (2,5) where they are in form (x,y)
    The midpoint of the x's is (-2 + 2)/2
    The midpoint of the y's is (8 + 5)/2

    So the midpoint is: (0,13/2)

    And the midpoint is the center of the circle.

    Given what you've been taught about cartesian form, the answer is therefore:

    (x-0)^2 + (y-13/2)^2 = (5/2)^2
    x^2 + (y-13/2)^2 = 25/4
    or
    4(x^2) + 4(y-13/2)^2 = 25

    I personally prefer the second way better.
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