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Math Help - New! coordinate Geometry

  1. #1
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    New! coordinate Geometry

    Could someone show me how to work out these please

    1. The circle x^2(y+3)^2=25 meets the x axis at p and q, find the coordinates of p and q?

    2. The points A(7,4) B(8,-7) and C(-4,3) lie on a circle. Find the equation of the circle?

    3. Find the equation of the tangent at the point (3,1) on the circle
    x^2+y^2-4x+10y-8=0

    Thanks for your help
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  2. #2
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    Quote Originally Posted by Colin_m View Post
    1. The circle x^2(y+3)^2=25 meets the x axis at p and q, find the coordinates of p and q?
    x^2(y+3)^2=25 isn't a circle. Write it properly.

    Quote Originally Posted by Colin_m View Post
    2. The points A(7,4) B(8,-7) and C(-4,3) lie on a circle. Find the equation of the circle?
    Plug these points in (x-a)^2 + (y-b)^2 = r^2 and find a, b and r.

    Quote Originally Posted by Colin_m View Post
    3. Find the equation of the tangent at the point (3,1) on the circle
    x^2+y^2-4x+10y-8=0
    Use implicit differentiation. Or write the circle as two functions of x, then use one of them and find the slope. Then use the point given to find the equation.
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  3. #3
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    thanks.
    isn't a circle. Write it properly.
    my maths teacher is think then lol
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  4. #4
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    Hello, Colin!

    2. The points A(7,4),\: B(8, \text{-}7),\: C(\text{-}4,3) lie on a circle.
    Find the equation of the circle.
    On a hunch, I found the slopes of the sides of this triangle.

    m_{AB} \:=\:\frac{\text{-}7-4}{8-7} \:=\:-11

    m_{AC} \:=\:\frac{3-4}{\text{-}4-7} \:=\:\frac{1}{11}

    Hence: . AB \perp AC\quad\hdots \text{ We have a right triangle!}


    The center is the midpoint of the hypotenuse BC.

    . . Center: . O\left(\frac{8-4}{2},\:\frac{-7+3}{2}\right) \;=\;(2,-2)


    The radius is the length of OA (or OB or OC).

    . . r^2 \;=\;OA^2 \;=\;(7-2)^2 + (4+2)^2 \;=\;61


    The equation is: . (x-2)^2 + (y+2)^2 \:=\:61

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