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math

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Solve the system

x^2+y^2-4x+2y=20
4x +3y =5

  • math -

    The first one is a circle centred at (2,-1), or
    (x-2)^2+(y+1)^2=20+4+1=25
    or
    C: (x-2)^2+(y+1)^2=5^2

    The second equation is a straight line.
    L: 4x+3y=5
    To solve the system, set
    y=(5-4x)/3
    and substitute into equation C
    or
    (x-2)^2+((5-4x)/3+1)^2=25
    Solve for x to get x=5 or x=-1
    Substitute into y=(5-4x)/3
    to get
    y=-5, y=3
    So the intersections are
    (5,-5), or (-1,3)

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