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Posted by on Thursday, January 31, 2013 at 8:39am.

Find the equation of the circle, with radius 1, tangent to the line 3x+4y=5 and having its center on the line x+2y=0

  • Analytic Geometry - , Thursday, January 31, 2013 at 9:30am

    let the centre be (a,b)
    but it must lie on x+2y=0, so a+2b=0
    a = -2b

    we can call our centre (-2b, b)
    its distance to 3x + 4y - 5 must be 1

    | 3(-2b) + 4b - 5 }/√(3^2+4^2) = 1

    case 1:
    (-6b+4b-5)/√25 = 1
    -2b - 5 = 5
    b = -5
    a= 10 ----> equation: (x-10)^2 + (y+5)^2 = 1

    case 2:

    6b - 4b + 5 = 5
    2b = 0
    b=0
    a=0 ----------> x^2 + y^2= 1

    make a sketch to see if both are possible
    What do you think?

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