pre calc

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Find the values of b such that the system has one solution.

x^2 + y^2 = 36
y= x+b

and then they give me an answer box

b=_________ (smaller value)

b=_________ (larger value)

i have absolutely no idea how to even begin this and what they are asking. could some please help and explain what im supposed to do with this. thanks so much!!!

  • pre calc -

    sub in the straight line into the circle

    x^2 + (x+b)^2 = 36
    x^2 + x^2 + 2bx + b^2 - 36 = 0

    for one solution the discriminant, that is
    b^2 - 4ac = 0

    (2b)^2 - 4(2)(b^2-36) = 0
    4b^2 - 8b^2 + 288=-
    -4b^2 = -288
    b^2 = 72
    b = ± √72 or ± 6√2

  • pre calc -

    Draw two tangents to the circle x^2+y^2=6^2 parallel to the line y=x in the second and fourth quadrants.
    b is y-intercept of these tangents.
    Сonsider the algebraic method.

    Find the coordinates of common points of the line and circle:

    x^2+(x+b)^2=36
    2x^2+2bx+b^2-36=0

    Since the line is tangent then the equation has unique solution =>

    the discriminant (2b)^2-4*2(b^2-36)=0
    4b^2-8b^2+288=0
    b^2=72
    b=6sqrt(2) or b=-6sqrt(2)

  • pre calc -

    Consider the geometric method.

    b(b>0)- is the leg in a isosceles rectangular triangle with height=6

    6^2+6^2=b^2

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