How to find the equation of a circle?

tangent to 2x+3y+13 = 0 and 2x-3y-1=0 contains (0,4)

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  • 8 years ago
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    Let's first find the intersection point of the tangent lines:

    subtract the LHS of the second eqn from that of the first eqn.

    3y + 13 + 3y + 1 = 0

    6y = - 14 => y = -7/3

    x = (3y - 1)/2 = -4

    The slopes of the tangents are -2/3 and 2/3,

    so the bisectors of the angles between the tangents are

    parallel to the x and y axes; in particular, the center of

    the circle must lie either on the line x = -4 or the line y = -7/3.

    The point (0,4) lies "above" both tangents, so the circle

    specified has its center on the line x = -4. The equation of the

    circle is of the form

    (x+4)^2 + (y-k)^2 = r^2, [ Equation 0 ]

    where we still need to determine k and r.

    Since the specifed circle must contain (0,4), we know that

    16 + k^2 = r^2. [ Equation A ]

    The distance from (-4, k) to the line 2x - 3y - 1 = 0 is r, and

    that distance is measured along a line whose slope is -3/2

    because it's perpendicular to 2x-3y-1=0. This particular radius

    is the hypotenuse of a right triangle whose legs have the ratio

    -3/2 and satisfy the Pythagorean Theorem: their lengths are

    2 r /sqrt(13) and -3 r /sqrt(13). The coordinates

    of the point of tangency are -4 + 2 r / sqrt(13) and

    k - 3 r / sqrt(13). The requirement that this point lie on the

    line 2x - 3y - 1 = 0 gives us

    -8 + 4 r /sqrt(13) - 3k - 9 r /sqrt(13) - 1 = 0 [ Equation B ]

    Now, Equations A and B can be simultaneously solved

    to obtain k and r:

    From Eqn A, k = sqrt( r^2 - 16 ), so Eqn B becomes

    -8 + 4 r /sqrt(13) - 3 sqrt( r^2 - 16 ) - 9 r /sqrt(13) - 1 = 0

    We need to eliminate the unknown radicand, so we rearrange:

    3 sqrt( r^2 - 16 ) = - 9 - 5 r /sqrt(13)

    Square both sides:

    9 r^2 - 144 = 81 + 90 r sqrt(13) + (25/13) r^2

    Collect terms:

    (92/13) r^2 - 90 r sqrt(13) - 225 = 0 [ Equation C ]

    At this point, you should review my work in detail

    and correct any arithmetic errors.

    Then use the quadratic formula to solve Equation C,

    obtaining r. Plug that back into Eqn A to get k.

    Plug r and k into [ Equation 0 ] to obtain the

    explicit equation of the required circle.

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