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Math Help - The Tangent Line

  1. #1
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    The Tangent Line

    The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the POINT OF TANGENCY.

    If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show that:

    (a) r^2(1 + m^2) = b^2

    HINT GIVEN FOR PART A:

    The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly ONE SOLUTION.

    (b) The point of tangency is the point (-r^2m/b, r^2/b).

    (c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

    (d) Draw a tangent line to a circle in the xy-plane.
    Use r to represent the radius.
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  2. #2
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    Quote Originally Posted by symmetry View Post
    The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the POINT OF TANGENCY.

    If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show that:

    (a) r^2(1 + m^2) = b^2

    HINT GIVEN FOR PART A:

    The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly ONE SOLUTION.

    (b) The point of tangency is the point (-r^2m/b, r^2/b).

    (c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

    (d) Draw a tangent line to a circle in the xy-plane.
    Use r to represent the radius.
    Hello,

    solve the given quadratic equation for x:

    x^2 + (mx + b)^2 = r^2 \Longleftrightarrow x^2+m^2 x^2 + 2mbx + b^2-r^2=0  \Longleftrightarrow (1+m^2)x^2+2mbx+b^2-r^2=0. Use the formula to solve a quadratic equation:

    x=\frac{-2mb \pm \sqrt{4m^2 b^2 - 4(1+m^2)(b^2-r^2)}}{2(1+m^2)}

    x=\frac{-2mb \pm 2 \sqrt{m^2 b^2 -b^2 + r^2  - m^2 b^2 + m^2 r^2}}{2(1+m^2)}. There exists only one solution if the radicand equals zero:

    -b^2+r^2+m^2 r^2=0 \Longleftrightarrow r^2(1+m^2)=b^2

    From this equation follows:

    1+m^2=\frac{b^2}{r^2}

    to b)

    If the radicand equals zero, than the x-value of the tangent point is

    x=\frac{-mb}{1+m^2}. Now use the term to substitute (1+mē) and you'll get:

    x=\frac{-mb}{1+m^2}=\frac{-mb}{\frac{b^2}{r^2}}=\frac{-m r^2}{b}. Plug in this term into the equation of the tangent:

    y=\frac{-m r^2}{b} \cdot m + b=\frac{-m^2 r^2 + b^2}{b}

    From the equation of part (a) follows:

    -b^2+r^2+m^2 r^2=0 \Longleftrightarrow r^2=-m^2 r^2 + b^2 Use this result to simplify the y-value to: y=\frac{r^2}{b}

    to (c):

    The slope of the line from the centre to the tangent point is:

    \frac{y}{x}=\frac{\frac{r^2}{b}}{\frac{-m r^2}{b}}=-\frac{1}{m} Two lines are perpendicula if their slopes hold the equation:

    m_2=-\frac{1}{m_1}. Thus the tangent and the line containing centre and tangent point are perpendicular.

    to (d).

    There are 2 different situations:

    (i) You already know the tangent point. Then draw the line centre-tangentpoint and afterwards a perpendicular line through the tangent point.

    (ii) The tangent passes through a point outside the circle. Then you have to construct a right triangle using the circle of Thales.

    EB
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  3. #3
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    ok

    Absolutely amazing reply with GREAT detail.

    Thanks!
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