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Math Help - TANGENT and CIRCLE

  1. #1
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    Mar 2008
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    Exclamation TANGENT and CIRCLE

    I'm having a problem with the following question. If anyone can help, that would be very much appreciated.

    (1) Suppose that (a,b) is on the circle (x^2)+(y^2)=(r^2). Show that the line ax+by=(r^2) is tangent to the circle at (a,b).

    PLEASE, IF ANYONE CAN HELP, I WOULD LOVE IT.

    Thank you in advance
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    after an implicit differentiation of the circles equation we get:


    \begin{gathered}<br />
  2x + 2y\frac{{dy}}<br />
{{dx}} = 0 \hfill \\<br />
   \Leftrightarrow \frac{{dy}}<br />
{{dx}} =  - \frac{x}<br />
{y} \hfill \\ <br />
\end{gathered}

    thus the general equation of the slope of any tangent to the circle is:

    <br />
\frac{{dy}}<br />
{{dx}} =  - \frac{x}<br />
{y}

    the general equation of a line passing through a coordinate (a,b) with a slope m is:

    <br />
y - b = m(x - a)<br />

    the slope m can be obtained from:

    <br />
m =  - \left. {\frac{x}<br />
{y}} \right|_{\left( {x,y} \right) = (a,b)}  =  - \frac{a}<br />
{b}<br />

    thus the equation of the tangent is:

    <br />
\begin{gathered}<br />
  y - b =  - \frac{a}<br />
{b}(x - a) \hfill \\<br />
   \hfill \\<br />
   \Leftrightarrow by - b^2  =  - ax + a^2  \hfill \\<br />
   \Leftrightarrow ax + by = a^2  + b^2  \hfill \\ <br />
\end{gathered} <br />

    if we substitute (x,y) = (a,b) in the equation of the given circle we get:

    <br />
a^2  + b^2  = r^2 <br />

    thus the equation of the tangent is:

    <br />
ax + by = r^2 <br />
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