Question 653881
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Complete the square for both variables in your equation for the circle:


1.  Divide by the lead coefficient


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ -\ 6x\ +\ 8y\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 6x\ +\ 9\ +\ y^2\ +\ 8y\ +\ 16\ =\ 25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)^2\ +\ (y\ +\ 4)^2\ =\ 25]


Note that the center of the circle is at *[tex \LARGE (3, -4)] and that the radius is 5.


2.  Find the slope of the line containing the radius segment from the center of the circle to the point of tangency.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_r\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2}\ =\ \frac{-4\ -\ (-8)}{3\ -\ 6}]


You do your own arithmetic


3.  Use the fact that any tangent to a circle is perpendicular to the radius at the point of tangency plus the fact that the slopes of perpendicular lines are negative reciprocals to calculate the slope of the desired line.  *[tex \LARGE m\ =\ -\frac{1}{m_r}]


4.  Use the point-slope form of an equation of a line to derive the desired equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the slope calculated in step 3. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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