Question 237326
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The general equation of a circle with center at *[tex \Large \left(h,k\right)] and radius *[tex \Large r] is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - h)^2 + (y - k)^2 = r^2] 


Hence the circle


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ =\ 25]


Must be centered at the origin.


The line tangent to the point *[tex \Large \left(3,4\right)] is perpendicular to the radius segment with endpoints *[tex \Large \left(0,0\right)] and *[tex \Large \left(3,4\right)]


Determine an equation of the line that contains the aforementioned radius segment by using the two point form of the equation of a line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y - y_1 = \left(\frac{y_1 - y_2}{x_1 - x_2}\right)(x - x_1) ]


Where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the given points.


Solve the resulting equation for *[tex \Large y] to put this equation into slope-intercept form.  Then determine the slope of the radius segment by inspection of the coefficient on *[tex \Large x].


Next use the fact that the slopes of two perpendicular lines are negative reciprocals, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1 \perp L_2 \ \ \Leftrightarrow\ \ m_1 = -\frac{1}{m_2} \text{ and } m_1, m_2 \neq 0]


Calculate the negative reciprocal of the slope of the radius segment to get the slope of the tangent.


Then use the point-slope form of the equation of a line with the calculated slope and the point *[tex \Large \left(3,4\right)] to derive the desired equation:


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


Where *[tex \Large m] is the calculated slope and *[tex \Large \left(x_1,y_1\right)] is the point *[tex \Large \left(3,4\right)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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