Question 653877
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The equation of a circle centered at *[tex \LARGE (h,k)] with radius *[tex \LARGE r] is *[tex \LARGE (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2]


The vertex of a parabola described as *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c] is at the point *[tex \LARGE \left(\frac{-b}{2a},\rho\left(\frac{-b}{2a}\right)\right)]


Step 1: Rewrite your circle equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ (-2)\right)^2\ +\ \left(y\ -\ (-1)\right)^2\ =\ 5]


And then by inspection you can see that the first point for your line is *[tex \LARGE (-2,-1)]


Step 2:  Calculate the *[tex \LARGE x]-coordinate of your parabola vertex:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-b}{2a}\ =\ \frac{-12}{-6}\ =\ 2]


Step 3:  Evaluate the parabola function at the *[tex \LARGE x] coordinate of the vertex, i.e. calculate *[tex \LARGE f(2)]


Step 4:  Form an ordered pair from the results of steps 2 and 3.


Step 5:  Use the results of steps 1 and 4 and the two-point form of an equation of a line to write your desired equation:


*[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.


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