Question 597160
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This looks a good deal uglier than it actually is.  The fortunate placement of the the point (-12,9) makes things much easier, as you will see.


Step 1:  Take the given tangent line equation and put it into slope-intercept form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x\ +\ 5]


noting that the slope is 1.


Step 2: Use the fact that a radius to a point of tangency is perpendicular to the tangent line at that point, perpendicular lines have negative reciprocal slopes, and the point-slope form to write an equation of the line containing the radius from the center of the circle to the point of tangency.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 1\ =\ -1(x\ +\ 4)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -x\ -\ 3]


Step 3:  Here is the fortunate circumstance.  Since *[tex \LARGE 9\ =\ -(-12)\ -\ 3], the point (-12, 9) <i><b>is the other end of the diameter</b></i> contained in the line *[tex \LARGE y\ =\ -x\ -\ 3].  Therefore the center of the circle must be the midpoint of the segment with endpoints (-12, 9) and (-4, 1)!


Step 4:  Calculate the midpoint coordinates using the midpoint formulas:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_m\ = \frac{x_1 + x_2}{2}\ =\ \frac{-12 + (-4)}{2}\ =\ -8]
 

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_m\ = \frac{y_1 + y_2}{2}\ = \frac{9 + 1}{2}\ =\ 5]


Hence the midpoint of the diameter which must be the center of the circle is (-8, 5).


Step 5: The distance from either endpoint to the center is the radius, so using the distance formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ sqrt{(x_1\ -\ x_2)^2\ +\ (y_1\ -\ y_2)^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ sqrt{(-4\ -\ (-8))^2\ +\ (1\ -\ 5)^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ sqrt{32}\ =\ 4\sqrt{2}]


Step 6:  The equation of a circle with center at *[tex \LARGE (h,k)] and radius *[tex \LARGE r] is *[tex \LARGE (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2], so read your values directly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ 4\sqrt{2}]


And the final form of the equation of your circle is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 8)^2\ +\ (y\ -\ 5)^2\ =\ 32]


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