Question 358061
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In the first place, what makes you think this is a "Quick Question"?


Since there is a *[tex \Large y] term, but no *[tex \Large x] term, you only need to complete the square on *[tex \Large y].


Add the opposite of the constant term to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ -\ 4y\ =\ 1]


Divide the coefficient on *[tex \Large y] by 2, square the result, and add that result to both sides: (-4 divided by 2 is -2, -2 squared is 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ -\ 4y\ +\ 4=\ 5]


Factor the perfect square in *[tex \Large y]:


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


Rewrite:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 0)^2\ +\ (y\ -\ 2)^2=\ \left(\sqrt{5}\right)^2]


Hence the circle has a center at *[tex \Large (0, 2)] and a radius *[tex \Large \sqrt{5}]


The tangent to a circle at a given point is perpendicular to the line containing the radius through the given point.


Find the slope of the line containing the radius by using the slope formula on the given point and the coordinates of the center of the circle:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and  *[tex \Large \left(x_2,y_2\right)] are the coordinates of the circle center.


Since


*[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 you calculated above.


Finally use the point-slope form of an equation of a straight line:


*[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 calculated negative reciprocal of the radius slope.


Remember to put the result in the appropriate final form as specified by your instructor or text (if any).


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