Question 652668
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The equation of <b><i>any</i></b> circle centered at the origin is:


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


where *[tex \LARGE r] is a real number that represents the measure of the radius.  Our task for this particular problem is to determine the specific value for *[tex \LARGE r] that will make our circle tangent to *[tex \LARGE 4x\ +\ 3y\ =\ 10]


Fact:  Any line tangent to a circle is perpendicular to the radius at the point of tangency.


First, we need an equation of the line that contains the radius at the point of tangency.  Such a line must contain the point *[tex \LARGE \left(0,\,0\right)] and have a slope that is the negative reciprocal of the slope of the tangent line.


Since we have a line given in Standard Form, we calculate the slope by the opposite of the coefficent on *[tex \LARGE x] divided by the coefficient on *[tex \LARGE y], namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_T\ =\ -\frac{4}{3}]


Then, taking the negative reciprocal:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_r\ =\ \frac{3}{4}]


Secondly we need to apply the point-slope form of an equation of a line to find an equation that contains the radius at the point of tangency:


*[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 calculated slope.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 0\ =\ \frac{3}{4}(x\ -\ 0)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{3}{4}x]


Next, we need to find the point of tangency, said point being the intersection of the tangent line and the line containing the radius.  Solve the 2X2 system of equations by substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ +\ 3\left(\frac{3}{4}x\right)\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1.6]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 1.2]


Note:  Verification of the arithmetic is left as an exercise for the student.


Next you need the measure of the radius which is the distance from the center to the point of tangency, namely the distance from *[tex \LARGE \left(0,\,0\right)] to *[tex \LARGE \left(1.6,\,1.2\right)].  Use the distance formula.


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


I'll leave this last bit of arithmetic to you.  Once you know the value of *[tex \LARGE r], plug that value into:


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


to get your final answer.


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