Question 1019885
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Since the points *[tex \Large A:(2,5)] and *[tex \Large B:(4,3)] are on the desired circles, segment *[tex \Large AB] is a chord.  Using the two-point form of a linear equation, the derived slope-intercept form of the line passing through the two given points is *[tex \Large y\ =\ -x\ +\ 7].


Since the center of a circle lies on the perpendicular bisector of any chord of a circle and the mid-point of segment AB is the point *[tex \Large (3,4)], the equation of the perpendicular bisector of the chord, and hence the relationship between the coordinates of either of the center points of the two desired circles is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ x\ +\ 1]


Since the desired circles are tangent to the *[tex \Large y]-axis, the centers must lie on lines parallel to the *[tex \Large x]-axis through the point of tangency.  Referring to the diagram, the two centers are *[tex \Large O_1:(x_1,y_1)] and *[tex \Large O_2:(x_2,y_2)], and the centers must lie on the lines *[tex \Large y\ =\ x\ +\ 1] and *[tex \Large y\ =\ y_1] or *[tex \Large y\ =\ y_2].  Also, the points of tangency are *[tex \Large T_1:(0,y_1)] and *[tex \Large T_2(0,y_2)]


*[illustration circletangenttoYaxis2points_(3).jpg]


Note that segments *[tex \Large T_1O_1], *[tex \Large AO_1], and *[tex \Large BO_1] are all radii of the desired circles and must perforce be of equal measure.  Same thing for segments *[tex \Large T_2O_2], *[tex \Large AO_2], and *[tex \Large BO_2].


Using the distance formula without being specific about which of the centers we use gives us:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_{TO}\ =\ \sqrt{(x\ -\ 0)^2\ +\ (y\ -\ y)^2}\ =\ x]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_{AO}\ =\ \sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 5)^2}]


but we will find it convenient to use the relationship *[tex \Large y\ =\ x\ +\ 1] to substitute *[tex \Large x\ +\ 1] for *[tex \Large y] in this distance formula.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_{AO}\ =\ \sqrt{(x\ -\ 2)^2\ +\ (x\ -\ 4)^2}]


Since the two radii must be of equal measure:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{(x\ -\ 2)^2\ +\ (x\ -\ 4)^2}\ =\ x]


Square both sides, expand the squared binomials, and collect terms gives us:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 12x\ +\ 20\ =\ 0]


which factors to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 2)(x\ -\ 10)\ =\ 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ 2\ \Rightarrow\ y_1\ =\ 3]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ 10\ \Rightarrow\ y_2\ =\ 11]


And if *[tex \Large x\ =\ 2] then *[tex \Large T_1O_1\ =\ 2], hence *[tex \Large r_1\ =\ 2].  Whereas if *[tex \Large x\ =\ 10] then *[tex \Large T_1O_1\ =\ 10], hence *[tex \Large r_1\ =\ 10].


The standard form equation of a circle of radius *[tex \Large r] centered at *[tex \Large (h,k)] is *[tex \Large (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2].


Hence, your two equations are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ O_1:\ (x\ -\ 2)^2\ +\ (y\ -\ 3)^2\ =\ 4]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ O_2:\ (x\ -\ 10)^2\ +\ (y\ -\ 11)^2\ =\ 100]



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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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