Question 1019918
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I presume you mean *[tex \Large x^2\ +\ y^2\ -\ 10x\ -\ 16y\ +\ 88\ =\ 0] because *[tex \Large X^2\ +\ y]^*[tex \Large\ -\ 10x\ -\ 16y\ +\ 88\ =\ 0] is not the equation of a circle.  In the first place *[tex \Large X] and *[tex \Large x] are two different variables and you are either raising y to the 0 power or to the -10x power the way you wrote the equation.


Presuming


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ y^2\ -\ 10x\ -\ 16y\ +\ 88\ =\ 0]


Then rearrange:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ -\ 10x\ +\ y^2\ -\ 16y\ =\ -88]


Complete the square on *[tex \Large x] and *[tex \Large y]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ -\ 10x\ +\ 25\ +\ y^2\ -\ 16y\ +\ 64\ =\ -88\ 89]


Factor the two perfect square trinomials and combine terms in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x\ -\ 5)^2\ +\ (y\ -\ 8)^2\ =\ 1]


Any equation of the form


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2]


Is the equation of a circle centered at *[tex \Large (h,\,k)] with radius *[tex \Large r], so you can read your center and radius directly from the re-formed equation.


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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