Question 261817
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 2x\ +\ y^2\ =\ 8]


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


Divide the coefficient on the first degree *[tex \Large x] term by 2 and square the result.  -2 divided by 2 is -1.  -1 squared is 1.  Add the result to both sides of the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 2x\ +\ 1\ +\ y^2\ =\ 9]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\,-\,1\right)^2 +\ y^2\ =\ 9]


Re-write *[tex \Large y^2] as *[tex \Large \left(y\,-\,0\right)^2]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\,-\,1\right)^2 +\ \left(y\,-\,0\right)^2\ =\ 9]


A circle centered at *[tex \Large \left(h,\,k\right)] and with radius *[tex \Large r] has an equation:


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


So find the point *[tex \Large \left(1,\,0\right)] and draw a circle with radius 3 centered there.


{{{drawing(
500, 500, -5, 5, -5, 5,
grid(1),
circle(1,0,3),
green(circle(1,0,.07)),
locate(1.05,-0.3,C(1,0)))}}}


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>