Respuesta :
[tex]x^2+6x+y^2-16y=23[/tex]
The standard form of the equation of a circle is:
[tex](x-h)^2+(y-k)^2=r^2[/tex][tex]\begin{gathered} x^2+6x+y^2-16y=23 \\ (x^2+6x+a)+(y^2-16y+b)=(23+a+b) \end{gathered}[/tex]To answer this we have to think of the numbers a and b to form a a trinomial (for x and y) that is reducible to the form (x-h)^2 and (y-k)^2 respectively. We call this completing the square. If you recall the special products below, here we will be doing that but in reverse.
[tex](x\pm y)^2=x^2\pm2xy+y^2[/tex]The value of a is always the square of the half of the coefficient of x ( which in this case is 6). Thus, a = 9. While b is always the square of the half of the coefficient of y ( which in this case is -16). Thus, b = 64. If we substitute the value of a and b , the resulting equation is:
[tex](x^2+6x+9)+(y^2-16y+64)=(23+9+64)[/tex]At this point , you can factor each of the trinomials (of x and y) into a form similar to (x-h)^2 and (y-k)^2 respectively.
[tex]\begin{gathered} (x^2+6x+9)+(y^2-16y+64)=(23+9+64) \\ (x+3)^2+(y-8)^2=96 \end{gathered}[/tex]The standard form of the equation the circle is :
[tex](x+3)^2+(y-8)^2=96[/tex]it gives us the following information about the coordinates of its center and the radius of the circle.
In this equation the center is at (-3,8) and the radius is sq.root of 96.
