Respuesta :
Answer:
Standard form:
[tex](x-11)^2+(y-6)^2=36[/tex]
or
[tex](x-11)^2+(y-6)^2=6^2[/tex]
The center is [tex](11,6)[/tex] and the radius is [tex]6[/tex].
Step-by-step explanation:
[tex]x^2+y^2-22x-12y+121=0[/tex]
We will group terms with [tex]x[/tex] together and also group terms with [tex]y[/tex] together.
[tex]x^2-22x+y^2-12y+121=0[/tex]
We will not subtract 121 on both sides.
[tex]x^2-22x+y^2-12y=-121[/tex]
We are about to complete the square both both the [tex]x[/tex] terms and then the [tex]y[/tex] terms.
Whatever we add on one side, we must add to the other.
[tex]x^2-22x+(\frac{-22}{2})^2+y^2-12y+(\frac{-12}{2})^2=-121+(\frac{-22}{2})^2+(\frac{-12}{2})^2
Now let's simplify the right hand side and write the equivalent perfect squares for the left hand side.
[tex](x+\frac{-22}{2})^2+(y+\frac{-12}{2})^2=-121+(-11)^2+(-6)^2[/tex]
[tex](x-11)^2+(y-6)^2=-121+121+36[/tex]
[tex](x-11)^2+(y-6)^2=0+36[/tex]
[tex](x-11)^2+(y-6)^2=36[/tex]
We can also write it as:
[tex](x-11)^2+(y-6)^2=6^2[/tex]
Now it it easy to compare to:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
to find the center [tex](h,k)[/tex] and the radius, [tex]r[/tex].
The center is [tex](11,6)[/tex] and the radius is [tex]6[/tex].
