Question

Write the equation of the circle in standard form. Then find the center x2 + y2 - 22x - 12y + 121 = 0

Answer 1

Answer:

Standard form:

$(x-11)^2+(y-6)^2=36$

or

$(x-11)^2+(y-6)^2=6^2$

The center is $(11,6)$ and the radius is $6$.

Step-by-step explanation:

$x^2+y^2-22x-12y+121=0$

We will group terms with $x$ together and also group terms with $y$ together.

$x^2-22x+y^2-12y+121=0$

We will not subtract 121 on both sides.

$x^2-22x+y^2-12y=-121$

We are about to complete the square both both the $x$ terms and then the $y$ terms.

Whatever we add on one side, we must add to the other.

$x^2-22x+(\frac{-22}{2})^2+y^2-12y+(\frac{-12}{2})^2=-121+(\frac{-22}{2})^2+(\frac{-12}{2})^2Now let's simplify the right hand side and write the equivalent perfect squares for the left hand side.[tex](x+\frac{-22}{2})^2+(y+\frac{-12}{2})^2=-121+(-11)^2+(-6)^2$

$(x-11)^2+(y-6)^2=-121+121+36$

$(x-11)^2+(y-6)^2=0+36$

$(x-11)^2+(y-6)^2=36$

We can also write it as:

$(x-11)^2+(y-6)^2=6^2$

Now it it easy to compare to:

$(x-h)^2+(y-k)^2=r^2$

to find the center $(h,k)$ and the radius, $r$.

The center is $(11,6)$ and the radius is $6$.