Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation.
$x^2 + y^2 - 2x - 4y - 31 = 0$

the equation in standard form is \\(\square\\).
(simplify your answer.)

the center of the circle is \\(\square\\).
(type an ordered pair.)

the radius of the circle is \\(r = \square\\).

use the graphing tool to graph the circle.
\\(\text{click to enlarge graph}\\)

Explanation:

Step1: Group x and y terms

Rearrange the equation \(x^{2}+y^{2}-2x - 4y-31 = 0\) as \((x^{2}-2x)+(y^{2}-4y)=31\).

Step2: Complete the square for x

For \(x^{2}-2x\), take half of -2, which is -1, square it: \((-1)^{2}=1\). Add 1 to both sides: \((x^{2}-2x + 1)+(y^{2}-4y)=31 + 1\).

Step3: Complete the square for y

For \(y^{2}-4y\), take half of -4, which is -2, square it: \((-2)^{2}=4\). Add 4 to both sides: \((x^{2}-2x + 1)+(y^{2}-4y + 4)=31 + 1+4\).

Step4: Write as perfect squares

\((x - 1)^{2}+(y - 2)^{2}=36\). The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. So center is \((1,2)\) and radius \(r=\sqrt{36}=6\).

Answer:

The equation in standard form is \((x - 1)^{2}+(y - 2)^{2}=36\).
The center of the circle is \((1,2)\).
The radius of the circle is \(r = 6\).