Question

rewrite the equation of the circle $x^{2}+y^{2}-6x - 2y-54 = 0$ into standard form. use the keypad to enter the missing values into the boxes. be sure to include a plus or minus symbol where needed. $(xsquare)^{2}+(ysquare)^{2}=square$

Explanation:

Step1: Complete the square for x - terms

The x - terms in the equation \(x^{2}+y^{2}-6x - 2y-54 = 0\) are \(x^{2}-6x\). Using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), for \(x^{2}-6x\), we have \(a = x\) and \(2b=6\) (so \(b = 3\)), and \(x^{2}-6x=(x - 3)^{2}-9\).

Step2: Complete the square for y - terms

The y - terms are \(y^{2}-2y\). Using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), for \(y^{2}-2y\), we have \(a = y\) and \(2b = 2\) (so \(b = 1\)), and \(y^{2}-2y=(y - 1)^{2}-1\).

Step3: Rewrite the original equation

Substitute the completed - square expressions into the original equation:
\[

$$\begin{align*} x^{2}+y^{2}-6x - 2y-54&=0\\ (x - 3)^{2}-9+(y - 1)^{2}-1-54&=0\\ (x - 3)^{2}+(y - 1)^{2}&=9 + 1+54\\ (x - 3)^{2}+(y - 1)^{2}&=64 \end{align*}$$

\]

Answer:

\((x - 3)^{2}+(y - 1)^{2}=64\)