Question

rewrite the equation of the circle (x^{2}+y^{2}+4x - 6y-23 = 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

For the $x$ - terms in the equation $x^{2}+4x + y^{2}-6y - 23=0$, we use the formula $(a + b)^2=a^{2}+2ab + b^{2}$. For $x^{2}+4x$, we have $a = x$ and $2ab = 4x$, so $b = 2$. Then $x^{2}+4x=(x + 2)^{2}-4$.

Step2: Complete the square for y - terms

For the $y$ - terms $y^{2}-6y$, using the formula $(a - b)^2=a^{2}-2ab + b^{2}$ with $a = y$ and $2ab = 6y$ (so $b = 3$), we get $y^{2}-6y=(y - 3)^{2}-9$.

Step3: Rewrite the original equation

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

$$\begin{align*} (x + 2)^{2}-4+(y - 3)^{2}-9-23&=0\\ (x + 2)^{2}+(y - 3)^{2}&=4 + 9+23\\ (x + 2)^{2}+(y - 3)^{2}&=36 \end{align*}$$

\]

Answer:

$(x + 2)^{2}+(y - 3)^{2}=36$