Question

write (x^{2}+y^{2}-18x + 8y+5 = 0) in standard form.
1 group terms and move the constant to the other side of the equation.
(x^{2}-18x + y^{2}+8y=-5)
2 determine the values that need to be added to both sides of the equation.
((-18div2)^{2}=81) and ((8div2)^{2}=16)
3 add the values to both sides of the equation.
4 write each trinomial as a binomial squared, and simplify the right side.
what is the standard form of the equation of a circle given by (x^{2}+y^{2}-18x + 8y+5 = 0)?
((square)^{2}+(square)^{2}=square)

Explanation:

Step1: Group terms and move constant

Group $x$ - terms and $y$ - terms and move the constant to the right - hand side.
$x^{2}-18x + y^{2}+8y=-5$

Step2: Complete the square for $x$ and $y$ terms

For the $x$ - terms: The coefficient of $x$ is $-18$. Half of it is $\frac{-18}{2}=-9$, and $(-9)^{2}=81$. For the $y$ - terms: The coefficient of $y$ is $8$. Half of it is $\frac{8}{2}=4$, and $4^{2}=16$.
Add 81 and 16 to both sides of the equation:
$x^{2}-18x + 81+y^{2}+8y + 16=-5+81 + 16$

Step3: Write as binomial squares and simplify

$(x - 9)^{2}+(y + 4)^{2}=92$

Answer:

$(x - 9)^{2}+(y + 4)^{2}=92$