Question

put the equation in standard form.
x² + y² + 4x - 8y + 13 = 0
(x + ?)² + (y - )² =

Explanation:

Step1: Group x - terms and y - terms

$(x^{2}+4x)+(y^{2}-8y)+13 = 0$

Step2: Complete the square for x - terms

For $x^{2}+4x$, we know that $(a + b)^2=a^{2}+2ab + b^{2}$. Here $a = x$ and $2ab=4x$, so $b = 2$ and $x^{2}+4x=(x + 2)^{2}-4$.

Step3: Complete the square for y - terms

For $y^{2}-8y$, since $2ab=-8y$ and $a = y$, then $b=-4$ and $y^{2}-8y=(y - 4)^{2}-16$.

Step4: Substitute into the equation

$((x + 2)^{2}-4)+((y - 4)^{2}-16)+13 = 0$.

Step5: Simplify the equation

$(x + 2)^{2}+(y - 4)^{2}-4-16 + 13=0$.
$(x + 2)^{2}+(y - 4)^{2}-7 = 0$.
$(x + 2)^{2}+(y - 4)^{2}=7$.

Answer:

The first blank is $2$, the second blank is $4$, and the third blank is $7$.