Question

x^{2}+y^{2}+10x + 10y+46 = 0

Explanation:

Step1: Rewrite the equation in standard - form

The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius. Given $x^{2}+y^{2}+10x + 10y+46 = 0$. Complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}+10x=(x + 5)^{2}-25$.
For the $y$ - terms: $y^{2}+10y=(y + 5)^{2}-25$.
So the equation becomes $(x + 5)^{2}-25+(y + 5)^{2}-25+46 = 0$.

Step2: Simplify the equation

$(x + 5)^{2}+(y + 5)^{2}-25-25 + 46=0$, which simplifies to $(x + 5)^{2}+(y + 5)^{2}=4$.
The center of the circle is $(-5,-5)$ and the radius $r = 2$.

Answer:

The circle has center $(-5,-5)$ and radius $2$. To determine which graph is correct, look for a circle centered at $(-5,-5)$ with a radius of 2 units.