Question

write the equation in the form $(x - h)^2+(y - k)^2 = c$. then, if the equation represents a circle, identify the center and radius. if the equation represents the degenerate case, give the solution set.
$x^{2}+y^{2}+10x - 236 = 0$

Explanation:

Step1: Group the $x$ - and $y$ - terms

$x^{2}+10x + y^{2}=236$

Step2: Complete the square for the $x$ - terms

For the $x$ - terms in $x^{2}+10x$, we know that for $ax^{2}+bx$, to complete the square we add $(\frac{b}{2})^{2}$. Here $a = 1$, $b = 10$, so we add $(\frac{10}{2})^{2}=25$ to both sides of the equation.
$x^{2}+10x + 25+y^{2}=236 + 25$

Step3: Rewrite in standard form

Using the perfect - square formula $(a + b)^2=a^{2}+2ab + b^{2}$, where $a=x$ and $b = 5$ for $x^{2}+10x + 25$, we get $(x + 5)^2+y^{2}=261$.
The standard form of a circle is $(x - h)^2+(y - k)^2=r^{2}$, where $(h,k)$ is the center and $r$ is the radius. Comparing $(x + 5)^2+y^{2}=261$ with $(x - h)^2+(y - k)^2=r^{2}$, we have $h=-5$, $k = 0$, and $r=\sqrt{261}$.

Answer:

The equation in standard form is $(x + 5)^2+y^{2}=261$. The center is $(-5,0)$ and the radius is $\sqrt{261}$.