Question

in the xy - plane, the graph of the given equation (x^{2}+52x + y^{2}=0) is a circle. what are the coordinates ((x,y)) of the center of the circle? a ((0,26)) b ((0, - 26)) c ((26,0)) d ((-26,0))

Explanation:

Step1: Recall circle - equation form

The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center of the circle. Given the equation $x^{2}+52x + y^{2}=0$, we complete the square for the $x$ - terms.

Step2: Complete the square for $x$

For the $x$ - terms in $x^{2}+52x + y^{2}=0$, we use the formula $(x + m)^2=x^{2}+2mx+m^{2}$. In $x^{2}+52x$, we have $2m = 52$, so $m = 26$ and $x^{2}+52x=(x + 26)^2-26^{2}$. The equation becomes $(x + 26)^2-26^{2}+y^{2}=0$, or $(x + 26)^2+y^{2}=26^{2}$.

Step3: Identify the center

Comparing $(x + 26)^2+y^{2}=26^{2}$ with the standard form $(x - a)^2+(y - b)^2=r^2$, we have $a=-26$ and $b = 0$. So the center of the circle is $(-26,0)$.

Answer:

D. $(-26,0)$