Question

in the xy - plane, the graph of the given equation 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) x² + 52x + y² = 0

Explanation:

Step1: Recall circle - standard form

The general equation of a circle is \((x - a)^2+(y - b)^2=r^2\), and the general expanded form is \(x^{2}-2ax + a^{2}+y^{2}-2by + b^{2}-r^{2}=0\). For the given equation \(x^{2}+52x + y^{2}=0\), we need to complete the square for the \(x\) - terms.

Step2: Complete the square for \(x\)

We know that \(x^{2}+52x=(x + 26)^{2}-676\). So the equation \(x^{2}+52x + y^{2}=0\) can be rewritten as \((x + 26)^{2}-676+y^{2}=0\), or \((x + 26)^{2}+y^{2}=676\).

Step3: Identify the center

Comparing \((x + 26)^{2}+y^{2}=676\) with the standard - form \((x - a)^2+(y - b)^2=r^2\), where the center of the circle is \((a,b)\). Here \(a=-26\) and \(b = 0\). So the center of the circle is \((-26,0)\).

Answer:

d. \((-26,0)\)