Question

a circle in the xy - plane has its center at (-5,2) and has a radius of 9. an equation of this circle is $x^{2}+y^{2}+ax + by + c = 0$, where $a,b$, and $c$ are constants. what is the value of $c$?

Explanation:

Step1: Write standard - form of circle equation

The standard - form of the equation of a circle with center \((h,k)\) and radius \(r\) is \((x - h)^2+(y - k)^2=r^2\). Here, \(h=-5\), \(k = 2\), and \(r = 9\). So the equation is \((x+5)^2+(y - 2)^2=81\).

Step2: Expand the equation

Expand \((x + 5)^2+(y - 2)^2=81\).
\((x + 5)^2=x^{2}+10x + 25\) and \((y - 2)^2=y^{2}-4y+4\).
So \(x^{2}+10x + 25+y^{2}-4y + 4=81\).

Step3: Rearrange to general form

Rearrange the equation \(x^{2}+y^{2}+10x-4y+25 + 4-81=0\), which simplifies to \(x^{2}+y^{2}+10x-4y-52=0\).

Answer:

\(-52\)