The origin lies within the circle with equation \[x^2+ax+y^2+by=c\] precisely when
\(c > 0\)
\(a^2 + b^2 > c\)
\(a^2 + b^2 < c\)
\(a^2 + b^2 > 4c\)
\(a^2 + b^2 < 4c\).
We can write the equation of the circle as
\[\left(x+\dfrac{a}{2}\right)^2 -\left(\dfrac{a}{2}\right)^2+\left(x+\dfrac{b}{2}\right)^2 -\left(\dfrac{b}{2}\right)^2 = c,\]
or
\[\left(x+\dfrac{a}{2}\right)^2 +\left(x+\dfrac{b}{2}\right)^2 = c + \left(\dfrac{a}{2}\right)^2 +\left(\dfrac{b}{2}\right)^2.\]
Thus the region inside the circle is denoted by
\[\left(x+\dfrac{a}{2}\right)^2 +\left(x+\dfrac{b}{2}\right)^2 < c + \left(\dfrac{a}{2}\right)^2 +\left(\dfrac{b}{2}\right)^2.\]
Does \((0,0)\) satisfy this inequality? If and only if
\[\left(0+\dfrac{a}{2}\right)^2 +\left(0+\dfrac{b}{2}\right)^2 < c + \left(\dfrac{a}{2}\right)^2 +\left(\dfrac{b}{2}\right)^2,\]
which is true if and only if \(0 < c\), and the answer is (a).
