You can complete the square to find the equation of the circle in standard form.
\(0=x^2+y^2+Cx+Dy+E\)
\(\dfrac{C^2+D^2}4=x^2+Cx+\dfrac{C^2}4+y^2+Dy+\dfrac{D^2}4+E\)
\(\dfrac{C^2+D^2}4-E=\left(x+\dfrac C2\right)^2+\left(y+\dfrac D2\right)^2\)
So the circle is centered at \(\left(-\dfrac C2,-\dfrac D2\right)\) and has radius \(\sqrt{\dfrac{C^2+D^2}4-E}\).
A horizontal shift to the left can be done by adding a positive number to the \(x\) term, as in
\(\dfrac{C^2+D^2}4-E=\left(x+\dfrac C2+\mathbf k\right)^2+\left(y+\dfrac D2\right)^2\)
Expanding, you end up with
\(x^2+(C+2k)x+y^2+Dy+E+Ck=0\)
So the coefficient \(C\) is increased by twice the horizontal shift, while \(D\) remains unchanged.
