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