By translating the coordinate axes, reduce to canonical form…
By translating the coordinate axes, reduce to canonical form and identify: $1^\circ$ $x^2+y^2-2x+4y+1=0$; $2^\circ$ $9x^2+4y^2+54x-8y+49=0$; $3^\circ$ $4x^2-16y^2+4x+16y-19=0$.
Explanation of Method
Moving the origin to the centre reduces a circle, an ellipse and a hyperbola to canonical form.
Step 1: Translation of the axes
A translation of the axes to the new origin $O_1(x_0,y_0)$ is
\[ x=x'+x_0,\qquad y=y'+y_0 . \]
We choose $O_1$ by completing the squares; then the linear terms vanish.
Step 2: $1^\circ$ Circle
\[ (x^2-2x+1)+(y^2+4y+4)=-1+1+4\ \Longrightarrow\ (x-1)^2+(y+2)^2=4. \]
With $O_1(1,-2)$, i.e. $x=x'+1,\ y=y'-2$: $x'^2+y'^2=4:\quad\text{circle, centre }(1,-2),\ r=2$.
Step 3: $2^\circ$ Ellipse
\[ 9(x^2+6x+9)+4(y^2-2y+1)=-49+81+4=36 \]
\[ 9(x+3)^2+4(y-1)^2=36\ \Longrightarrow\ \frac{(x+3)^2}{4}+\frac{(y-1)^2}{9}=1. \]
With $O_1(-3,1)$: $\dfrac{x'^2}{4}+\dfrac{y'^2}{9}=1$. Here $b=3>a=2$, so the major axis is parallel to the $y$-axis; $c=\sqrt{9-4}=\sqrt5$, $e=\sqrt5/3$.
\[\frac{x'^2}{4}+\frac{y'^2}{9}=1:\quad\text{ellipse, centre }(-3,1),\ F_{1,2}(-3,\,1\pm\sqrt5)\]
Step 4: $3^\circ$ Hyperbola
\[ 4\left(x^2+x+\tfrac14\right)-16\left(y^2-y+\tfrac14\right)=19+1-4=16 \]
\[ 4\left(x+\tfrac12\right)^2-16\left(y-\tfrac12\right)^2=16\ \Longrightarrow\ \frac{\left(x+\frac12\right)^2}{4}-\left(y-\tfrac12\right)^2=1. \]
With $O_1\left(-\tfrac12,\tfrac12\right)$: $a=2$, $b=1$, $c=\sqrt5$, asymptotes $y'=\pm\tfrac12x'$.
\[\frac{x'^2}{4}-\frac{y'^2}{1}=1:\quad\text{hyperbola, centre }\left(-\tfrac12,\tfrac12\right)\]
Result
\[\boxed{\,x'^2+y'^2=4:\quad\text{circle, centre }(1,-2),\ r=2\,}\]
\[\boxed{\,\frac{x'^2}{4}+\frac{y'^2}{9}=1:\quad\text{ellipse, centre }(-3,1),\ F_{1,2}(-3,\,1\pm\sqrt5)\,}\]
\[\boxed{\,\frac{x'^2}{4}-\frac{y'^2}{1}=1:\quad\text{hyperbola, centre }\left(-\tfrac12,\tfrac12\right)\,}\]