214. By transforming the coordinates through a parallel shift of the axes, simplify the equation of the circle

$$
x^{2}+y^{2}-4 x+6 y-68=0
$$

taking the new origin as the center of the circle.
Solution. Let's find the center of the circle, i.e., find the origin of the new coordinate system. We will convert the given equation of the circle to its standard form:

$$
(x-2)^{2}+(y+3)^{2}=81
$$

From this, it is clear that the center of the circle lies at the point $C(2, -3)$, i.e., we take the new origin as point $O_{1}$ with coordinates $\alpha=2$, $\beta=-3$. Using the formulas (4), we find:

$$
x=X+2, \quad y=Y-3 \quad \text{or} \quad X=x-2, \quad Y=y+3
$$

Substituting these values into the original equation of the circle or into the standard form, we respectively obtain

$$
X^{2}+Y^{2}=81
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-045.jpg?height=325&width=417&top_left_y=469&top_left_x=687)

Fig. 21