185. Reduce the equations of the circles to their standard form:
1) $x^{2}+y^{2}-4 x+6 y-3=0$
2) $4 x^{2}+4 y^{2}-8 x+4 y-11=0$
Solution. We will bring the first equation to the standard form as follows:

$$
\left(x^{2}-4 x+4\right)+\left(y^{2}+6 y+9\right)-4-9-3=0
$$

from which

$$
(x-2)^{2}+(y+3)^{2}=16
$$

consequently, the center of the circle is at point $C(2 ;-3)$, and the radius $R=4$.

We will bring the second equation to the standard form. By dividing all terms of the equation by the coefficient of $x^{2}$, which is 4, we get

$$
x^{2}+y^{2}-2 x+y-\frac{11}{4}=0
$$

Now, in the left part of the equation, we will add and subtract the squares of half the coefficients of $x$ and $y$ and write it as follows:

$$
\left(x^{2}-2 x+1\right)+\left(y^{2}+y+\frac{1}{4}\right)-1-\frac{1}{4}-\frac{11}{4}=0
$$

from which

$$
(x-1)^{2}+\left(y+\frac{1}{2}\right)^{2}=4
$$

Consequently, the coordinates of the center of the circle are $a=1$, $b=-\frac{1}{2}$, i.e., the center of the circle is at point $C\left(1 ;-\frac{1}{2}\right)$, and the radius $R=2$.