121. Construct the circle $x^{2}+y^{2}+6 x-4 y-3=0$.
Solution. To construct a circle, it is necessary to find its center and radius. For this, we need to complete the squares in the equation, i.e., transform the equation into the form $(x-a)^{2}+(y-b)^{2}=r^{2}$. We have

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

By adding the sum $9+4$ to both sides, we get $\left(x^{2}+6 x+9\right)+\left(y^{2}-4 y+4\right)=3+9+4$, i.e., $(x+3)^{2}+(y-2)^{2}=16$.

From this equation, it is clear that $a=-3, b=2, r=4$, i.e., the center of the circle is the point $O(-3 ; 2)$, and its radius is 4.