42. A circle is defined by its center $C$, which is determined by the radius vector $\overrightarrow{O C}=\vec{c}$, and its radius $R$. Write the equation of the circle in vector and coordinate form, if the coordinates of the point $C$ are $(a ; b)$. How do these equations look depending on the position of the center: a) the center lies on the $O x$ axis; b) the center lies on the $O y$ axis; c) the center coincides with the origin $O$, - and in the case where the circle passes through the origin $O$?
42. Let $\vec{r}$ be a variable vector. The magnitude of the distance $|\vec{r}-\vec{c}|$ is constant and equal to the radius $R$ of the circle. From this, we obtain:

$$
(\vec{r}-\vec{c})^{2}=R^{2}
$$

This is the equation of a circle with center $C(O \vec{C}==\vec{c})$ and radius $R$. If $\vec{r}=x \vec{i}+y \vec{j} ; \quad \vec{c}=a \vec{i}+b \vec{j}, \quad$ then $\quad \vec{r}-\vec{c}=(x-a) \vec{i}+(y-b) \vec{j}$, and therefore the equation of the circle in coordinate form is: $(x-a)^{2}+(y-b)^{2}=R^{2}$. If the center lies on the $O x$ axis, then $b=0$, and we get:

$$
(x-a)^{2}+y^{2}=R^{2}
$$

If the center lies on the $O y$ axis, then $a=0$, and we get:

$$
x^{2}+(y-b)^{2}=R^{2}
$$

If the center coincides with the origin, then $a=0$ and $b=0$, and the equation becomes:

$$
x^{2}+y^{2}=R^{2}
$$

If the circle passes through the origin, then the distance from the origin to the center is equal to the radius $R^{2}=a^{2}+b^{2}$. Substituting this value into the canonical equation of the circle, we get:

$$
x^{2}-2 a x+y^{2}-2 b y=0
$$

This is the equation of a circle passing through the origin.