$\left[\begin{array}{l}\text { Method of Coordinates on the Plane } \\ {[\text { Circles (other) }]}\end{array}\right]$

Find the radius and coordinates of the center of the circle given by the equation

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

b) $x^{2}+y^{2}-2(x-3 y)-15=0$

c) $x^{2}+y^{2}=x+y+\frac{1}{2}$.
a) A circle with radius $R$ and center at point $A(a ; b)$ has the equation

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

In this case, $a=3, b=-2, R=4$.

b)

$$
\begin{aligned}
x^{2}+y^{2}-2(x-3 y)-15= & 0 \Leftrightarrow x^{2}-2 x+1+y^{2}+6 y+9-1-9-15=0 \Leftrightarrow \\
& \Leftrightarrow(x-1)^{2}+(y+3)^{2}=25
\end{aligned}
$$

Therefore, $a=1, b=-3, R=5$.

c)

$$
\begin{aligned}
x^{2}+y^{2}=x+y+ & \frac{1}{2} \Leftrightarrow x^{2}-x+\frac{1}{4}+y^{2}-y+\frac{1}{4}+\frac{1}{4}+\frac{1}{2}+\frac{1}{2} \Leftrightarrow \\
& \Leftrightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=1
\end{aligned}
$$

Therefore, $a=b=\frac{1}{2}, R=1$.

## Answer

a) $(3 ;-2), R=4 ;$ b) $(1 ;-3), R=5 ;$ c) $\left(\frac{1}{2} ; \frac{1}{2}\right), R=1$.