Answer
Standard Form $(x - 2)^2 + (y + 3)^2 = 16$
General Form: $x^2 + y^2 - 4x + 6y - 3 = 0$
Refer to the graph below..
Work Step by Step
The standard form of acircle's equation is $(x-h)^2+(y-k)^2=r^2$ where $r$=radius and $(h, k)$ is the center.
Substitute $r=4$, $h=2$, and $k=-3$in the standard form above to obtain:
\begin{align*}
(x-2)^2+[y-(-3)]^2&=4^2\\
(x-2)^2+(y+3)^2&=16
\end{align*}
General form of the equation of a circle is $x^2+y^2 + ax+by+c=0$.
Subtract $16$ from each side of the equation above to obtain:
\begin{align*}
(x-2)^2+(y+3)^2-16&=16-16\\
x^2-4x+4+y^2+6y+9-16&=0\\
x^2+y^2-4x+6y-3&=0
\end{align*}
With $r=4$ and center at $(2, -3)$, plot the points $4$ units directly above, below, to the left, and to the right of the circle$. Then, connect these four points using a smooth curve to form a circle.
Refer to the graph above.
