Answer
Standard Form: $(x - 4)^2 + (y + 3)^2 = 25$
General Form $x^2 + y^2 - 8x + 6y = 0$
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 $5$ to $r$, $h=4$, and $k=-3$in the standard form above to obtain:
\begin{align*}
(x-4)^2+[y-(-3)]^2&=5^2\\
(x-4)^2+(y+3)^2&=25
\end{align*}
General form of the equation of a circle is $x^2+y^2 + ax+by+c=0$.
Subtract $25$ from each side of the equation above to obtain:
\begin{align*}
(x-4)^2+(y+3)^2-25&=25-25\\
x^2-8x+16+y^2+6y+9-25&=0\\
x^2+y^2-8x+6y&=0
\end{align*}
With $r=5$ and center at $(4, -3)$, plot the points $5$ 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.
