How to write this conic equation in standard form? $$x^2+y^2-16x-20y+100=0$$
Standard form? Circle or ellipse?
 A: \begin{align}x^2+y^2-16x-20y+100=0&\iff(x-8)^2-64+(y-10)^2-100+100=0\\&\iff(x-8)^2+(y-10)^2=8^2\end{align}
so it's the equation of a circle with center $(8,10)$ and radius equal 8
A: $$
\begin{align} 
x^2+y^2-16x-20y+100 & = 0 \\ \\
\color{blue}{\bf x^2 -16x} + \color{red}{\bf y^2 -20y } + 100 & = 0
\end{align}
$$
We complete the square: $$x^2+bx+(b/2)^2-(b/2)^2+c= (x+b/2)^2+c-(b/2)^2$$
$$\begin{align}
\color{blue}{\bf x^2 - 16x} + \underbrace{\bf \left(\frac{-16}{2}\right)^2}_{\color{green}{\bf \large +64}} + \color{red}{\bf y^2 - 20y} + \underbrace{\bf \left(\frac{-20}{2}\right)^2}_{\color{green}{\bf \large + 100}} + 100 {\color{green}{\bf - 64 -100}} & = 0 \\ \\
(x^2 - 16x + 64) + (y^2 - 20 y + 100) - 64 & = 0\\ \\
(x - 8)^2 + (y - 10)^2 - 64 & = 0\\ \\
(x - 8)^2 + (y - 10)^2 & = 64 = (8)^2 \\
\end{align}
$$
The equation of a circle with center $(a, b)$ and radius $r$ is given by $$(x - a)^2 + (y - b)^2 = r^2$$
In your case, we have a circle centered at $(8, 10)$ with radius $r = 8$.
A: Recall that one of the usual standard forms is:
$(x - a)^{2} + (y - b)^{2} = r^{2}$ where...


*

*(a,b) is the center of the circle

*r is the radius of the circle


Rearrange the terms to obtain:
$x^{2} - 16x + y^{2} - 20y + 100 = 0$
Then, by completing the squares, we have:
$(x^{2} - 16x + 64) + (y^{2} - 20y + 100) = 64$
$(x - 8)^{2} + (y - 10)^{2} = 8^{2}$
Thus, we have a circle with radius $r = 8$ and center $(a,b) = (8,10)$
