How do you graph x^2+y^2-6x-14y+42=0?

Sep 1, 2015

This formula describes a circle, which center is in $O = \left(3 , 7\right)$ and radius is $4$

Explanation:

To find this out you have to transform the formula to:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$.

In this formula $a$ and $b$ are coordinates of the center and $r$ is the radius

${x}^{2} + {y}^{2} - 6 x - 14 y + 42 = 0$

$\left({x}^{2} - 6 x \textcolor{red}{+ 9}\right) + \left({y}^{2} - 14 y \textcolor{red}{+ 49}\right) \textcolor{red}{- 9 - 49} + 42 = 0$

${\left(x - 3\right)}^{2} + {\left(y - 7\right)}^{2} - 16 = 0$

${\left(x - 3\right)}^{2} + {\left(y - 7\right)}^{2} = 16$