How do you graph (x-3)^2 + (y-2)^2 = 9?

Jun 27, 2016

circle: centre (3 ,2), radius= 3

Explanation:

The standard form of the equation of a circle is.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (a ,b) are the coordinates of the centre and r, the radius.

The equation: ${\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = 9$ is in this form and therefore is a circle.

Comparison with the standard equation gives a = 3 , b = 2, r = 3

Thus circle centre (3 ,2) and radius = 3
graph{y^2-4y+x^2-6x+4=0 [-10, 10, -5, 5]}