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

1 Answer
Jun 27, 2016

Answer:

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

Explanation:

The standard form of the equation of a circle is.

#color(red)(|bar(ul(color(white)(a/a)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(a/a)|)))#
where (a ,b) are the coordinates of the centre and r, the radius.

The equation: #(x-3)^2+(y-2)^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]}