How do you find the center and radius of the circle #x^2+y^2-2x+6y-26=0#?

1 Answer
Jun 24, 2018

#"centre "=(1,-3)" radius "=6#

Explanation:

#"the equation of a circle in standard form is"#

#color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))#

#"where "(a,b)" are the coordinates of the centre and r"#
#"is the radius"#

#"to obtain this form "color(blue)"complete the square"#
#"on both the x and y terms"#

#x^2-2x+y^2+6y=26#

#x^2+2(-1)xcolor(red)(+1)+y^2+2(3)ycolor(magenta)(+9)=26color(red)(+1)color(magenta)(+9)#

#(x-1)^2+(y+3)^2=36larrcolor(blue)"in standard form"#

#"with centre "=(1,-3)" radius "=sqrt36=6#
graph{((x-1)^2+(y+3)^2-36)=0 [-20, 20, -10, 10]}