How do you identify the center and radius of the circle #x^2+(y+12)^2=24#?

2 Answers
Feb 23, 2017

#"centre "(0,-12)" radius "r=2sqrt6#

Explanation:

#"The standard equation of a circle with centre "(a,b)" and radius " r " is"#

#(x-a)^2+(y-b)^2=r^2#

#"We have "x^2+(y+12)^2=24#

#"comparing this with the standard eqn."#

#a=0, b=-12#

#r^2=24=>r=sqrt24=2sqrt6#

#"centre "(0,-12)" radius "r=2sqrt6#

Feb 23, 2017

#"centre "=(0,-12)," radius "=2sqrt6#

Explanation:

The standard form of the #color(blue)"equation of a circle"# 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, the radius.

#x^2+(y+12)^2=24" is in this form"#

by comparison with the standard form equation.

#a=0,b=-12" and "r^2=24#

#r^2=24rArrr=sqrt24=2sqrt6#

#rArr"centre "=(a,b)=(0,-12)" and "r=2sqrt6#