Question

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

Answer 1

`"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`

Answer 2

`"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`