Question

How do you write the equation `x^2+x+y^2-6/5y=1` in standard form and find the center and radius?

Answer 1

`(x+1/2)^2+(y-3/5)^2=161/100`

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.

`"Rearrange the given equation into this form"`

`"Using the method of "color(blue)"completing the square"`

`(x^2+xcolor(red)(+1/4))+(y^2-6/5ycolor(magenta)(+9/25))=1color(red)(+1/4)color(magenta)(+9/25)`

`rArr(x+1/2)^2+(y-3/5)^2=161/100`

`"here " a=-1/2,b=3/5" and " r^2=161/100`

`rArr"centre "=(-1/2,3/5)`

`"and " r=sqrt(161/100)=sqrt161/10~~1.27" to 2 dec. places"`
graph{(x+1/2)^2+(y-3/5)^2-161/100=0 [-10, 10, -5, 5]}