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

1 Answer
Apr 8, 2017

Answer:

#(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]}