How do you find the equation of the circle given Center at the origin; r=squareroot 5?

1 Answer
Sep 1, 2016

Answer:

#x^2+y^2=5#

Explanation:

The standard form of the equation of a circle is.

#color(red)(|bar(ul(color(white)(a/a)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(a/a)|)))#
where (a ,b) are the coordinates of the centre and r, the radius.

here the centre is at the origin, hence a = 0 , b = 0
and #r=sqrt5#

substitute into the standard equation.

#(x-0)^2+(y-0)^2=(sqrt5)^2" simplifying to"#

#x^2+y^2=5" the equation of the circle"#

However, it is worth noting and remembering that when the circle is centred at the origin, the standard equation reduces to.
#color(red)(|bar(ul(color(white)(a/a)color(black)(x^2+y^2=r^2)color(white)(a/a)|)))#