How do you write the standard from of the equation of the circle given center C ( 6, - 2 ), radius 2?

2 Answers
Jul 23, 2018

Answer:

#(x-6)^2+(y+2)^2=4#

Explanation:

#"the equation of a circle in standard form is"#

#•color(white)(x)(x-a)^2+(y-b)^2=r^2#

#"where "(a,b)" are the coordinates of the centre and r is"#
#"the radius"#

#"here "(a,b)=(6,-2)" and "r=2#

#(x-6)^2+(y+2)^2=4larrcolor(blue)"equation of circle"#

Jul 23, 2018

Answer:

#(x-6)^2+(y+2)^2=4#

Explanation:

Recall that the equation of a circle is given by

#bar( ul(|color(white)(2/2)(x-h)^2+(y-k)^2=r^2color(white)(2/2)|))#, with center #(h,k)# and radius #r#

We know that we are centered at #(6,-2)# and have a radius of #2#, so #h=6, k=-2# and #r^2=4#.

We can plug these into our equation to get

#(x-6)^2+(y+2)^2=4#

Hope this helps!