What is the center and radius of the circle with equation #2(x-2)^2+2(y+5)^2=28#?

1 Answer
Jan 25, 2016

Answer:

Center #(x,y)=(2,-5)#
Radius: #sqrt(14)#

Explanation:

#2(x-2)^2+2(y+5)^2=28#
#color(white)("XXX")#is equivalent to
#(x-2)^2+(y+5)^2=14 # (after dividing by #2#)
or
#(x-2)^2+(y-(-5))^2=(sqrt(14))^2#

Any equation of the form
#color(white)("XXX")(x-a)^2+(y-b)2 = r^2#
is a circle with center #(a,b)# and radius #r#

So the given equation
is a circle with center #(2,-5)# and radius #sqrt(14)#
graph{2(x-2)^2+2(y+5)^2=28 [-7.78, 10, -8.82, 0.07]}