How do you find the center and radius for #2(x-2)^2+2(y+5)^2=28#?

1 Answer
May 17, 2018

Answer:

The equation #2(x-2)^2+2(y-5)^2=28# represents a circle with center #(2,5)# and radius #sqrt14#.

Explanation:

We can write #2(x-2)^2+2(y-5)^2=28#

as #(x-2)^2+(y-5)^2=14#

or #(x-2)^2+(y-5)^2=(sqrt14)^2#

or #sqrt((x-2)^2+(y-5)^2)=sqrt14#

this means the point #(x,y)# on the curve moves so that its distance #sqrt((x-2)^2+(y-5)^2)# from point #(2,5)# is always #sqrt14# i.e. it forms a circle with center #(2,5)# and radius #sqrt14#.

Hence, the equation #2(x-2)^2+2(y-5)^2=28# represents a circle with center #(2,5)# and radius #sqrt14#.

graph{(2(x-2)^2+2(y-5)^2-28)((x-2)^2+(y-5)^2-0.03)=0 [-8.79, 11.21, -0.6, 9.4]}