Find the center and radius of circle #(x-3)^3+(y+4)^2=10#?

1 Answer
Feb 7, 2017

Coordinates of the centre of the circle are #(3,-4)# and circle's radius is #sqrt10#.

Explanation:

Perhaps you mean the equation to be #(x-3)^2+(y+4)^2=10#

As #(x-3)^2+(y+4)^2=10# is equivalent to

#(x-3)^2+(y-(-4))^2=(sqrt10)^2#

or #sqrt((x-3)^2+(y-(-4))^2)=sqrt10#

The Left hand side shows that the distance of #(x,y)# from point #(3,-4)#

and as the equation indicates that this is always #sqrt10#

Equation #(x-3)^2+(y+4)^2=10# indicates that #(x,y)# is always at a distance of #sqrt10# from #(3,-4)#.

Hence, coordinates of the centre of the circle are #(3,-4)# and circle's radius is #sqrt10#.