What is the center of the sphere with the equation #(x-2)^2+(y+3)^2+(z-6)^2=36#?

2 Answers
Sep 1, 2015

Answer:

#(2,-3,6)#

Explanation:

The center of any sphere written in the form:
#(x-a)+(y-b)+(z-c)=r^2" "#, is #(a,b,c)#

#(x-2)^2+(y+3)^2+(z-6)^2=36# can be written as:

#(x-2)^2+(y-(-3))^2+(z-6)^2=(6)^2#

hence center is at #color(blue)((2","-3","6))#

Sep 1, 2015

Answer:

The centre has coordinates: #(+2,-3,+6)#

Explanation:

If an equation of a circle (a sphere) is given in such form, then:

1) The coordinates of the centre are values next to #x,y# (and #z#) with changed signs,

2) Radius is the square root of the value on right hand side