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

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Answer 1 · 2015-09-01T18:08:32

#(2,-3,6)#

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))#

Answer 2 · 2015-09-01T18:10:47

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

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