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## Homework Statement

I know that x+y+z=0 is a tangent plate to my sphere at (0,0,0).

I want to find the tangent plane for my sphere at (1,-2,3).

## Homework Equations

Suppose that his is my sphere [tex](x-a)^2+(y-b)^2+(z-c)^2=R^2[/tex]

(a,b,c) is the center of my sphere while R is the radius of it.

## The Attempt at a Solution

First thing I did is to find the radius: Because (0,0,0) is on my sphere i know that [tex](0-a)^2+(0-b)^2+(0-c)^2=a^2+b^2+c^2=R^2[/tex]

So [tex]R=\sqrt {a^2+b^2+c^2}[/tex]

No I tried to find the center of my sphere:

First thing is to spot that [tex]((\frac {1} {\sqrt{3}},\frac {1} {\sqrt{3}},\frac {1} {\sqrt{3}})[/tex] is orthogonal to the plane x+y+z=0 so it's direction is towards the radius. [I understand that there are 2 orthogonal directions to the plane but the "second" direction is not relevant because it contradicts point (1,-2,3)]

It means that [tex](a,b,c)=(0,0,0)+\sqrt{a^2+b^2+c^2} (\frac {1} {\sqrt{3}},\frac {1} {\sqrt{3}},\frac {1} {\sqrt{3}})[/tex] from this I concluded that a=b=c.

Now when i plug a=b=c=t into the sphere equation I get: -t+2t-3t+14=0 so t = 7/2 =a=b=c.

It means that the tangent plane to my sphere at (1-2,3) should be z=-5x-11y-14

but when i plug it into my Grapher [program which translate equations into graphics] it showing that i completely missed with my sphere equation and with the tangent plane.

What I did wrong?