# the equation of the tangent plane at the point (1,-1,2) to the sphere x^2 + y^2 + z^2 - 2x + 4y + 6z -12 = 0

## Expert Answers You need to write the standard form of equation of sphere completing the squares such that:

`(x^2 - 2x + 1) + (y^2 + 4y + 4) + (z^2 + 6z + 9) - 1 - 4 - 9 - 12 = 0`

`(x - 1)^2 + (y + 2)^2 + (z + 3)^2 = 26`

Hence, the center of sphere is at `(1,-2,-3)`  and the radius is `r=sqrt26` .

You need to rewrite the equation of sphere as a function z of x and y such that:

`(z + 3)^2 = 26 - (x - 1)^2- (y + 2)^2`

`z + 3 = sqrt(26 - (x - 1)^2 - (y + 2)^2)`

`z = sqrt(26 - (x - 1)^2 - (y + 2)^2) - 3`

You need to check if the point `(1,-1,2)`  lies on sphere, hence you need to substitute 1,-1,2 for x,y,z in equation `z = sqrt(26 - (x - 1)^2 - (y + 2)^2) - 3`  such that:

`2 = sqrt(26 - (1-1)^2 - (-1+2)^2)`

`2 = sqrt(26-1) - 3 =gt 2 = sqrt25 - 3`

`2 = 5 - 3 =gt 2 = 2 `

This last line proves that the point (1,-1,2) lies on sphere, hence the equation of the tangent plane is:

`z - f(1,-1) = (x - 1)*f_x (1,-1) + (y+1)*f_y (1,-1)`

You need to find `f_x (x,y),`  hence you need to differentiate the equation `z = sqrt(26 - (x - 1)^2 - (y + 2)^2) - 3 ` with respect to x only, taking y as constant such that:

`f_x (x,y) = ((26 - (x - 1)^2 - (y + 2)^2)')/(2sqrt(26 - (x - 1)^2 - (y + 2)^2))`

`f_x (x,y) = (-2(x-1))/(2sqrt(26 - (x - 1)^2 - (y + 2)^2))`

`f_x (x,y) = (1-x)/(sqrt(26 - (x - 1)^2 - (y + 2)^2))`

`f_x (1,-1) = 0`

You need to find `f_y (x,y), ` hence you need to differentiate the equation `z = sqrt(26 - (x - 1)^2 - (y + 2)^2) - 3`  with respect to y only, taking x as constant such that:

`f_y (x,y) = (-2(y+2))/(2sqrt(26 - (x - 1)^2 - (y + 2)^2))`

`f_y (x,y) = (-(y+2))/(sqrt(26 - (x - 1)^2 - (y + 2)^2))`

`f_y (1,-1) =(-(-1+2))/(sqrt(26 - (1 - 1)^2 - (-1 + 2)^2))`

`f_y (1,-1) = (-1)/sqrt(26-1)`

`f_y (1,-1) = -1/5`

Substituting 0 for `f_x (1,-1)`  and -1/5 for `f_y (1,-1)`  in equation of tangent plane yields:

`z - 2 = (x - 1)*0 + (y+1)*(-1/5)`

`z = -y/5 - 1/5 + 2 =gt `

`=gt z = -y/5 + 9/5 =gt`

`=gt z = (9-y)/5`

Hence, the equation of tangent plane at the point (1,-1,2) to the sphere `x^2 + y^2 + z^2 - 2x + 4y + 6z -12 = 0`  is `z = (9-y)/5` .

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