Thread: Help with equation of plane tangent to surface

x^2 + y^2 +z^2 = 1
Let (a,b,c) be a point on the surface, find the equation of plane tangent to this surface at (a, b, c)

Do I put it in terms of z?
f(x,y) = z = sqroot( 1-x^2-y^2)
Solve for fx(x,y) and fy(x,y) and plugging in 'a' and 'b' for x and y

z= (fx(a,b)(x-a) + (fy(a,b)(y-b) + c

Is this correct? Thanks in advance.

Hi

Here is a simple way but I do not know if it is in the spirit of the exercise
x² + y² + z² = 1 is the sphere (center O, radius 1)
A plane tangent to the sphere at point M(a,b,c) is perpendicular to vector OM whose coordinates are (a,b,c)
Therefore an equation of the plane is ax + by + cz + k = 0

To find k you just have to substitute (a,b,c) to (x,y,z), M being on the plane
a² + b² + c² + k = 0 therefore k = - a² - b² - c²

An equation of the plane is ax + by + cz - a² - b² - c² = 0