# Find the equation of tangent plane to the surface x^2+y^2+z^2=25 at the point P(-3,0,4) .

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The general equation for a plane is defined by any vector n normal to the plane:

**n **dot (**r** - **r0**) = 0

where **r** is the position vector, and **r0** is going to be the point P in the question (that it, it is any point in the plane).

**n** is defined by the gradient of the function. The gradient is normal to the surface defined by the equation.

grad [ x^2+y^2+z^2 - 25 ] = ( 2x, 2y, 2z ) = **n**

so n = (-6, 0, 8 ) at P

**n **dot (**r** - **r0**) = 0

nx(x - x0) + ny(y - y0) + nz(z - z0) = 0

-6(x + 3) + 0 (y - y0) + 8(z - 4) = 0

**-6x + 8z - 50 = 0**

We know that x^2+y^2+z^2 = 25 is a sphere with radius 5 and centre at the origin O(0,0,0) from the solid geometry.

The point (-3,0, 4) is in the (x,z) plane (x,z) = (3,4)

The the tangent plane to the sphere, therefore , cuts the plane (x,z), along the tangent line to the circle x^2+z^2 = 25 in (x,z) plane.

The equation of the tanget to the circle x^2+z^2 = 25 at point (3,4) is given by:

z-4 = m(3,4)(x-(-3)), where m(3,4) is the value of dz/dx at (-3,4).

To find dz/dx , we write z^2 = 25-x^2. Therefore, differentiating z^2= 25-x^2 we get:

2z dz/dx = (25-x^2)' = -2x.

Therefore dz/dx = -x/z .

m(3,4) = (dz/dx at x=-3, z = 4) = -(-3)/4 = 3/4.

Therefore the equation of tangent at (-3,4) in (x,z) plane is given by:

z-4 = (3/4)(x+3).

4(z-4) = 3x+9.

4z-3x- 25 = 0 is the tangent line in the (x,z) plane.

Or -3x+0y +4z = 25 is the tangent plane to the sphere x^2+y^2+z^2= 25 in the 3 dimensional space at the point (3,0,4).

Hope this helps.