# Consider the sphere S of radius 5 centred at (1, 0, 2). Find the equation of the plane tangent to S at the point (5,-3, 2).Hint : To get started, draw this situation in 2D with a circle.

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The radius vector `vec r` from the center of the sphere to the point of tangency will be normal (perpendicular) to the tangent plane. This vector is

`vec r=(5,-3,2)-(1,0,2)=(4,-3,0).`

The equation of the tangent plane with normal vector `vec r` that passes through the point `(a,b,c)` is `vec r * (x-a,y-b,z-c)=0.` In this case, this equation becomes

`(4,-3,0)*(x-5,y+3,z-2)=4(x-5)-3(y+3)=0.` **Simplifying gives**

**`4x-3y=29` as the equation of the plane. **

Below is an analogous case in two dimensions. Note that the red tangent line, in this case, is vertical because the radius vector has zero y-coordinate. Similarly in three dimensions, the tangent plane is vertical because the radius vector has zero z-coordinate. Also note that the radius vector is normal to the tangent line.