Question

Find the standard equation of the sphere with center (-3,2,4) that is tangent to the plane given by \(2 x+4 y-3 z=8\).

Short answer

The standard equation of the sphere tangent to the given plane is \((x + 3)^2 + (y - 2)^2 + (z - 4)^2 = 4\).

Step-by-step solution

Setup Distance Formula for Plane

Substitute the normal vector and the constants of the plane into the formula for finding the distance D from a point to a plane. The plane equation becomes \(x+2y-\frac{3}{2}z-4=0\) after dividing through by 2, so the normal vector is (1, 2, -3/2) and the constant d is -4.

Calculate Radius

Substitute the coordinates of the sphere's center and the d value into the distance formula: r = \( \frac{|1*(-3) + 2*2 - \frac{3}{2}*4 + 4|} {\sqrt{1^2+2^2+(-3/2)^2}} \). This gives r = 2 after simplifying.

Substitute Radius into Sphere's Equation

Substitute r = 2 into the equation of the sphere. Therefore, the final equation is \((x + 3)^2 + (y - 2)^2 + (z - 4)^2 = 2^2\), which simplifies to \((x + 3)^2 + (y - 2)^2 + (z - 4)^2 = 4\).