Equation of sphere

Question

Find equation of the sphere if $$\frac{3}{2}x-\frac{1}{2}y+\frac{7}{2}z=14 $$ is a tangent plane at the point $(2,-1,3)$.

What I have done so far.

from the tangent plane, the normal is $(\frac{3}{2},-\frac{1}{2},\frac{7}{2})$ and the radius is $\sqrt{14}$

I think what I need now is the center point but how can I find that

Answer 1

Tangent plane to $f(x,y,z) = k$ at the point $P = (a,b,c)$ is given by $$ f_x (x-a) + f_y(y-b) + f_z(z-c) = 0 $$ where each partial derivatives of $f$ is evaluated at the point $P$.

Let $f(x,y,z) = (x-p)^2 + (y-q)^2+ (z-s)^2 = r^2$ then we have $f_x = 2(x-p)$, $f_y = 2(y-q)$ and $f_z = 2(z-s)$, at the point $P$ we have $f_x = 4 - 2p$ , $f_y = -2 - 2q$ and $f_z = 6 -2s$. Then the tangent plane equation becomes $$ (4-2p)(x-2) + (-2-2q)(y+1) + (6-2s)(z-3) = 0 $$ Compare this with the given tangen equation and determine the value of $p$,$q$ and $s$. Note that for any real number $t \ne 0$ , the equation $3tx-yt+7zt = 28t$ is still tangent equation (This explains why there are infinite such spheres!). Then it is straightforward to find the value of $p$,$q$ and $s$. To evaluate $r$, observe that the point $(2,-1,3)$ is part of the sphere.

I hope you can finish rest of the calculations.

Answer 2

Take any real number $t\ne0$ and consider the point $(x_0,y_0,z_0)=(2,1,-3)+t\left(\frac32,-\frac12,\frac72\right)$. Then the plane $\frac32x-\frac12y+\frac72z=14$ is tangent to the sphere$$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=t^2\left(\left(\frac32\right)^2+\left(-\frac12\right)^2+\left(\frac72\right)^2\right)$$at $(2,1,-3)$. The radius of such a sphere is $|t|\sqrt{\left(\frac32\right)^2+\left(-\frac12\right)^2+\left(\frac72\right)^2}$.