How do I can simplify the below intersection?

Question

let $S$ be a sphere defined by this equation :$ \left(x-\frac{4}{3}\right)^2+\left(y-\frac{1}{3}\right)^2+\left(z-\frac{5}{3}\right)^2=\frac{25}{36}$ and $(p)$ the plane defined by the following equation :$2x+y-2z+4=0$ ,

My question here is : How do I can get the intersection between $(p)$ and $(S)$? By the way I want to get the center and the radius of the circle which it is the geometrical intersection of $(p)$ and $(S)$ .

Attempt : I took $y=2(z-x-2)$ from the equation of $(p)$ and by substitution in $(S)$ I have got this complicated form

Answer 1

\begin{align*} \text{Radius of the sphere} &= \frac{5}{6} \\ \text{Distance of the centre from the plane} &= \frac{2\left( \frac{4}{3} \right)+ \left( \frac{1}{3} \right)- 2\left( \frac{5}{3} \right)+4} {\sqrt{2^2+1^2+2^2}} \\ &= \frac{11}{3} \\ &> \frac{5}{6} \end{align*}

Hence, no intersections between plan and sphere.

Answer 2

The perpendicular distance from the centre of the sphere to the plane is $$\frac{\frac 83+\frac 13-\frac {10}{3}+4}{3}=\frac{11}{9}$$ but the radius of the sphere is only $\frac 56$ so there is no intersection.

Answer 3

It is worth it to try to solve the general problem, instead of this specific one.

Since we know the center of the sphere $(x_C,y_C,z_C)$, we can transform the coordinates such that the origin is at the sphere center: $x=x'+x_C,\,y=y'+y_C,\,z=z'+z_C$.

Sphere: $$ x'^2 + y'^2 + z'^2 - R^2 = 0$$

Plane: $$ a x' + b y' + c z' + d' = 0 $$ where $ d'= a x_C + b y_C + c z_C +d$

So for every value of $x'$ and $y'$ the value of $z'$ is calculated such that it lies on the plane $$z' = - \frac{a x'+b y'+d'}{c} $$ which is used in the equation of the sphere to intersect with the plane

$$ C(x',y') = x'^2 + y'^2 + \left(- \frac{a x'+b y'+d'}{c} \right)^2 - R^2 =0 $$

Now we can find the center of the circle by solving this 2×2 system for $x'$ and $y'$

$$ \begin{Bmatrix} \frac{\partial }{\partial x'} C(x',y') =0 \\ \frac{\partial }{\partial y'} C(x',y') =0 \end{Bmatrix} $$

$$ \mbox{center} = \begin{Bmatrix} -\frac{a d'}{a^2+b^2+c^2} \\ -\frac{b d'}{a^2+b^2+c^2} \\ -\frac{c d'}{a^2+b^2+c^2} \end{Bmatrix} $$

The circle radius $r$ is found from the distance of the plane to the center of the circle. This distance is $$t = -\frac{d'}{\sqrt{a^2+b^2+c^2}}$$

and from trigonometry we know that $R^2 = r^2 + t^2$ so the radius of the circle is $$ r = \sqrt{R^2 -t^2}$$