Find sphere equation

Question

How to find sphere equation with center in plane $\pi =\{x-5y+z-2=0\}$ and tangent to plane $\alpha =\{ 2x+3y-4z-1=0 \}$ at $M(1,1,1)$

Where I've stopped:

Knowing the sphere equation is: $$(X-a)^2 + (Y-b)^2 + (Z-c)^2 = r^2$$ Center of sphere is $C(a,b,c)$ and can be found in plane $\pi$

$r^2$ is radius and can be found doing distance from center of sphere to tangent point M ( $d(C,M)$ ) My problem is center point, how can I find this? My first thought was something using plane $\pi$ but then plane $\alpha$ was given in question and never used.

Answer 1

Hint: Radius and tangent are perpendicular. Where does the perpendicular to $\alpha$ at $M$ hit $\pi$?

A different way: Parametrize the plane $\pi$, then find for which value of the parameters the vector $\vec{MP}$ (with $P\in\pi$) is orthogonal to $\alpha$.

Answer 2

$$MC: (1,1,1)+t(2,3,-4).$$ Let $C(1+2t,1+3t,1-4t).$

Thus, $$1+2t-5(1+3t)+1-4t-2=0,$$ which gives $t=-\frac{5}{17},$ $$C\left(\frac{7}{17},\frac{2}{17},\frac{37}{17}\right),$$ $$MC=\sqrt{\frac{725}{289}}$$ and the answer is $$\left(x-\frac{7}{17}\right)^2+\left(y-\frac{2}{17}\right)^2+\left(z-\frac{37}{17}\right)^2=\frac{725}{289}.$$

Answer 3

The center of the sphere is the intersection of the plane $\pi $ with the line passing through $M$ and perpendicular to $\alpha$

The equation of the line passing through M and perpendicular to $\alpha$ is $$x=1+2t, y=1+3t, z=1-4t$$

The point of intersection with $\pi $ is $$ C(7/{17},2/{17},37/{17})$$which is the center of the sphere.

The radius is simply $CM$ the distance between $C$ and $M$.