Through the line x+28y-2z+17=0, 5x+8y-z+1=0 , find the plane…

Through the line $x+28y-2z+17=0,\ 5x+8y-z+1=0$, find the plane that $1^\circ$ is tangent to the sphere $x^2+y^2+z^2=1$, and $2^\circ$ cuts it in a great circle.

Step 1: Pencil

\[ (1+5\lambda)x+(28+8\lambda)y-(2+\lambda)z+(17+\lambda)=0 . \]

The member $\lambda\to\infty$, $5x+8y-z+1=0$, is at distance $\tfrac1{\sqrt{90}}$ from $O$. It is neither tangent nor through $O$, so the $\lambda$-form misses nothing here.

Step 2: $1^\circ$ Tangent planes

\[ (17+\lambda)^2=(1+5\lambda)^2+(28+8\lambda)^2+(2+\lambda)^2\ \Longrightarrow\ 89\lambda^2+428\lambda+500=0, \]

\[ \lambda_1=-2,\qquad \lambda_2=-\tfrac{250}{89}. \]

\[3x-4y-5=0\qquad\text{and}\qquad 387x-164y-24z-421=0\]

Checks: $\tfrac5{\sqrt{9+16}}=1$ and $\tfrac{421}{\sqrt{177241}}=\tfrac{421}{421}=1$. $\checkmark$

The tangent points are $\left(\tfrac35,-\tfrac45,0\right)$ and $\tfrac1{421}(387,-164,-24)$.

Step 3: $2^\circ$ Great circle

The plane must pass through $O$: $17+\lambda=0$, so $\lambda=-17$, which gives $-84x-108y+15z=0$.

\[28x+36y-5z=0\]