Find the equation of the tangent line to the circle $3x^2+3y^2+3z^2-2x-3y-4z=22;3x+4y+5z=26$ at the point $(1,2,3)$.

Question

I try take the centre of the sphere, then equation of the perpendicular line joining the centre of circle and the centre of sphere, then find the centre of circle.Tell me if I am going right way?

Answer 1

You are doing well. Once you have the center $O$ of the circle, find the equation of the plane through $P=(1,2,3)$ and perpendicular to $OP$. The intersection between that plane and the plane of the circle is the tangent you want.

Answer 2

Let $(x,y,z)=(1,2,3)+t(a,b,c)$ is an equation of out tangent line.

Thus, $3a+4b+5c=0$.

Rewrite an equation of our sphere in the following form: $$\left(x-\frac{1}{3}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{2}{3}\right)^2=\frac{293}{36}$$

Thus, $\left(1-\frac{1}{3}\right)a+\left(2-\frac{1}{2}\right)b+\left(3-\frac{2}{3}\right)c=0$ or $4a+9b+14c=0$,

which with $3a+4b+5c=0$ gives $b=-2c$ and $a=c$ and we get the answer: $$(x,y,z)=(1,2,3)+t(1,-2,1)$$