Show that the locus of the point from which equal tangents may be drawn to the spheres $x^2 +y^2 +z^2 =1$, $x^2 + y^2 +z^2 + 2x – 2y +2z -1 = 0$ , $x^2 + y^2 +z^2 – x +4y -z – 2 = 0$ is the straight line $\frac{x – 1}{2} = \frac{y-2}{5} = \frac{z – 1}{3}$
I have tried :
Let $P(x ,y , z)$ be the point whose locus is required. Since length of the tangents to the three spheres are equal
$$ \sqrt{x^2 +y^2 +z^2 -1} = \sqrt{x^2 + y^2 +z^2 + 2x – 2y +2z -1 } = \sqrt{x^2 + y^2 +z^2 – x +4y -z – 2}$$
from the last to members,we get
$$x- y + z = 0,$$
from the last two members , we get
$$ 3x – 6y +8z +1 = 0 ,$$
from the first and last member, we get
$$x – 4y + 6z +1 = 0$$
Please tell me how to find the equation of line.
any help would be appreciated, Thank you
Best Answer
Notice that two equation are enough to specify a line, so for instance you could give the solution as the line given by $x-y+z=0$ and $x-4y+6z=-1$ (your first and third equation, but every other pair would do), and that would be perfectly legitimate. The other equation is a combination of these, so it is not necessary.
These can be rewritten in several different ways. For instance, you can eliminate $x$ by subtracting the second from the first one and get $3y-5z=1$, that is $3(y-2)=5(z-1)$. Similarly, you can eliminate $y$ and get $3(x-1)=2(z-1)$. These two equations are equivalent to yours and are those given by the textbook.