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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

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  • $\begingroup$ How have you found your first equation? $\endgroup$ Sep 8 '15 at 17:09
  • $\begingroup$ @Aretino From squaring the first 2. $\endgroup$
    – Guy
    Sep 8 '15 at 17:10
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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.

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Rewrite the equations of the spheres: $$ x^2+y^2+z^2=1 $$ $$ (x+1)^2+(y-1)^2+(z+1)^2=4 $$ $$ \left(x-\frac{1}{2}\right)^2+(y+2)^2+\left(z-\frac{1}{2}\right)^2=\frac{13}{2} $$ So: $$ \begin{array}{ccc} \text{sphere}&\text{radius}&\text{center}\\ 1.& 1 & (0,0,0)^T\\ 2.& 2 & (-1,1,-1)^T\\ 3.& \sqrt{13/2} &(1/2,-2,1/2)^T \end{array} $$ The length of a tangent to a sphere centered at $(x,y,z)$ with radius $r$ from $(a,b,c)$ is: $$ \sqrt{(x-a)^2+(y-b)^2+(z-c)^2-r^2} $$ From here, it's not hard to finish.

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