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Determine the equation of a sphere of radius $\sqrt{5}$, whose center belongs to the line $$d : {{x}\over1} = {{y-1}\over -1} = {{z+2}\over 1} $$ and it is passing through the point $A(0,2,-1)$

My solution: If the center belongs to the line $d$, it means that the line is tangent to the sphere and so the distance from the center $C(x,y,z)$ to the line $d$ is equal with R = $\sqrt{5}$. After doing all the computation I got the following result: $$2x^2+2y^2+2z^2-6x-6y+2xy-2xz+2yz+6 = \sqrt{15}$$ Now, this isn't really the equation of the sphere I'm looking for because the radius here is $\sqrt{15}$. Could you tell me where I went wrong in my solution?

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  • $\begingroup$ You say "If the center belongs to the line d, it means that the line is tangent to the sphere". I rather think that this means that there is a plane tangent to the sphere orthogonal to the line. You may try to understand the situation by looking at a circle through a point, a line passing through the center. $\endgroup$ Feb 8 at 13:35
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From the equations $$d : {{x}\over1} = {{y-1}\over -1} = {{z+2}\over 1} $$ we get $$y=1-x;\;z=x-2$$ So the line $d$ has parametric equations $$ \begin{cases} x=t\\ y=1 - t\\ z= -2 + t\\ \end{cases} $$ Thus the center of the sphere has coordinates $C(t,1-t,-2+t)$

The sphere passes through $A(0,2,-1)$ and radius is $r=\sqrt 5$

Therefore $AC^2=5$ that is $$t^2+(t+1)^2+(t-1)^2=5$$ which gives $t=\pm 1$

The center of the sphere can be $$C_1(1,0,-1);\;C_2(-1,2,-3)$$

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