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?
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 $$C_1(1,0,-1);\;C_2(-1,2,-3)$$