# The equation of a sphere of a given radius, whose center belongs to a given line and it is passing through a point

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?

• 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. Feb 8 at 13:35

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