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I'm learning analytic geometry, specifically the theory of quadratic surfaces, and need help with the following exercise:

Find the equation of the sphere of radius $R = \frac{3\sqrt{14}}{7}$ which touches the plane $E:3x_1-x_2+2x_3+6=0$ at the point $P(1,9,0)$.

Even though the problem looks simple, I can't seem to figure it out. The point $P$ lies on the sphere. Is there a way I can find the centre of the sphere given the informations of the problem? Any help would be much appreciated.

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    Hint: any radius of a sphere is normal to the tangent plane. – 2017-01-26
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    @lulu Thank you for the hint. I understand what it means but it is still unclear to me how I should continue from here. From the problem we also know that the normal vector of the plane is given by $N=(3, -1, 2)$. Do I have to use this information to solve the problem and how it is useful in this context? – 2017-01-26
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    Let $C$ be the center of the circle (if we knew the coordinates of $C$, the problem would be easy, yes?). The vector $\vec v=P-C$ is a radius for the sphere. But that means that $\vec v$ is normal to the plane $E$, hence $\vec v$ must be a multiple of the normal vector you (correctly) provided. As we know the length of $\vec v$.... – 2017-01-26
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    Worth noting: the problem says "the" sphere but of course there are two, one on either side of the plane. In the context of what I wrote, that happens because writing $\vec v=\lambda \vec n$ and knowing the length of $\vec v$ only determines $\lambda$ up to sign. – 2017-01-26

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