<s>[INST] Coordinate Geometry of circles; Radical Axis question 
If one of the diameters of the circle $x^2+y^2-2x-6y+6=0$ is a chord to the circle with center at $(2, 1)$, then the radius of the second circle is?

Apparently the solution is $3$, with the cryptic reasoning that $r^2= 2^2+ (3−1)^2+ (2−1)^2 ⇒ r =3$.
I have no idea what this means and here is how I solved this problem. The equation of the second circle is $(x-2)^2+ (y-1)^2 = r^2$. Therefore, we can subtract the equation of first circle with that of the second circle to get the equation to the radical axis (which in this case is also the chord they are talking about). 
Equation of chord (radical axis) is $2x -3y +1-r^2=0$
The condition we haven't used so far is that, the center of the first circle $(2,3)$ obtained from the equation lies on the radical axis, because it says the diameter of the first circle is a chord to the second. substituting $(2,3)$ in this equation and solving for $r$, I get $r=2$. Which doesn't match the solution.
(and common sense says if if the diameter is the chord, then the second circle likely has a radius greater than $2$) 
What am I missing? Can someone explain the the cryptic solution? 
 [/INST] Explanation of cryptic solution.
First let us rewrite the given equation in standard form:
$$x^2+y^2-2x-6y+6=0\implies (x-1)^2+(y-3)^2=4$$
Now we know the original circle has radius $r=2$ and is centered at $(1,3)$
Now we can make a convenient right triangle with $r$ the radius of the original circle, $d$ the distance between the centers, and $R$ the radius of the unknown circle:

Hence we have $r^2+d^2=R^2$. The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. Hence $d^2=(1-2)^2+(3-1)^2$, which gives us the final step
$$R^2=r^2+d^2=2^2+(1-2)^2+(3-1)^2=9\implies R=3$$
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