1. Write the given circle in standard form

The circle is $x^2+y^2-2x-6y+6=0.$

Complete the squares: $x^2-2x+y^2-6y+6=0$ $(x-1)^2-1+(y-3)^2-9+6=0$ $(x-1)^2+(y-3)^2=4.$

So the given circle has:

• Center $O=(1,3)$
• Radius $r=2$

2. Understand the diameter condition

Any diameter of this circle has length $2r=4.$

The question says that one of the diameters of this circle is a chord of another circle $C$ whose center is at $(2,1)$.

So, in circle $C$:

• a chord has length $4$
• the center is $P=(2,1)$

3. Use the property of a chord

A line segment can be a chord of circle $C$ only if its two endpoints lie on circle $C$. Since the segment is a diameter of the first circle, its midpoint is the center of the first circle, i.e. $O=(1,3)$.

Hence the chord of circle $C$ has midpoint $O=(1,3)$.

In any circle, the perpendicular from the center to a chord passes through its midpoint. Therefore the distance from the center $P=(2,1)$ of circle $C$ to this chord equals the distance from $P$ to its midpoint $O$: $PO=\sqrt{(2-1)^2+(1-3)^2}=\sqrt{1+4}=\sqrt{5}.$

4. Relate chord length, distance from center, and radius

If a circle has radius $R$, and a chord is at distance $d$ from the center, then $\left(\frac{\text{chord length}}{2}\right)^2+d^2=R^2.$

Here,

• chord length $=4$, so half-chord $=2$
• $d=\sqrt{5}$

Thus, $R^2=2^2+(\sqrt{5})^2=4+5=9.$ So, $R=3.$

5. Final answer

The radius of circle $C$ is $\boxed{3}.$