Given that, one of the diameter of the circle (x−2)2+(y−32)2=6 is a chord of another circle (x−22)2+(y−22)2=r2.
Let, C and C1 be the centres of the circles (x−2)2+(y−32)2=6 and (x−22)2+(y−22)2=r2 respectively.
So, C(\sqrt{2},3\sqrt{2})&{C}_{1}(2\sqrt{2},2\sqrt{2})
The position of the two circles as shown in figure.

Now, CC1=(22−2)2+(22−32)2=2+2=2
It is clear from the above diagram, ΔACC1 is a right angled triangle.⇒r2=(AC)2+(CC1)2=(6)2+22 ⇒r2=10