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 (2, 1), then the radius of the circle is:
$\sqrt3,\sqrt2,3,2$
I have no clue as to where to begin this question. I tried drawing a diagram, but that didn't help me.
So, the radius of the given circle $=\sqrt{1^2+3^2-6}=2$
Using Perpendicular Bisector of Chord Passes Through Centre
Observe that the distance between the two centres , this radius($=2$) & the other radius(hypotenuse) form a Pythagorean triangle