Let (x1,y1) and (x2,y2) be end point of diameter of circle, then equation of circle be (x−x1)(x−x2)+(y−y1)(y−y2)=0 ...(i) Given, equation of circle is x2+y2−2x−6y+6=0 ⇒x2−2x+1+y2−6y+5=0 ⇒(x−1)2+(y−1)(y−5)=0 ⇒(x−1)(x−1)+(y−1)(y−5)=0 ...(ii) Compare Eqs. (i) and (ii), we obtain end points of diameter be (1,1) and (1,5). Now, we have to find radius of circle having centre at (2,1) and one of its chord having end point (1,1) and (1,5)
From given circle,x2+y2=r2 x=√(2−1)2+(1−3)2=√5 y=√(1−1)2+(3−1)2=√4 ⇒r2=x2+y2=5+4=9 ⇒r=3