Question

The radius of the circle, having centre at $(2,1)$ whose one of the chord is a diameter of the circle $x^2+y^2-2x-6y+6=0$

Answer 1

Given Circle equation is $x^2+y^2-2x-6y+6=0$

So the centre and radius is $(1,3)$ and $\sqrt{1^2+3^2-6}=\sqrt{1+9-6}=2$

Let the radius of the required circle be $r$

Distance between the two centres is $d=\sqrt{(2-1{)}^2+(1-3{)}^2}=\sqrt{1+4}=\sqrt{5}$

So the radius of the circle $r=\sqrt{(\sqrt{5}{)}^2+2^2}=\sqrt{5+4}=3$