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 is

Answer 1

Equation of a circle with it's center at (2,1), and radius rr is x^2+y^2-4x-2y+5-r^2=0$

The other given circle is x2+y2−2x−6y+6=0x2+y2−2x−6y+6=0 (Note: it's center can be worked out to be =(1,3)=(1,3)

The equation of the common chord can be got by subtracting the above two equations, →→ 2x−4y+1=r22x−4y+1=r2

This chord is the diameter of the second circle, which means that it's center (1,3) must be a solution of the above equation.

⇒2×1−4×3+1=r2→r2=9→r=3