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If one of the diameters of the circle $A$ with equation $x^2+y^2-2x-6y+6=0$ is a chord to the circle $B$ with centre $(2, 1)$ then the radius of the circle $B$ is

$(1)\ \sqrt 3$

$(2)\ \sqrt 2$

$(3)\ 3$

$(4)\ 2$

If I draw a perpendicular from $(2,1)$ to that diameter of $A$ then since it is the chord of $B$ the foot of the perpendicular should be $(1,3)$ since the centre of $A$ is the midpoint of that diameter.So the perpendicular distance of that chord of $B$ (equivalently the diameter of $A$) from the centre of $B$ is $\sqrt 5$ and hence the radius of $B$ is $\sqrt {2^2 + (\sqrt 5)^2} =\sqrt 9 = 3$. So $(3)$ is the correct answer.

Is it correct at all? Please verify it.

Thank you in advance.

Arnab Chattopadhyay.
  • 2,570

1 Answers1

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Yes, it is correct. Note that the equation $x^2+y^2-2x-6y+6=0$ can be rewritten as $$(x-1)^2+(y-3)^2=1+9-6=2^2$$ so the circle $A$ has centre $c_A=(1,3)$ and radius $r_A=2$. Therefore the radius of $B$ is given by $$r_B=\sqrt{r_A^2+d(c_A,c_B)^2}=\sqrt {2^2 + (1-2)^2+(3-1)^2} =\sqrt 9 = 3.$$

Robert Z
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