Question
If one of the diameters of the circle $${x^2} + {y^2} - 2x - 6y + 6 = 0$$ is a chord to the circle with centre (2, 1), then the radius of the circle is-
A.
$$\sqrt 3 $$
B.
$$\sqrt 2 $$
C.
$$3$$
D.
$$2$$
Answer :
$$3$$
Solution :
The given circle is $${x^2} + {y^2} - 2x - 6y + 6 = 0$$ with centre $$C\left( {1,\,3} \right)$$ and radius $$ = \sqrt {1 + 9 - 6} = 2$$
Let $$AB$$ be one of its diameter which is the chord of other circle with centre at $${C_1}\left( {2,\,1} \right)$$
Then in $$\Delta {C_1}CB,$$
$$\eqalign{ & {C_1}{B^2} = CC_1^2 + C{B^2} \cr & {r^2} = \left[ {{{\left( {2 - 1} \right)}^2} + {{\left( {1 - 3} \right)}^2}} \right] + {\left( 2 \right)^2} \cr & \Rightarrow {r^2} = 1 + 4 + 4 \cr & \Rightarrow {r^2} = 9 \cr & \Rightarrow r = 3 \cr} $$
The given circle is $${x^2} + {y^2} - 2x - 6y + 6 = 0$$ with centre $$C\left( {1,\,3} \right)$$ and radius $$ = \sqrt {1 + 9 - 6} = 2$$
Let $$AB$$ be one of its diameter which is the chord of other circle with centre at $${C_1}\left( {2,\,1} \right)$$
Then in $$\Delta {C_1}CB,$$
$$\eqalign{ & {C_1}{B^2} = CC_1^2 + C{B^2} \cr & {r^2} = \left[ {{{\left( {2 - 1} \right)}^2} + {{\left( {1 - 3} \right)}^2}} \right] + {\left( 2 \right)^2} \cr & \Rightarrow {r^2} = 1 + 4 + 4 \cr & \Rightarrow {r^2} = 9 \cr & \Rightarrow r = 3 \cr} $$