If one of the diameters of the circle is a chord to the circle with centre (2, 1), then the radius of the circle is [2004]

Question

If one of the diameters of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord to the circle with centre (2, 1), then the radius of the circle is [2004]

Smart Achievers · Accepted Answer

(3)

The given circle is $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ with centre C(1, 3) and radius $= \sqrt{1 + 9 - 6} = 2$ .

Let AB be one of its diameter which is the chord of other circle with centre at $C_{1} (2, 1)$ .

$Then in ∆ C_{1} C B,$

$C_{1} B^{2} = C C_{1}^{2} + C B^{2}$

$= {(2 - 1)}^{2} + {(1 - 3)}^{2} + {(2)}^{2}$

$= 1 + 4 + 4 = 9 \Rightarrow C_{1} B = 3$

Solution

Q.
If one of the diameters of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord to the circle with centre (2, 1), then the radius of the circle is [2004]

1 $\sqrt{3}$

2 $\sqrt{2}$

3 $3$

4 $2$

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