The diameter of the circle x^2 + y^2 - 2x - 6 | vedclass

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(C) The given circle is $x^2 + y^2 - 2x - 6y + 6 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -3, c = 6$.The center of this

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$3$

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(C) The given circle is $x^2 + y^2 - 2x - 6y + 6 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1, f = -3, c = 6$.The center of this circle is $(-g, -f) = (1, 3)$ and its radius $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$.The diameter of this circle is a chord $AB$ of another circle with center $D(2, 1)$.The length of the chord $AB$ is the diameter of the first circle,so $AB = 2 	imes r_1 = 2 	imes 2 = 4$.Let $C$ be the midpoint of the chord $AB$. Since $AB$ is the diameter of the first circle,its center is $(1, 3)$. Thus,the midpoint $C$ of the chord is $(1, 3)$.The distance $d$ from the center $D(2, 1)$ to the chord $AB$ is the distance between $D(2, 1)$ and $C(1, 3)$.$d = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.In the right-angled triangle $ACD$,the radius $R$ of the second circle is the hypotenuse $AD$.$R^2 = AC^2 + d^2$.Since $AB = 4$,$AC = \frac{AB}{2} = 2$.$R^2 = 2^2 + (\sqrt{5})^2 = 4 + 5 = 9$.$R = \sqrt{9} = 3$.

The diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is a chord of another circle with center $(2, 1)$. Find the radius of this circle.

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