Let a circle C pass through the points (4, 2) | vedclass

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(B) Let the centre of the circle be $O(h, k)$. Since the circle passes through $A(4, 2)$ and $B(0, 2)$,the perpendicular bisector of $AB$ must pass th

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$2 \sqrt{3}$

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(B) Let the centre of the circle be $O(h, k)$. Since the circle passes through $A(4, 2)$ and $B(0, 2)$,the perpendicular bisector of $AB$ must pass through the centre $O$.The midpoint of $AB$ is $M(\frac{4+0}{2}, \frac{2+2}{2}) = (2, 2)$.Since $A$ and $B$ have the same $y$-coordinate,$AB$ is a horizontal line. Its perpendicular bisector is the vertical line $x = 2$.Thus,the $x$-coordinate of the centre is $h = 2$.Given that the centre lies on $3x + 2y + 2 = 0$,we substitute $x = 2$:$3(2) + 2k + 2 = 0$ $\Rightarrow 6 + 2k + 2 = 0$ $\Rightarrow 2k = -8$ $\Rightarrow k = -4$.So,the centre is $O(2, -4)$.The radius $r$ is the distance from $O(2, -4)$ to $A(4, 2)$:$r^2 = (4 - 2)^2 + (2 - (-4))^2 = 2^2 + 6^2 = 4 + 36 = 40$.Let the chord have midpoint $N(1, 2)$. The distance $ON$ from the centre $O(2, -4)$ to $N(1, 2)$ is:$ON^2 = (1 - 2)^2 + (2 - (-4))^2 = (-1)^2 + 6^2 = 1 + 36 = 37$.The length of the chord is $2 \sqrt{r^2 - ON^2} = 2 \sqrt{40 - 37} = 2 \sqrt{3}$.

Let a circle $C$ pass through the points $(4, 2)$ and $(0, 2)$,and its centre lie on the line $3x + 2y + 2 = 0$. Then the length of the chord of the circle $C$ whose midpoint is $(1, 2)$ is:

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