A circle passes through the points (2, 3) and | vedclass

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(D) Let the center of the circle be $(h, k)$.Since the center lies on the line $y - 4x + 3 = 0$,we have $k - 4h + 3 = 0$,or $k = 4h - 3$.The distance

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(D) Let the center of the circle be $(h, k)$.Since the center lies on the line $y - 4x + 3 = 0$,we have $k - 4h + 3 = 0$,or $k = 4h - 3$.The distance from the center $(h, k)$ to the points $(2, 3)$ and $(4, 5)$ must be equal (radius $r$).$(h - 2)^2 + (k - 3)^2 = (h - 4)^2 + (k - 5)^2$$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 8h + 16 + k^2 - 10k + 25$$-4h - 6k + 13 = -8h - 10k + 41$$4h + 4k = 28 \Rightarrow h + k = 7$.Substituting $k = 4h - 3$ into $h + k = 7$:$h + (4h - 3) = 7$ $\Rightarrow 5h = 10$ $\Rightarrow h = 2$.Then $k = 4(2) - 3 = 5$.The center is $(2, 5)$.The radius $r$ is the distance between $(2, 5)$ and $(2, 3)$:$r = \sqrt{(2 - 2)^2 + (5 - 3)^2} = \sqrt{0^2 + 2^2} = 2$.

$A$ circle passes through the points $(2, 3)$ and $(4, 5)$. If its centre lies on the line $y - 4x + 3 = 0$,then its radius is equal to

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