If a chord of the circle $x^{2} + y^{2} - 4 x - 2 y - c = 0$ is trisected at the points (1/3, 1/3) and (8/3, 8/3), then

Length of the chord=

7 \sqrt{2}

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c=20

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Radius of the circle 25

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c=25

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Solution

The correct option is C Radius of the circle 25
Equation of the line joining the given points is y=x and the distance between them is
$\sqrt{(7 / 3)^{2} + (7 / 3)^{2}} i . e ., 7 \sqrt{2} / 3.$

If PQ is the chord of the given circle which is trisected at these points then equation of PQ is y=x and the length of this chord is $3 \times \frac{7 \sqrt{2}}{3} = 7 \sqrt{2} .$
Let C(2, 1) be the centre of the given circle and CL be the perpendicular from C to PQ, then the radius of the given circle is
$\sqrt{P L^{2} + C L^{2}} = (7 / \sqrt{2})^{2} + (1 / \sqrt{2})^{2} = 25$

$\Rightarrow (2)^{2} + (1)^{2} + c = 25$
$\Rightarrow c = 20.$

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Similar questions

If a chord of the circle $x^{2} + y^{2} - 4 x - 2 y - c = 0$ is trisected at the points (1/3, 1/3) and (8/3, 8/3), then

x^{2} + y^{2} - 4 x - 2 y - c = 0

(\frac{1}{3}, \frac{1}{3})

(\frac{8}{3}, \frac{8}{3}),

x^{2} + y^{2} + 4 x + 1 = 0

x^{2} + y^{2} + 6 x + 2 y + 3 = 0

x^{2} + y^{2} - 2 x - 2 y - 2 = 0

120^{\circ}

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