If a straight line through C(-sqrt{8}, sqrt{8 | vedclass

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(C) The equation of the line passing through $C(-\sqrt{8}, \sqrt{8})$ with an angle of inclination $	heta = 135^\circ$ is given by $y - y_1 = m(x - x

Vedclass · Accepted Answer

$10$

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(C) The equation of the line passing through $C(-\sqrt{8}, \sqrt{8})$ with an angle of inclination $	heta = 135^\circ$ is given by $y - y_1 = m(x - x_1)$.Here,$m = 	an(135^\circ) = -1$.So,$y - \sqrt{8} = -1(x + \sqrt{8})$,which simplifies to $y - \sqrt{8} = -x - \sqrt{8}$,or $x + y = 0$.The circle is given by $x = 5\cos	heta, y = 5\sin	heta$,which corresponds to $x^2 + y^2 = 25$.The distance from the center $(0, 0)$ to the line $x + y = 0$ is $d = \frac{|0 + 0|}{\sqrt{1^2 + 1^2}} = 0$.Since the distance from the center to the line is $0$,the line passes through the center of the circle.Therefore,the line $x + y = 0$ is a diameter of the circle.The length of the diameter $AB$ is $2r = 2 	imes 5 = 10$.

If a straight line through $C(-\sqrt{8}, \sqrt{8})$ making an angle of $135^\circ$ with the $x$-axis cuts the circle $x = 5\cos\theta, y = 5\sin\theta$ at points $A$ and $B$,then the length of $AB$ is

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