If the straight line 4x + 3y + lambda = 0 tou | vedclass

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(D) The given equation of the circle is $2(x^2 + y^2) = 5$,which can be rewritten as $x^2 + y^2 = \frac{5}{2}$.Comparing this with the standard form $

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$\frac{5\sqrt{10}}{2}$

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(D) The given equation of the circle is $2(x^2 + y^2) = 5$,which can be rewritten as $x^2 + y^2 = \frac{5}{2}$.Comparing this with the standard form $x^2 + y^2 = r^2$,the radius $r = \sqrt{\frac{5}{2}}$.The center of the circle is $(0, 0)$.$A$ line $ax + by + c = 0$ touches a circle with center $(h, k)$ and radius $r$ if the perpendicular distance from the center to the line equals the radius.Thus,$\frac{|4(0) + 3(0) + \lambda|}{\sqrt{4^2 + 3^2}} = \sqrt{\frac{5}{2}}$.$\frac{|\lambda|}{\sqrt{16 + 9}} = \sqrt{\frac{5}{2}}$.$\frac{|\lambda|}{5} = \sqrt{\frac{5}{2}}$.$|\lambda| = 5 	imes \frac{\sqrt{5}}{\sqrt{2}} = \frac{5\sqrt{5}}{\sqrt{2}} 	imes \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{10}}{2}$.Therefore,$\lambda = \pm \frac{5\sqrt{10}}{2}$.

If the straight line $4x + 3y + \lambda = 0$ touches the circle $2(x^2 + y^2) = 5$,then $\lambda$ is

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