For what value of lambda does the line 3x - 4 | vedclass

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(B) The given circle equation is $x^2 + y^2 - 4x - 8y - 5 = 0$.Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) The given circle equation is $x^2 + y^2 - 4x - 8y - 5 = 0$.Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -4$,and $c = -5$.The center of the circle is $(-g, -f) = (2, 4)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-4)^2 - (-5)} = \sqrt{4 + 16 + 5} = \sqrt{25} = 5$.$A$ line $Ax + By + C = 0$ touches a circle if the perpendicular distance from the center to the line equals the radius.The line is $3x - 4y - \lambda = 0$.The distance $d = \frac{|3(2) - 4(4) - \lambda|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 - 16 - \lambda|}{\sqrt{9 + 16}} = \frac{|-10 - \lambda|}{5}$.Setting $d = r$,we get $\frac{|-10 - \lambda|}{5} = 5$,which implies $|-10 - \lambda| = 25$.This gives two cases: $-10 - \lambda = 25 \Rightarrow \lambda = -35$ or $-10 - \lambda = -25 \Rightarrow \lambda = 15$.Comparing with the given options,the correct value is $15$.

For what value of $\lambda$ does the line $3x - 4y = \lambda$ touch the circle $x^2 + y^2 - 4x - 8y - 5 = 0$?

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